Physics for Youfebruaty 2013

physics for you | February ‘13 5

Page 5

india progress in science

Many Indian scientists are working in NaSa and many reputed universities

and laboratories in uSa, england and europe including CerN. One works

there, gets a lot of experience and settles there because of better facilities for

working. In Darmstadt, Germany, GSI Helmholtz Centre for Heavy Ion research,

is being upgraded, to become Facility for antiproton and Ion research (FaIr).

FaIr will feature a double-ring synchrotron with a circumference of 1100 metres

connected to the existing facility at GSI. We are very much involved in this work,

not for just working there but as a partner. We have a variable energy Cyclotron

Centre Salt-Lake, Calcutta. Our scientists have ample experience in this field.

Our scientists are making advanced components for FaIr as India is also a

partner in this venture. This facility is more powerful than even the one at cern,

in Geneva – The large Hadron Collider which is trying to detect the Higgs boson

or ‘God particle’. although the energy of the particles attained in FaIr, is much

lower than that of the Geneva facility, the intensity attained will be higher than

that of any other facility. To create heavy particles by collision, one needs very

high energy. This is the reason why there is a race for making more and more

powerful cyclotrons. energy is a very important factor. If energy is less, one

cannot do the experiments that are done by very powerful accelerators. With

the available energy, if one can make a very intense beam, the measurements

made will be more reliable in the given energy range. The new detector can

also detect collisions of the order of 10 million per second whereas the other

ones can detect 10,000 collisions per second. If the collisions are more, more

the new particles produced and less the error of determination. both energy

and intensity are important. These facilities can also be used for irradiation and

other uses ranging from agriculture to atomic physics.

We wish all the scientists working with FaIr a very bright future. The work

done by every scientist in every cormer of the world is for the benefit of the

whole humanity.

Anil Ahlawat, Editor

Vol. XXI No. 2 February 2013

Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (Hr), Tel : 0124-4951200

Regd. Office406, Taj apartment, Near Safdarjung Hospital, ring road, New Delhi - 110029.e-mail : [email protected] website : www.mtg.in

Managing Editor : Mahabir SinghEditor : anil ahlawat (be, Mba)Hony. Advisor : Dr. S. Malhotra Director, Delhi Public School, Faridabad (Hr)

contents

Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only.Editor : Anil AhlawatCopyright© MTG Learning Media (P) Ltd.All rights reserved. Reproduction in any form is prohibited.

Full Length JEE Advanced Practice Paper : 2013 6

Examiner’s Mind 17

Thought Provoking Problems 20

(SHM and Waves)

Interview 24

NEET/PMTs Practice Paper : 2013 25

NCERT Xtract Questions for NEET 29

CBSE Board Sample Paper 2013

Class XII 36

JEE Main Practice Paper : 2013 42

Brain Map 48

Exam prep : Chapterwise MCQs 51

for Practise

NEET 2013 MBBS Practice Paper 59

Essential Formulae for Competitive Exams 77

Class XI

rialedit

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6 Physics for you | feBruary ‘13

Page 6

sEcTioN - istraight objective Type

This section contains 6 multiple choice questions numbered 1 to 6. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

1. Two cylinders of same cross-section and length L but made of two material of densities d1 and d2 are cemented together to form a cylinder of length 2L. The combination floats in a liquid of density d with a length L/2 above the surface of liquid. If d1 < d2 then

(a) d d134

< (b) d

d2 1>

(c) dd

4 1> (d) d < d1

2. A plank of mass m is placed over smooth inclined plane and a sphere is placed on plank as shown.

There is sufficient friction between sphere and plank to prevent slipping.

If system is released from rest, the frictional force

on sphere is (a) up the plane

(b) down the plane(c) zero (d) horizontal

3 A transverse wave y = 0.05sin(20px – 50pt) m, is propagating along +ve x-axis on a string (x is in cm). A light insect starts crawling on the string with velocity 5 cm s–1 at t = 0 along +ve x-axis from a point where x = 5 cm. The difference in phase between its position at t = 5 s and its position at t = 0 is

(a) 15p (b) 250p (c) –24p (d) –5p

4. Two point charges q and 2q are placed (a, 0) and (0, a). A point charge q1 is placed at point P on the quarter circle of radius a as shown in figure.

If electric field at origin is zero, then

(a) point P is aa

323

,

a q

q1P

2q

O

(b) point P is a a

52

5,

(c) q1 = –3q (d) none of these

5. Consider a Young’s double slit set up as shown. The slits have equal width. Take O as origin. If

average intensity between y Dd1 4

= −λ and y D

d2 4= +

λ

equals n times the intensity of maxima, then n is

(a)

12

1 2+

π

(b)

2 1 2+

π

(c)

1 2+

π

(d) 12

1 2−

π

6. Optic axis of a thin equiconvex lens is the x-axis, the co-ordinate of a point object and its image are (– 40 cm, 1 cm) and (50 cm, –2 cm) respectively. Lens is located at x = (a) + 20 cm (b) – 30 cm (c) – 10 cm (d) 0 cm

physics

By Momentum : JABALPUR : (0761) 2400022, NAGPUR : (0712) 3221105, GWALIOR : (0751) 3205976.


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sEcTioN - iiMultiple correct Answer Type

This section contains 4 multiple choice questions numbered 7 to 10. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

7. The least count of a stop watch is 110

s. Two

persons A and B use this watch to measure the time period of an oscillating pendulum. Person A takes the time period of 50 oscillations, while B takes the reading of 100 oscillations. Neglecting all other sources of error we can say that(a) accuracy in measurement by A is equal to

that of B.(b) accuracy in measurement by A is less than

that of B.(c) absolute error in measurement by A is greater

than that of B.(d) absolute error in measurement by A is equal

to that of B.

8. A light rope passes over a light frictionless pulley attached to

the ceiling. An object with a large mass is tied to one end

and an object with a smaller mass is t ied to the other end.

Both masses are released from rest. Which of the following statement(s) is/are false for the system consisting of the two moving masses while string remains taut?(a) The center of mass remains at rest. (b) The net external force is zero.(c) The velocity of the center of mass is constant.(d) The acceleration of the center of mass is g

downward.

9. An artificial satellite is moving in a circular orbit around the earth. If the universal gravitational constant start decreasing at time t = 0 at a constant rate with respect to time t. Then satellite has its(a) path gradually spiraling out away from the

centre of the earth.(b) path gradually spiraling in towards the centre

of the earth.(c) angular momentum about the centre of the

earth remains constant.(d) potential energy increases.

10. In the network shown in figure, points A, B and C are at potential 70 V, 0 V and 10 V respectively. Choose the correct alternatives.

70 V

10 V

0 V

C

D

B

A10

30 20

(a) Point D is at a potential of 40 V.(b) The currents in the sections AD, DB, and DC

are in the ratio 3 : 2 : 1.(c) The currents in the sections AD, DB and DC

are in the ratio 1 : 2 : 3.(d) The network draws a total power of 200 W.

sEcTioN - iiiLinked comprehension Type

This section contains 2 paragraphs P11-13 and P14-16. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

P11-13 : Paragraph for Question Nos. 11 to 13You are on a summer tour to a remote hill station. You do not have facility of an electric heater and you want some ice. Unfortunately, the air temperature drops to only 6°C during night. Being a physicist you know that a clear, moonless night sky acts like a black body radiator at a temperature of TS = – 23°C, and decide to make ice by letting water radiate energy to such a sky. You take a cylindrical container, thermally insulated from ground and pour 4.5 g of water in it. The cross-section of container is 10 cm2. Assume that absorptivity and emissivity of water surface are same and neglect the presence of atmosphere. Also assume the average temperature of water to be 2°C.Stefan’s constant s = 5.0 × 10–8 W m–2 K–4

Emissivity of water e = 1.0Specific heat of water, s = 4190 J kg–1 K–1

Latent heat of fusion of ice L = 3.33 × 105 J kg–1

Take 254 = 4 × 105 and 114 – 104 = 5000

11. What is the total loss in heat for the above sample of water to freeze ?

(a) 1612 J (b) 1512 J (c) 1132 J (d) 1499 J

12. What is the approximate rate of energy loss by the water sample ?(a) 100 mJ s–1 (b) 8 mJ s–1 (c) 20 mJ s–1 (d) None of these

13. Is it possible to freeze the water sample during one night ?(a) Yes (b) No (c) Cannot be predicted (d) None of these


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P14-16 : Paragraph for Question Nos. 14 to 16

A block of mass M2

is kept over another block of

mass M and the heavier block is kept over smooth

horizontal surface as shown in figure. The heavier block is long enough such that the smaller block does not fall from it during the subsequent motion and the coefficient of friction between the blocks is

m. Suddenly block of mass M2

is given an impulse J in horizontal direction.

14. Which of the following is correct?(a) The friction between the blocks will be

impulsive.(b) The friction between the blocks will be

non-impulsive.(c) Normal between the b locks wi l l be

impulsive.(d) None of these.

15. The work done by impulsive force will be

(a) JM

2 (b)

2 2JM

(c) 12

2JM (d) none of these

16. The amount of heat evolved during the subsequent motion of the blocks is

(a) JM

2 (b)

JM

2

2

(c) 23

2JM (d) none of these

sEcTioN - iVinteger Answer Type

This section contains 7 questions numbered X Y Z0123456789

0123456789

0123456789

W0123456789

17 to 23 The answer to each of the questions is a single digit integer, ranging from 0 to 9. for example, if the correct answers to question numbers X, y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following.17. A barrier AB of length 12 m is hinged

at A. At the lower end a horizontal spring keeps the barrier closed. The height of water is 6 m and the width of the barrier is 5 m.

Water level is 4 m below the hinge A. If spring constant is 6 × 105 N m–1, find elongation of spring (in metre).

4 m

6 m

5 m

A

B

Water

18. A wire loop ABCD is divided into two parts. AB = AD = 2 m, BC = CD = BD = 1 m. The wire loop is lying on x-y plane. The resistance per unit length of the wire is 0.2 W m–1. There exist time dependent magnetic field

B t k= ( . )

^2 5 tesla. Find

the approximate value of current (in A) through part AB. Take 3 1 68= . and 45 7= .

B

C

D

A

19. Electrons in H-like atom (Z = 3) make transitions from 4th excited state to 3rd excited state and from 3rd excited state to 2nd excited state. The resulting radiation are incident on a metal plate to eject photoelectrons. The stopping potential for photoelectrons ejected by the shorter wavelength is 4.20 V. What is stopping potential (in volts) for the photoelectrons ejected by longer wavelength ?

20. In a circus act, a 4 kg dog is trained to jump from B cart to A cart and then immediately back to the B cart. The carts each have a mass of 20 kg and they are initially at rest. In both cases the dog jumps at 6 m s–1 relative to the cart. If the cart moves along the same line with negligible friction, If the final magnitude of velocity of cart B with respect to the floor is x/36 then find the value of x.

21. Glycerine is filled in 25 mm wide space between

two large plane horizontal surfaces. A thin plate of area 0.75 m2 at a distance of 10 mm from one of the surfaces is in horizontal position between the plates inside the glycerine. It is dragged horizontally at a constant speed of 0.5 m s–1. Take coefficient of viscosity h = 0 .5 N s m –2 . I f the force required to drag the p la te a t constant speed i s 125X / 4 newton, find the value of X.


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Physics for you | feBruary ‘13 11

Page 11

0.5 m s–1

Fixed

surface

10 mm

15 mm

22. A cylinder of height h, diameter h/2 and mass M and with a homogeneous mass distribution is placed on a horizontal table. One end of a string running over a pulley is fastened to the top of the cylinder, a body of mass m is hung from the other end and the system is released. Friction is negligible everywhere. At what minimum ratio m/M will the cylinder tilt ?

m

M

h/2

h

23. Water temperature 20°C flows from a tap T into a heated container C. The container has a heating element (a resistor R) which is generating heat at

the rate of P, that may be varied. The rate of water in flow from tap is, F = 0.021 litres per minute. The heat generated is sufficient so that the water in the container is boiling and getting converted into steam at a steady rate. What is the minimum power P (in kW) that must be generated as heat in the steady state in resistor R so that the amount of liquid water in the container neither increases nor decreases with time? (Neglect other losses of heat, such as conduction from the container to the air and heat capacity of container)

For water, specific heat s = 4.2 kJ kg–1 K–1, latent heat of vaporisation Lvap = 2.3 MJ kg–1, density r = 1000 kg m–3. Mark your answer in nearest integer.

R

T

chemistry



24. The passage of a constant current through a solution of dilute H2SO4 with ‘Pt’ electrodes liberated 336 cm3 of a mixture of H2 and O2 at S.T.P. The quantity of electricity that was passed is (a) 96500 C (b) 965 C (c) 1930 C (d) 100 faraday

25. Amongst [Co(ox)3]3–, [CoF6]3– and [Co(NH3)6]3+ (a) [Co(ox)3]3– and [CoF6]3– are paramagnetic and

[Co(NH3)6]3+ is diamagnetic.(b) [Co(ox)3]3– and [Co(NH3)6]3+ are paramagnetic

and [CoF6]3– is diamagnetic.(c) [Co(ox)3]3– and [Co(NH3)6]3+ are diamagnetic

and [CoF6]3– is paramagnetic. (d) [Co(NH3)6]3+ and [CoF6]3– are paramagnetic

and [Co(ox)3]3– is diamagnetic.

26. PhPh

OHC CHO

PhPh

Conc. NaOH, Conc. H2SO4

The final product is

(a)

Ph

Ph

HOH C2HO

COOHPh

Ph

(b)

PhPh

PhPh

CH2H2C

O

(c)

Ph

Ph

PhPh

OO

(d)

OO

Ph

Ph

PhPh

O

27. Two 1st order reactions have half-lives in the ratio 3 : 2. Then the ratio of time intervals t1 : t2, will be?

Where t1 is the time period for 25% completion of the first reaction and t2 is the time required for 75% completion of the second reaction. [log 2 = 0.3, log 3 = 0.477](a) 0.199 : 1 (b) 0.420 : 1 (c) 0.273 : 1 (d) 0.311 : 1

28. Identify the anomers.

(a)

O

OHH

H OH

HO H

H OH

H

CH – OH2

O

HHO

H

HO H

H OH

H

CH – OH2

and

HO


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Page 12

(b)

MeHHH OH

OHOH

Me

MeHOHOHO H

HH

Me

and

(c)

CH OH2

C OOHOH

HH

CH OH2

CH OH2

C OOHH

HHO

CH OH2

and

(d)

and

DHHH

Et

OHOHOH

DHO

HH

HOHOH

Et

29. Which amongst the following metal carbonyls are inner orbital complexes with diamagnetic property?(I ) Ni(CO)4; (II) Fe(CO)5 ; (III) V(CO)6 (IV) Cr(CO)6

Select the correct answer from the codes given below:(a) I and II only (b) II , III and IV only(c) II and IV only (d) I , II and IV only



30. Which are correct against property mentioned? (a) CH3COCl > (CH3CO)2O > CH3COOEt >

CH3CONH2 (Rate of hydrolysis)

(b) CH3–CH2–COOH > CH3 CH COOHCH3

>

COOH

CH3

CH3

CH3

(Rate of esterification)

(c)

OH

ON2

OH OH> >

(Rate of esterification)

(d) CH3 – CH3 –C C– COOH >

— —— —

–

O O

– CH2 COOH >

Ph – CH2 – COOH (Rate of decarboxylation)

31. The correct statement(s) regarding defects in solids is (are)(a) Schottky defect is usually favoured by small

difference in the sizes of cation and anion.(b) Schottky defect lowers the density of

solids.(c) C o m p o u n d s h a v i n g F - c e n t r e s a r e

diamagnetic.(d) Frenkel defect is a dislocation defect.

32. Which of the following products is/are correctly mentioned in the following reactions? (a) HCHO NaOD HCOONa + CH3OD

(b) HCDO NaOH DCOONa + CH2DOH(c) HCDO NaOEt DCOOEt + DCH2ONa

(d) D2CO NaOD DCOONa + CD3OD

33. The van der Waals equation of state for a non-ideal

gas can be rearranged to give PVRT

VV b

aVRT

=−

−

for 1 mol of gas.

PVRT

1.0

0 40 60 80P, atm

The constants ‘a’ and ‘b’ are

positive numbers. When applied to H2 at 80 K, the equation gives the curve as shown in the figure. Which one of the following statements is(are) correct?(a) At 40 atm the two terms V/(V – b) and

a/VRT are equal.(b) At 80 atm the two terms V/(V – b) and

a/VRT are equal.(c) At a pressure greater than 80 atm, the term

V/(V – b) is greater than a/VRT.(d) At 60 atm the term V/(V – b) is smaller than

1 +a

VRT




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P34-36 : Paragraph for Question Nos. 34 to 36A fuel cell is a cell that is continuously supplied with an oxidant and a reductant so that it can deliver a current indefinitely.Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work. Internal combustion engines at best convert only the fraction (T2 – T1)/T2 of the heat of combustion into mechanical work.While the thermodynamic efficiency of the fuel cell

is given by, η =∆∆

GH

, where DG is the Gibbs energy

change for the cell reaction and DH is the enthalpy change of the cell reaction. A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte.

Pt|H2(g)|H+(aq.)||H2O(l)|O2(g)| Pt; 2 303 0 06. .RT

F=

The above fuel cell is used to produce constant current supply under constant temperature and 30 atm constant total pressure condition in a cylinder. If 10 mol of H2 and 5 mol of O2 were taken initially. Rate of combustion of O2 is 10 millimoles per minute. The half cell reactions are12

2 2 1 2462 2O H H O V( ) ( ) ( ) ; .g aq le E+ + → ° =+ −

2H+(aq) + 2e–H2(g); E° = 0

To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized. Since fused salts have lower electrolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications 34. Calculate e.m.f. of the given cell at t = 0. (log 2 = 0.3).

(a) 1.255 V (b) 1.35 V (c) 1.3 V (d) 1.246 V

35. The above fuel cell is used completely as an electrolytic cell with Cu voltameter of resistance 26.94 using Pt electrodes. Initially Cu voltameter contains 1 litre solution of 0.05 M CuSO4.[H+] in solution after electrolysis (Assuming no change on volume of solution).(a) 0.015 M (b) 0.03 M (c) 0.025 M (d) 0.01 M

36. If (Cu2+) = 0.01S m2 mole–1, (H+) = 0.035 S m2 mole–1 and (SO4

2–) = 0.016 S m2 mole–1, specific conductivity of resulting solution left in sopper voltameter after above electrolysis is

(a) 2.57 S m–1 (b) 1.75 S m–1 (c) 1.525 S m–1 (d) 2.25 S m–1

P37-39 : Paragraph for Question Nos. 37 to 39A black coloured (A) on reaction with dil. H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed in an acidified solution of a compound (E) gives a black precipitate (F) which is soluble in hot concentrated (C). After boiling this solution when excess of ammonia solution is added, a blue coloured compound (G) is formed. To this solution of (E), on addition of acetic acid and aqueous potassium ferrocyanide, a chocolate brown precipitate (H) is formed. On addition of an aqueous solution of BaCl2 to an aqueous solution of (E) white precipitate insoluble in HNO3 is obtained.

37. Black coloured compound (A) is(a) PbS (b) CuS (c) Ag2S (d) All of these.

38. The gas (B) on passing through an acid (C) gives a white turbidity (D) because(a) gas (B) acts as an oxidising agent (b) gas (B) acts as an reducing agent(c) acid (C) acts as an oxidising agent (d) both (b) and (c)

39. To which of the following property, the compound (E) will respond?(a) It gives white precipitate with (CH3COO)2

Pb solution soluble in ammonium acetate.(b) It gives dirty white precipitate with KI.(c) Its hydrated salt effloresces.(d) All of these.

sEcTioN - iVinteger Answer Type

This section contains 7 questions numbered X Y Z0123456789

0123456789

0123456789

W0123456789

40 to 46. The answer to each of the questions is a single digit integer, ranging from 0 to 9. for example, if the correct answers to question numbers X, y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following.

40. 5.6 g of an oxide of a hypothetical metal required 2.4 g coke for its complete reduction to metal along with the production of CO gas. Find the equivalent weight of metal in case of given metal oxide.


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41. How many of the given statements are correct?I : Molar entropy of a substance follows the

order (Sm)Solid < (Sm)liquid < (Sm)gas II: Entropy change for the reaction H2(g) → 2H(g)

is +ve.III: Molar entropy of a non-crystalline solid will

be zero at absolute zero.IV: If the path of an irreversible process is

reversed, then both system and surroundings shall be restored to their orginal states.

V : Refractive index and molarity are intensive properties.

42. Sulphide ions reacts with sodium nitropruside giving a coloured solution. In the reaction, the change in oxidation state of iron is

43. Two solids, A and B are present in two different container having same volume and same temperature following equilibrium are established :

In container (1) A(s) D(g) + C(g) PT = 2 atm at equilibrium

In container (2) B(s) E(g) + F(g) PT = 4 atm at equilibrium

If excess of A and B are added to a third container having double the volume and at same temperature then, the total pressure of this container at equilibrium will be

44. What is the free energy change (DG) in kcal, when 144g of water at 100°C and 1 atm pressure in converted into steam at 100°C and 4 atm pressure? (Take R = 2 cal/K mole, ln2 = 0.7).

45. Number of ‘H’ that can takes part in tautomerism in given compound are

O

O O

O

HNN

46. How many of the following compounds will give white precipitate with aqueous AgNO3?

ClCl Cl

Cl

||

||Cl O

Cl

ClPh PhPhPh

OH ClC C

mathematics



47. Let g A: ,π π2

→ defined by g x x

x( ) sin

sin= +

−42 be

invertible function, then the set A is equal to (a) [–5, –2] (b) [2, 5] (c) [–5, 2] (d) [–3, –2]

48. If f x f xf x

x N( ) ( )( )

,+ = +

∈1 1

29 and f(x) > 0 for

x ∈ N all then lim ( )x

f x→∞

=

(a) 3 (b) 4 (c) 5 (d) 6

49. The value of [tan ]−∫ 1

0

10

x dx (where [⋅] denotes greatest

integer) is equal to

(a) 104

−

π (b) (10 – tan1)

(c) 10 (d) None of these

50. If C0, C1, C2, ....., Cn are binomial coefficients then

lim .... ( )n n n n

nn

C C C C→∞ − −−

+

+ + −

23

23

1 231

2

2 0

=

(a) 0 (b) 1 (c) –1 (d) 2

51. I f the common tangents to the c i rc les x2 + y2 – 2x – 4y + 1 = 0 and x2 + y2 – 14x – 4y + 52 = 0 intersect at A and B on the line joining their centres then the equation of the ellipse with A,

B as foci and eccentricity 12

will be

(a) x y−( )+

−( )=

948

264

12 2

(b) x y−( )

+−( )

=9

642

481

2 2

(c) x y−( )

+−( )

=2

489

641

2 2

(d) x y−( )+

−( )=

924

232

12 2


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52. Two lines whose equations are x y z− = − = −32

23

1λ

and x y z− = − = −23

32

23

lie in the same plane, then

the value of sin–1 sinl is equal to (a) 3 (b) p – 3 (c) 4 (d) p – 4



53. Let unit vectorsa and

b are perpendicular & unit

vector c is inclined at an angle ‘q’ to both a and

b . If c a b a b= + + ×α β γ( ) then (a) a = b (b) 1 – 2a2 = g2

(c) α θ2 1 22

= + cos (d) a2 – b2 = g2

54. The equation of the curve passing through (3, 4) & satisfying the differential equation

y dydx

x y dydx

x

+ − − =

2

0( ) can be

(a) x – y + 1 = 0 (b) x2 + y2 = 25 (c) x2 + y2 – 5x – 10 = 0 (d) x + y – 7 = 0

55. The image of point having abscissa = a on y = x – 1 w.r.t. the line mirror 3x + y = 6a is the point on x = y2 + 1 with ordinate = a. Then the value of a is

(a) –1 (b) 13

(c) 2 (d) none of these

56. If in a triangle ABC, sin4A + sin4B + sin4C + 8cosA = 0, then triangle may be(a) Right angled (b) isosceles(c) Equilateral (d) Right angled isosceles



P57-59 : Paragraph for Question Nos. 57 to 59The diagram shows

x

the graph of the d e r i va t i ve o f a function y = f(x) for 0 ≤ x ≤ 5 with f(0) = 0 .

57. Tangent line to y = f(x) at x = 0 makes an angle of sec–1 k with the x-axis then k =

(a) 2 (b) 5 (c) 10 (d) 17

58. f is increasing in the interval (a) [1, 3] (b) [0, 4] (c) [0, 1] (d) none of these

59. f x dx( )0

1∫ =

(a) − 2120

(b) 2120

(c) –1 (d) none of these

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P60-62 : Paragraph for Question Nos. 60 to 62 A bag contains 6 balls of 3 different colour namely White, Green and Red, atleast one ball of each different colour. Assume all possible probability distributions are equally likely.60. The probability that the bag contains 2 balls of

each colour, is (a) 1/3 (b) 1/5 (c) 1/10 (d) 1/461. Three balls are picked up at random from the bag

and found to be one of each different colour. The probability that the bag contained 4 Red ball is

(a) 1/14 (b) 2/14 (c) 3/14 (d) 4/14

62. Three balls are picked at random from the bag and found to be one of each different colour. The probability that the bag contained equal number of White and Green balls, is

(a) 4/14 (b) 3/14 (c) 2/14 (d) 5/14sEcTioN - iV

integer Answer TypeThis section contains 7 questions numbered 63 to 69. The answer to each of the questions is a single digit integer, ranging from 0 to 9.

X Y Z0123456789

0123456789

0123456789

W0123456789

for example, if the correct answers to question numbers X, y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following.

63. If A −−

1 1 11 3 2

1 2 1 then det [adj adj adj A ] is

equal to

64. If (1 + ax + bx2)4 = a0 + a1x + a2x2 + .... + a8x8 where a, b, a0, a1, ..... a8 ∈ R such that a0 + a1 + a2 ≠ 0 and aaa

aaa

aaa

0

1

2

1

2

0

2

0

1

0= then 32(a – b) is

65. If roots of equation x2 – x(1 + 4a) + a + 3a2 = 0 are a1, b1 & that of x2 – ax – 2a2 = 0 are a2 and

b2. Given 0 14

< <a and a2 < a1 < b2 < b1 then

[|b1| – |a1|] + [|b2| – |a2|] is equal to (where [⋅] denotes greatest integer)

66. The product of maximum & minimum distance

between two curves zz

zz

−−

= −−

=12

2 21

2and is

67. The sum of infinite terms of a decreasing G.P. is equal to the greatest value of the function f(x) = x3 + 3x – 9 in the interval [–2, 3] and the difference between the first two terms is f ′(0). If the common ratio of the G.P. is k/3 then k =

68. If the area bounded by the curve f x x x x x( ) tan cot tan cot= + − − between the

lines x x= =02

, π and the x-axis is logK then K is

69. The sum of the factors of 7! , which are odd & are of the form 3t + 1, where ‘t’ is a whole number is

ANswEr kEys

PHYSICS1. (a) 2. (c) 3. (b) 4. (b) 5. (a) 6. (c) 7. (b,d) 8. (a,b,c,d)9. (a,c,d) 10. (a,b,d) 11. (a) 12. (a) 13. (a) 14. (b) 15. (a) 16. (c) 17. (1) 18. (3) 19. (1)20. (5) 21. (1) 22. (1) 23. (1)

CHEMISTRY24. (c) 25. (c) 26. (c) 27. (d) 28. (d) 29. (c) 30. (a,b) 31. (a,b,d)32. (a,b,c,d) 33. (c,d) 34. (c) 35. (a) 36. (b) 37. (d) 38. (d) 39. (d) 40. (6) 41. (3) 42. (0)43. (6) 44. (8) 45. (4) 46. (4)

MATHEMATICS47. (a) 48. (a) 49. (b) 50. (a) 51. (b) 52. (d) 53. (a,b,c) 54. (a,b)55. (c) 56. (a,d) 57. (d) 58. (b) 59. (a) 60. (c) 61. (a) 62. (b) 63. (1) 64. (3) 65. (1) 66. (3) 67. (2) 68. (4) 69. (8)

nn


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physics FOR yOU | February ‘13 17

seeking Unity in diversity and diversity in UnityPassage- 1

Differentiation and Integration – they are the right and left hands of science. We shall study a few cases of motion and cases similar to various types of motion in different fields.Water flow and current flow are similar. Resistance in narrow tubes will be more. although the same potential energy is applied.As the quantity of water flowing is the same, water will be faster in the narrow tube than in the wider one because the same quantity has to flow.The same current will be flowing in resistances in series.

R1 R2

I

R3

A B

CD + –I

I

In the given circuit same current is flowing in all the three resistances in series. – IR1 – IR2 – IR3 + e = 0 Take "up the current"-positive, down the flow, negative. The potential from the positive of the battery to its negative plate is negative and negative to positive is positive.

m From any point, back to I2

I3

I1I5

I4

the same point, the total potential difference should be zero.

m The total current entering a junction will be equal to the current leaving the junction.

I1 + I2 – I3 –I4 – I5 = 0 These are Kirchhoff's Laws.1. The total current flowing in the circuit is

12 V

2

2 2

2 2

(a) 1 A (b) 2 A (c) 3 A (d) 4 A

2. If the given bridge is balanced and no current is flowing through the galvanometer, what is the value of S?

GA

B

C

D

PQ

R S

4 2

8

(a) 4 W (b) 2 W(c) 8 W (d) None of these

3. When a straight conductor is held perpendicular to the plane of the paper and the current is outwards, what is the direction of the associated magnetic lines by Oersted's law?(a) It is parallel to the conductor and outwards.(b) It is parallel to the conductor and inwards.(c) It is circular, perpendicular to the direction

of current, in the plane of the paper, anticlockwise.

(d) Perpendicular to the plane of the paper, clockwise.

4. What is the essential difference in the direction of action of the magnetic field on a charge and the electric field?

5. Three capacitors are connected in series as shown in figure.

C1C2

C3

V+ –

(a) The same charge is flowing in all the three capacitors

(b) Total equivalent capacitance is C = C1 + C2 + C3(c) The equivalent capacitance C is given by

1 1 1 1

1 2 3C C C C= + +

(d) None of these


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Page 19

physics FOR yOU | February ‘13 19

6. The relation current, resistance and potential difference is given by

V IR I VR

= ⇒ =

The corresponding relation for capacitance is

(a) Q VC

=2 (b) V = QC

(c) Q = VC (d) none of these 7. The total resistance when R1, R2, R3 are in parallel is

1 1 1 1

1 2 3R R R R= + +

When capacitors are in parallel, the total capacitance, C is given by

(a) 1 1 1 1

1 2 3C C C C= + + (b) C = C1 + C2 + C3

(c) none of these

Passage-II In a CR circuit, when a capacitor is getting

charged, the voltage increases exponentially and when the current is observed, initially the current is high; till the capacitor gets charged fully, the current decreases exponentially till it reaches zero exponentially.

Charging and discharging a capacitor has its analogy in nuclear physics in the studies of radioactivity, production and decay of radioactive materials. The oscillations of mass and spring, L, C, R circuit are all analogous to problems in radioactivity.

Students are advised to prepare Charts for comparision.

8. The functions of spring and mass in electrical oscillations are given by(a) R and L (b) L and C(c) L and R (d) none of these

9. What is common in a leaking water tank, radioactive decay and discharge of a fully charged capacitor?

sOLUtiOns

1. (c) :

12 V

2

2 2

2 2 P Q R

S

In P and Q, 2 W and 2 W are in parallel.

i.e.,

1 1 1 1 12

12

11 2R R R R

RPQ= + ⇒ = + ⇒ = Ω

RQR is also 1 W. They are in series with S = 2 W.

Therefore, total resistance is, 1 W + 1 W + 2 W = 4 W.

The total current in the circuit = =124

3V

AΩ

2. (a) : This is a Wheatstone bridge where the resistances are so balanced that no current flows through the branch BD, as the galvanometer shows no deflection.

Here, PQ

RS

S QRP

= ⇒ = =×

=2 8

44 Ω

3. (c) : When current is outwards, ^r to the plane of the paper, the direction of the magnetic field is anticlockwise, perpendicular to the direction of the current.

4. The magnetic field always acts perpendicular to the direction of the current. The electric field always acts parallel to the motion of the charge.

5. (a, c): At any instant, the charges accumulated on all the capacitors will be equal (a) but the sign are different.

i.e., The total capacitance is C, where C is given by,

1 1 1 1

1 2 3C C C C= + +

6. (c) : Q = VC

7. (b) : C1 C2 C3V+–

The total charge, Q = Q1 + Q2 + Q3 The potential applied across all the three capacitors

in parallel is the same. Q = VC = VC1 + VC2 + VC3 ⇒ C = C1 + C2 + C3

8. (b) : L and C

d Qdt LC

QLC

2

221 1

= − =where ω

The period of oscillation, T =2πω

Frequency of oscillation =1

2π LC9. They all obey Y

Xtime

the exponential decay law. Let quantity of water, number of decaying nuclei

and the charge in the capacitor be Y and time denoted by X. The decay diagram is as shown.

Seek unifined approach to solve diverse problems.

vvv


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20 physics FOR yOU | february ‘13

1. A hole of radius R4 is drilled through a cylinder

of radius R to form a body as shown in figure. Assuming that the body is in pure rolling, find the time period T for small oscillations.

2. A cylinder of mass M and radius R is kept on a rough horizontal platform at one extreme end of the platform at t = 0. Axis of the cylinder is parallel to z-axis. The platform is oscillating in the xy plane and its displacement from the origin is represented as x = 2cos(4pt) m.

There is no slipping between the cylinder and the platform.

Find the acceleration of the centre of mass of the cylinder as a function of time.

3. A particle simultaneously participates in two mutually perpendicular oscillations : x = cospt m and

yt

= cos π2

. Find the trajectory of the particle.

4. When a tuning fork of frequency 262 Hz is struck, it loses half of its energy after 4 s. (a) What is the decay time t ? (b) What is the Q-factor for this tuning fork? (c) What is the fractional energy loss per cycle?

5. A metallic rod of length 1 m is rigidly clamped at its midpiont. Longitudinal stationary waves are set up in such a way that there are two nodes on either side of the midpoint. The amplitude of an antinode is 2 × 10–6 m. Write the equation of motion at a point 2 cm from the midpoint and those of constituent waves in the rod.

(Y = 2 × 1011 N m–2 and r = 8 × 103 kg m–3)6. The first overtone of an organ pipe beats with the

first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the length of the pipes.

7. Two identical wires are stretched by the same tension of 100 N and each emits a note of frequency 200 cycles per second. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked.

8. By what factor does the sound intensity increase if a sound level increases by 3 dB.

SOLUTIONS

1.

Let s be the mass density of the cylinder and x be the position of centre of mass. Then

SHM and Waves

Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh

By : Prof. Rajinder Singh Randhawa*


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Page 21

physics FOR yOU | february ‘13 21

016 2 16

22 2

= −

+x R

R R Rσ π σ π

or xR

= −30

Let M be the mass of the cylinder without cavity. Moment of inertia about contact point

I MR123

2=

And moment of inertia of an imaginary cylinder occupying the cavity is

IM R M R

2

2 212 16 4 16

32

=

×

+

So, the moment of inertia of the system about the contact point,

I I I MR MR= − = − −

=1 22 23

21

512964

695512

When the system is turned through a small angle clockwise, the restoring torque on it is

τ θ= −

1516 30

Mg Rsin

695512

1516 30

22

2MR ddt

Mg Rθ θ= −

sin

For small q, sinq ≈ q,ddt

gR

2

2512

32 6950θ θ+

×

=

Comparing it with the standard equation of SHM, we get,

ω π=×

=512

32 6952g

R T

or T Rg

Rg

= × =2 32 695512

41 4π .

2. From given equation x = 2 cos(4pt) m,angular frequency, w = 4p rad s–1.\ Acceleration of the platform when displacement is x, a = –w2x

= – 32p2cos(4pt) m s–2

Fictitious force acting on the cylinder = Ma = M 32p2 cos(4pt)From Newton's second law, 32 p2 M cos(4pt) – f = MaCM ...(i)Also, taking torque about C.M. of the cylinder,

fR MR=12

2α ...(ii)

For pure rolling; aCM = R a ...(iii) Solving above equations, we get,

f M t=323

42π πcos( ) N

And acceleration of C.M. of the cylinder relative

to platform = −643

42 2π πcos( )t m s

Acceleration of C.M. of the cylinder relative to

ground = −

643

32 42π πcos( )t

= − −323

42 2π πcos( )t m s

3. The given equations are x = cospt ... (i) y

t= cos π

2

∴ =+

− =yt

y t1

22 12cos cosπ

πor .... (ii)

Put (i) in (ii), we get, 2y2 – 1 = x or 2y2 = x + 1 represents the equation of a parabola.

4. Using E = E0e–t/t and EE

= 02

. The Q value can be

calculated from decay time and the frequency.(a) The energy at time, t = 4 s is equal to half the

original energy;

EE e e0

04 4

22= =− / /τ τor

By taking loge both sides, we get

4 2 4

25 77

ττ= = =ln ,

ln. s

(b) Q - factor = w0t = (2pu)t = 2p × 262 × 5.77 = 9.5 × 103.(c) The fractional energy loss in a period is given

by

| |

..∆E

ET

= = =×

= × −τ υτ

1 1262 5 77

6 6 10 4

5. When a longitudinal wave is set in the rod, its free end is antinode and clamped point is node. Two

consecutive nodes are separated by λ2

and a node

and an antinode by λ4

.

∴

+

= = ⇒ =4

22

41 0 4λ λ

λL m m.

Velocity of a transverse wave in a rod is, vY

=ρ

∴ =×

×=

−

−−v

2 108 10

500011 2

3 31N m

kg mm s

Also v = ul, υ

λ=

v

∴ = =−

υ5000

0 412500

1m sm

Hz.


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22

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22 physics FOR yOU | february ‘13

Let the incident wave on the rod be y1 = A sin(wt – kx). The reflected wave is y A t kx2 = + +sin( ).ω φ

From principle of superposition, y = y1 + y2 = A sin(wt – kx) + Asin (wt + kx + f)

=− + + +

22

At kx t kxsin ( ) ( )ω ω φ

cos ( ) ( )ω ω φt kx t kx− − + +

2

= +

+

2

2 2A kx tcos sin .φ

ωφ

The free end of the rod is an antinode, here

amplitude is maximum, cos .kx +

=

φ2

1

At x = 0, φ2

0= , hence f = 0

Amax = 2A = 2 × 10–6 m.

The equation of the resultant standing wave is

y x t= ×

−2 10 2 26 cos sinπλ

πυ

= 2 × 10–6cos(5px)sin(25000 pt).

For a point 2 cm from the midpoint, x = 0.5 ± 0.02

\ y = 2 × 10–6 cos5p (0.5 ± 0.02) sin(25000pt).

The equation of component waves can be obtained by applying 2cosA sinB = sin(A + B) – sin(A – B).

\ y = 10–6[sin(5px + 25000pt) – sin(5px – 25000pt)]

= 10–6 sin(25000pt + 5px) + 10–6 sin(25000pt – 5px)

The two terms in the expression for y represent the two component waves.

6. The frequency in nth mode of vibration of one end closed and an open organ pipe are given by;

For closed pipe, υc

c

n vl

=−( ) ,2 1

4where n = 1, 2, 3,....

For open organ pipe, υoo

nvl

=2

, where n = 1, 2, 3,...

Fundamental frequency of closed organ pipe,

vlc4

110= Hz or lc =×

=330

4 1100 75. m

First overtone of closed organ pipe =34

vlc

First overtone of open organ pipe =22

vlo

These two produce beats frequency of 2.2 Hz when sounded together, the expressions for beat frequency are

34

22

2 2 22

34

2 2vl

vl

vl

vlc o o c

− = − =. .or

3 110 330 2 2 330 3 110 2 2× − = − × =l lo o

. .or

\ lo = 1.0067 m or lo = 0.9934 m

7. The frequency of the fundamental note emitted by each wire before the tension change occurs is

υµ

=1

2LT

.... (i)

Differentiate equation (i) w.r.t. T, we get

ddT L

TL

Tυµ µ µ µ

=

=

⋅

− −12

12

1 14

11 2 1 2/ /

ddT LT

T

T LTTυ µ

µ µ=

=

14

14

2

2

1 2 1 2/ /

ddT T

υ υ=

2 (Using (i))

Hence, ∆

∆υ

υ=

2

TT

∆υ = × =2002

1100

1 cycle per second

8. Intensity level b is given by

β = 10(dB)log10II0

or II0

1010= β/( )

.... (i)

Let I′ be the intensity when the sound level is (b + 3) dB. Hence,

β + =

′3 100

log II

II′

= +

0

3 1010( )/β .... (ii)

Divide (ii) by (i), we get

II′

=+10

10

3 10

10

( )/

/

β

β

II′

= =10 20 3( . )

\ An increase in sound level of 3 dB increases the intensity by a factor of 2.

vvv


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24 physics for you | february ‘13

Page 24

Gone are the days when students used to believe in self study and pass their exams with flying colors. Today, in this highly

competitive world where children outdo each other in a bid to score 90 percent, professional training and coaching institutes have become a part and parcel of the modern age education.The purpose of competitive exams is to filter students with the right aptitude for a particular professional field. Simply mugging up the study material may not prove useful. It is important to have a strategy in place. Thus a mapped out plan is the need of the hour and coaching institutes play a very vital role in jotting out a good plan for the students.The coaching institutes play a very significant role in the enhancement of student’s preparation. They enhance student’s opportunities by providing a high standard of academic learning, environment and discipline. The institutes help students in developing skills in various academic fields. With the growing span of education in India, the students have become alert and conscious about their careers, they tend to seek the best quality education and for that they seek the best coaching institute in the town.Good coaching institute helps students to identify their weak areas and helps to strengthen them with well built strategies. Aspirants working hard to crack JEE Main, JEE Advanced, CAT, MAT, TOFEL, IAS etc opt for these coaching institutes for proper guidance and grooming. The growing demands of high level education have led to a fiercely competitive environment for students seeking admissions in prestigious institutes.The genesis of growth in competition lies in the education imparted in the schools. The school board exams have diverse levels of difficulty and it usually doesn’t match the competency level of the competitive exams. Schools basically focus on imparting syllabus education while entrance exams focus on application based education. The inability of traditional school set up in meeting the needs of students in preparing for

competitive exams and choosing the right career has ushered in the concept of coaching centers.Earlier students were unaware of the various new career options available due to lack of guidance and they used to rely on their parents, siblings, friends and other such sources for information. With the emergence of media penetration and online support systems, people have become more aware. Coaching institutes have generated greater awareness of the various options available for the students.The experienced and qualified faculty at a good coaching institute helps the students to understand and analyze the concepts and test patterns and guide them to work accordingly. They clarify all the related doubts of the students and provide them with relevant suggestions notes and handouts which help students utilize their time optimally.The pattern of competitive entrance exams such as JEE, CAT etc undergo frequent change and at times this change happens on yearly basis. The coaching centers understand the changes and lead the students in the desired manner. The Mock tests taken in these coaching classes are very helpful in giving fairly good exposure to the students and encourage them to put in focused effort and to work hard. These tests give ample practice as they teach the students to respond promptly to tricky questions and manage speed and time.Coaching centers are very important but one needs to keep a few things in mind before joining a coaching institute. The students should inquire about the faculties and the reputation of the coaching institute before joining. Also, the timing should be suitable and it should not be too far from your place as then it will be tiresome to travel and you will have no time to study at home. At the end of the day, irrespective of whether you go to a coaching class or not, it is your effort that will fetch you the kind of results you are longing for.

– Contributed by FIITJEE

nn

Role of Coaching Institutesin One’s Success


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physics for you | february ‘13 25

Page 25

1. A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

(a) fIand4

(b) 34 2f Iand

(c) fIand 3

4 (d) f I

2 2and

2. The radius of the rear wheel of bicycle is twice that of the front wheel. When the bicycle is moving, the angular speed of the rear wheel compared to that of the front is(a) greater (b) smaller(c) same (d) exact double

3. When a particle oscillates in simple harmonic motion, both its potential energy and kinetic energy vary sinusoidally with time. If u be the frequency of the motion of the particle, the frequency associated with the kinetic energy is

(a) 4u (b) 2u (c) u (d) υ2

4. A wooden block is dropped from the top of a cliff 100 m high and simultaneously a bullet of mass 10 g is fired from the foot of the cliff upwards with a velocity of 100 m s–1. The bullet and wooden block will meet each other after a time (a) 10 s (b) 0.5 s (c) 1 s (d) 7 s

5. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 W and an output impedance of 200 W. The power gain of the amplifier is(a) 500 (b) 1000 (c) 1250 (d) 50

6. A satellite is launched into a circular orbit of radius R around earth while a second satellite is launched into an orbit of radius 1.02R. The percentage difference in the time period is(a) 0.7 (b) 1.0 (c) 1.5 (d) 3.0

7. Keeping the banking angle same, to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent. The radius of curvature of the road has to be changed from 20 m to(a) 6 m (b) 18 m (c) 24.2 m (d) 30.5 m

8. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is(a) 3.6 A (b) 2.8 A(c) 2.5 A (d) 5.0 A

9. A block of mass m is attached with massless spring of spring constant k. The block is placed over a fixed rough inclined surface for which the

coefficient of friction is µ =34

. The block of mass m

is initially at rest. The block of mass M is released

from rest with spring in unstretched state. The minimum value of M required to move the block up the plane is (Neglect mass of string and pulley and friction in

pulley) ( sin )Take 37 35

° =

(a) 35

m (b) 45

m

(c) 65

m (d) 32

m


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Page 26

10. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. If the frequency of incident light is halved and the intensity is doubled, the photocurrent becomes(a) one fourth (b) doubled(c) halved (d) zero

11. The Poisson’s ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is(a) 3% (b) 2.5%(c) 1% (d) 0.5%

12. The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (2, 2, 2). The position of the fourth mass of 4 kg to be placed in the system so that the new centre of mass is at (0, 0, 0) is(a) (–3, –3, –3) (b) (–3, 3, –3)(c) (2, 3 –3) (d) (2, –2, 3)

13. A body moves with uniform acceleration, then which of the following graphs is correct?

(a)

t

v

(b)

(c)

t

v

(d)

t

S

14. A charged bead is capable of sliding freely through a string held vertically in tension. An electric field is applied parallel to the string, so that the bead stays at rest at middle of the string. If the electric field is switched off momentarily and switched on(a) the bead moves downwards and stops as

soon the field is switched on.(b) the bead moves downwards when the field

is off and moves upwards when the field is switched on.

(c) the bead moves downwards with constant acceleration till it reaches the bottom of the string.

(d) the bead moves downwards with constant velocity till it reaches the bottom of the string.

15. When a battery connected across a resistor of 16 W, the voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10 W, voltage across it is 11 V. The internal resistance of the battery is

(a) 107

Ω (b) 207

Ω (c) 257

Ω (d) 307

Ω

16. A satellite of mass ms revolving in a circular orbit of radius rs round the earth of mass M has a total energy E. Then its angular momentum will be(a) (2Emsrs 2)1/2 (b) (2Emsrs 2)(c) (2Emsrs)1/2 (d) (2Emsrs)

17. The equations of motion of a projectile are given by x = 36t m and 2y = 96t – 9.8t2 m. The angle of projection is

(a) sin–1 45

(b) sin–135

(c) sin–1 43

(d) sin–1 34

18. A body is moving forward and backward. Change in frequency observed by the body of a source is 2%. What is velocity of the body? (Speed of sound is 300 m s–1)(a) 6 m s–1 (b) 2 m s–1

(c) 2.5 m s–1 (d) 3 m s–1

19. The refractive index of a material of a plano concave lens is 5/3, the radius of curvature is 0.3 m. The focal length of the lens in air is(a) – 0.45 m (b) – 0.6 m(c) – 0.75 m (d) – 1.0 m

20. The average force that is necessary to stop a hammer with 25 N s–1 momentum in 0.05 s is(a) 500 N (b) 125 N (c) 50 N (d) 25 N

21. The dimensional formula for latent heat is(a) [MLT–2] (b) [ML2T–2](c) [M0L2T–2] (d) [MLT–1]

22. A charge Q situated at a certain distance from a short electric dipole in the end on opposite experiences a force F. If the distance of the charge from the dipole is doubled, force acting on the charge will be

(a) 2F (b) F2

(c) 8F (d) F8

23. The magnetic induction and the intensity of magnetic field inside an iron core of an electromagnet are 1 Wb m–2 and 150 A m–1 respectively. The relative permeability of iron is

(m0 = 4p × 10–7 H m–1)

(a) 106

4π (b) 106

6π (c)

105

4π (d) 105

6π


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Page 27

24. The equivalent resistance between the terminals A and D in the following circuit is

A B

C D

5 10 5

5 10 5

10 10

(a) 10 W (b) 20 W (c) 5 W (d) 30 W

25. A rain drop of radius 0.3 mm falling vertically downwards in air has a terminal velocity of 1 m s–1. The viscosity of air is 18 × 10–5 poise. The viscous force on the drop is(a) 101.73 × 10–4 dyne (b) 101.73 × 10–5 dyne(c) 16.95 × 10–5 dyne (d) 16.95 × 10–4 dyne

26. A stone is projected vertically up to reach maximum height h. The ratio of its kinetic energy

to its potential energy, at a height 45

h, will be

(a) 5 : 4 (b) 4 : 5 (c) 1 : 4 (d) 4 : 1

27. Refer to the arrangement of logic gates. For A = 0, B = 0 and A = 1, B = 0, the values of output Y are, respectively

AB

Y

(a) 0 and 1 (b) 1 and 0(c) 1 and 1 (d) 0 and 0

28. Four massless springs whose spring constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane as shown in figure. If the mass M is displaced in the horizontal direction, then the frequency of the system is

M2k 2k

k

2k

(a) 12 4π

kM

(b) 1

24

πk

M

(c) 1

2 7πkM

(d) 12

7π

kM

29. Oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is 0.5 m, the kinetic energy of a proton, which is accelerated by the cyclotron is(a) 10.2 MeV (b) 20.1 MeV (c) 5.1 MeV (d) 1.5 MeV

30. If the length of stretched string is shortened by 40% and the tension is increased by 44 %, then the ratio of the final and initial fundamental frequencies is (a) 2 : 1 (b) 3 : 2 (c) 3 : 4 (d) 1 : 3

31. Sodium lamps are used in foggy conditions because(a) yellow light is scattered less by the fog

particles.(b) yellow light is scattered more by the fog

particles.(c) yellow light is unaffected during its passage

through the fog.(d) wavelength of yellow light is the mean of the

visible part of the spectrum.

32. In Young’s double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 Å coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1P – S2P in microns is(a) 0.75 (b) 1.5 (c) 3.0 (d) 4.5

33. Photoelectric effect is an example of(a) elastic collision (b) inelastic collision(c) two dimensional collision(d) oblique collision

34. The angle of minimum deviation in an equilateral prism of refractive index 1.414 is(a) 60° (b) 30° (c) 90° (d) 45°

35. A black body of mass 34.38 g and surface area 19.2 cm2 is at an initial temperature of 400 K. It is allowed to cool inside an evacuated enclosure kept at constant temperature 300 K. The rate of cooling is 0.04°C per second. The specific heat of the body in J kg–1 K–1 is

(Stefan’s constant s = 5.73 × 10–8 Wm–2 K–4)(a) 2800 (b) 2100 (c) 1400 (d) 1200

36. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

(a) ln 25

(b) 52ln

(c) 5 log10 2 (d) 5 ln 2


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Page 28

37. Two concentric coils of 10 turns each are placed in the same plane. Their radii are 20 cm and 40 cm and carry 0.2 A and 0.3 A current respectively in opposite directions. The magnetic field (in tesla) at the centre is

(a) 34 0µ (b) 5

4 0µ

(c) 74 0µ (d) 9

4 0µ

38. When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal g–1 is(a) 273 cal K–1 (b) 8 × 104 cal K–1 (c) 80 cal K–1 (d) 293 cal K–1

39. A coil has 1,000 turns and 500 cm2 as its area. The plane of the coil is placed at right angles to a magnetic field of 2 × 10–5 Wb m–2. The coil is rotated through 180° in 0.2 s. The average emf induced in the coil is(a) 5 mV (b) 10 mV (c) 15 mV (d) 20 mV

40. Number of neutrons in 612C and 614C are(a) 8 and 6 (b) 6 and 8(c) 6 and 6 (d) 8 and 8

41. Pressure P, volume V and temperature T for a

certain gas are related by P = AT BTV− 2

, where

A and B are constants. The work done by the gas as its temperature changes from T1 to T2 while pressure remains constant is

(a) A B T T− −( )2 2 1

(b) A(T2 – T1) – B(T2 2 – T1 2)

(c) AT

T T B T T22

12

23

13

3−( ) − −( )

(d) A(T2 – T1)2 – B3

(T2 – T1)3

42. When 1530P decays to become 14

30Si the particle released is(a) electron (b) a-particle(c) neutron (d) positron

43. A network of six identical capacitors, each of value C is made as shown in the figure. Equivalent capacitance between points A and B is

A

B

(a) C4

(b) 34C

(c) 43C (d) 3C

44. A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If its equation of motion is y = bx2 (b is a constant), its velocity component in the x-direction is

(a) 2ba (b)

ab2 (c)

ab (d)

ba

45. A galvanometer, having a resistance of 50 W, gives a full scale deflection for a current of 0.05 A. The length of a resistance wire of area of cross-section 2.97 × 10–2 cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is

(Specific resistance of the wire = 5 × 10–7 W m)(a) 9 m (b) 6 m (c) 3 m (d) 1.5 m

46. In the figure, the velocity v3 will be

(a) zero (b) 4 m s–1

(c) 1 m s–1 (d) 3 m s–1

47. A travelling wave is represented by the equation

y t x= +1

1060 2sin( ), where x and y are in metres

and t is in seconds. This represents a wave(1) travelling with a velocity of 30 m s–1

(2) of frequency 30π

Hz

(3) of wavelength p m(4) of amplitude 10 cm(5) moving in the positive x directionPick out the correct statements from the above.(a) 1, 2, 4 (b) 3, 4, 5(c) 1, 2, 3, 4 (d) All

48. If 4 moles of an ideal monoatomic gas at temperature 400 K is mixed with 2 moles of another ideal monoatomic gas at temperature 700 K, the temperature of the mixture is(a) 550°C (b) 500°C (c) 550 K (d) 500 K

Cont. on Page No. 70


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Physics for you | february ‘13 29

Page 29

1. A ball is travelling with uniform translatory motion. This means that(a) it is at rest.(b) the path can be a straight line or circular and

the ball travels with uniform speed.(c) all parts of the ball have the same velocity

(magnitude and direction) and the velocity is constant.

(d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

2. Which one of the following statements is not true about Newton’s second law of motion

F ma= ?(a) The second law of motion is consistent with

the first law.(b) The second law of motion is a vector law.(c) The second law of motion is applicable to a

single point particle.(d) The second law of motion is not a local law.

3. A body of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is

(a) 1 m s–2 at an angle of tan–1 34

w.r.t. 8 N force.

(b) 0.2 m s–2 at an angle of tan–1 34

w.r.t. 8 N force.

(c) 1 m s–2 at an angle of tan–1 54

w.r.t. 8 N force.

(d) 0.2 m s–2 at an angle of tan–1 54

w.r.t 8 N force.

4. Conservation of momentum in a collision between particles can be understood from(a) conservation of energy.(b) Newton’s first law only.

(c) Newton’s second law only.(d) both Newton’s second and third law.

5. Which one of the following is not a contact force?(a) Viscous force (b) Magnetic force(c) Friction (d) Buoyant force

6. A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 m s–1, q = 4 m s–2 and r = 5 m s–3.

The force acting on the body at t = 2 s is (a) 136 N (b) 134 N (c) 158 N (d) 68 N7. A person of mass 50 kg stands on a weighing

scale on a lift. If the lift is descending with a downward acceleration of 9 m s–2, what would be the reading of the weighing scale?

(Take g = 10 m s–2) (a) 50 kg (b) 5 kg (c) 95 kg (d) 100 kg

8. Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 m s–1 each, collide and rebound with the same speed. The impulse imparted to each ball is

(a) 0.25 kg m s–1 (b) 0.5 kg m s–1

(c) 0.1 kg m s–1 (d) 0.125 kg m s–1

9. A body with mass 5 kg is acted upon by a force F i j= − +( )^ ^3 4 N. If its initial velocity at t = 0 is u i j= −( )^ ^6 12 m s–1, the time at which it will just

have a velocity along the y-axis is (a) never (b) 10 s (c) 2 s (d) 15 s10. In figure, the coefficient of

friction between the floor and the block B is 0.1. The coefficient of friction


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30 Physics for you | february ‘13

Page 30

between the blocks B and A is 0.2. The mass of A is m/2 and of B is m. What is the maximum horizontal force F can be applied to the block B so that two blocks move together?(a) 0.15mg (b) 0.05mg(c) 0.1mg (d) 0.45mg

11. A stream of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of cross-sectional area 10–2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?(a) 1.25 × 103 N (b) 2.25 × 103 N(c) 3.25 × 103 N (d) 4.25 × 103 N

12. Which one of the following statements is not true?(a) The same force for the same time causes the same

change in momentum for different bodies.(b) The rate of change of momentum of a body

is directly proportional to the applied force and takes place in the direction in which the force acts.

(c) A greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.

(d) The greater the change in the momentum in a given time, the lesser is the force that needs to be applied.

13. A block of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown in the figure. What is the angle

the rope makes with the vertical in equilibrium? Neglect the mass of the rope.

(Take g = 10 m s–2) (a) 30° (b) 40° (c) 60° (d) 45°14. A person in an elevator accelerating upwards with

an acceleration of 2 m s–2, tosses a coin vertically upwards with a speed of 20 m s–1. After how much time will the coin fall back into his hand?

(Take g = 10 m s–2) (a) 1.67 s (b) 2 s (c) 3.33 s (d) 5 s15. Two masses of 5 kg and 3 kg are

suspended with help of massless inextensible strings as shown in figure. The whole system is going upwards with an acceleration of 2 m s–2. The tensions T1 and T2 are

(Take g = 10 m s–2) 3 kg

5 kgT1

T2

(a) 96 N, 36 N (b) 36 N, 96 N(c) 96 N, 96 N (d) 36 N, 36 N

16. A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is m and the acceleration due to gravity is g, what is the minimum force required to be applied by the finger to hold the block against the wall?

(a) mMg (b) Mg (c) Mgµ

(d) 2mMg

17. There are four forces acting at a point P produced by strings as shown in figure, which is at rest. The forces F1 and F2 are

(a) 12

32

N N,

(b) 32

12

N N,

(c) 12

12

N N, (d) 32

32

N N,

18. A metre scale is moving with uniform velocity. This implies(a) the force acting on the scale is zero, but a

torque about the centre of mass can act on the scale.

(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.

(c) the total force acting on it need not be zero but the torque on it is zero.

(d) neither the force nor the torque need to be zero.

19. A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun.

(Take g = 10 m s–2)(a) 0.2 m s–1 (b) 0.4 m s–1

(c) 0.6 m s–1 (d) 0.8 m s–1

20. A large force is acting on a body for a short time. The impulse imparted is equal to the change in(a) acceleration (b) momentum(c) energy (d) velocity

21. Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s–2. What is the net force on the man? (Mass of the man = 65 kg)

2 N 1 N

P

45° 45°

90°

F1

F2


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Page 31

(a) 35 N (b) 45 N (c) 55 N (d) 65 N22. Figure shows (x, t), (y, t) diagram of a particle

moving in 2-dimensions.

If the particle has a mass of 500 g, the force acting on the particle is(a) 1 N along y-axis (b) 1 N along x-axis(c) 0.5 N along x-axis (d) 0.5 N y-axis

23. Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° as shown in the figure. A flexible cord attached to A passes over a frictionless pulley and is connected to

block B of weight W. Find the weight W for which the system is in equilibrium.

(a) 25 N (b) 50 N (c) 75 N (d) 100 N24. The position-time

graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 4 s?

(a) 23

kg m s–1

(b) −23

kg m s–1

(c) 32

kg m s–1 (d) −32

kg m s–1

25. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is

(a) T (b) T mvl

−2

(c) T mvl

+2

(d) 0

Where T is the tension in the string.26. Ten one-rupee coins are put on top of each other

on a table. Each coin has a mass m. The force on the 7th coin (counted from the bottom) due to all the coins on its top is(a) 3mg, vertically downwards(b) 3mg, vertically upwards(c) 7mg, vertically downwards(d) 7mg, vertically upwards

27. A car of mass m starts from rest and acquires a velocity along east v v i=

^ (v > 0) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is

(a) mv2

eastward and is exerted by the car engine.

(b) mv2

eastward and is due to the friction on

the tyres exerted by the road.

(c) more than mv2

eastward exerted due to the

engine and overcomes the friction of the road.

(d) mv2

exerted by the engine.

28. A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are

Lowest Point Highest Point(a) mg – T 1 mg + T 2(b) mg + T 1 mg – T 2

(c) mg + T 1 – mvR

12

mg – T 2 + mvR

22

(d) mg – T 1 – mvR

12

mg + T 2 + mvR

22

T 1 and v1 denote the tension and speed at the lowest point. T 2 and v2 denote corresponding values at the highest point.

29. A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race car and the road is 0.2, what is the maximum permissible speed to avoid slipping?

(Take tan15° = 0.27)(a) 18.2 m s–1 (b) 28.2 m s–1

(c) 38.2 m s–1 (d) 48.2 m s–1


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30. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is(a) frictional force along westward.(b) muscle force along southward.(c) frictional force along south-west.(d) muscle force along south-west.

solutions

1. (c)2. (d) : The second law of motion is a local law which

means that forceFat a point in space (location of

the particle) at a certain instant of time is related to a at that point at that instant. Acceleration here and now is determined by the force here and now, not by any history of the motion of the particle.

3. (a) : Here, m = 10 kg

6 N

8 N

FThe resultant force acting on the body is

F = + =( ) ( )8 6 102 2N N N

Let the resultant force F makes an q w.r.t. 8 N force.

From figure, tanθ = =68

34

NN

The resultant acceleration of the body is

a Fm

= =1010

Nkg

= 1 m s–2

The resultant acceleration is along the direction of the resultant force.Hence, the resultant acceleration of the body is

1 m s–2 at an angle of tan–1 34

w.r.t. 8 N force.

4. (d) : Newton’s second and third laws of motion lead to conservation of linear momentum.

5. (b) : Among the given forces magnetic force is a non-contact force.

6. (a) : Here, x(t) = pt + qt2 + rt3

where p = 3 m s–1, q = 4 m s–2 and r = 5 m s–3

m = 2 kg

Velocity, v dxdt

ddt

= = (pt + qt2 + rt3) = p + 2qt + 3rt2

Acceleration, a dvdt

= = 2q + 6rt

At t = 2 sa = 2(4 m s–2) + 6(5 m s–3)(2 s) = 8 m s–2 + 60 m s–2 = 68 m s–2

The force acting on the body of mass 2 kg isF = ma = (2 kg)(68 m s–2) = 136 N

7. (b) : The reading on the scale is a measure of the force on the floor by the person. By the Newton’s third law this is equal and opposite to the normal force N on the person by the floor.\ When the lift is descending downward with a acceleration of a m s–2, then50 × 10 – N = 50 × 9or N = 50 × 10 – 50 × 9 = 50 N\ The reading of weighing machine is 5 kg.

8. (b) :

Initial momentum of ball A = (0.05 kg)(5 m s–1) = 0.25 kg m s–1

As the speed is reversed on collision,Final momentum of the ball A = (0.05 kg)(–5 m s–1) = – 0.25 kg m s–1

Impulse imparted to the ball A = Change in momentum of ball A = Final momentum – Initial momentum = – 0.25 kg m s–1 – 0.25 kg m s–1

= – 0.5 kg m s–1

Similarly,Initial momentum of ball B = (0.05 kg)(–5 m s–1) = – 0.25 kg m s–1

Final momentum of ball B = (0.05 kg)(5 m s–1) = + 0.25 kg m s–1

Impulse imparted to ball B = (0.25 kg m s–1) – (– 0.25 kg m s–1) = 0.5 kg m s–1

Impulse imparted to each ball is 0.5 kg m s–1 in magnitude. The two impulses are opposite in direction.

9. (b) : Here, m = 5 kg, F i j= − +3 4

^ ^N

u i j= − −6 12 1^ ^m s

The acceleration of the body is

a Fm

i ji j= =

− += − + −( )

^ ^^ ^3 4

535

45

2Nkg

m s

Velocity of the body along x-axis at any time t is

vx = ux + axt = 6 35

i i t^ ^−

The body will have a velocity along y-axis, if its velocity along x-axis will be zero.i.e. vx = 0 ⇒ t = 10 s

10. (d) : Here, m A = m2

, mB = m

m A = 0.2, mB = 0.1Let both the blocks are moving with common acceleration a. Then,


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Page 33

am gm

g gA A

AA= = =

µµ 0 2.

and F – mB(mB + mA)g = (mB + mA)aF = (mB + mA)a + mB(mB + mA)g

= m m g m m g+

+ +

2

0 2 0 12

( . ) ( . )

= 32

0 2 32

0 1m g m g

+

( . ) ( . ) = 0 9

20 45. .mg mg=

11. (b) : Here, v = 15 m s–1

Area of cross section, A = 10–2 m2

Density of water, r = 103 kg m–3

Mass of water hitting the wall per second

= r × A × v= 103 kg m–3 × 10–2 m2 × 15 m s–1 = 150 kg s–1

Force exerted on the wall= Momentum loss of water per second

= 150 kg s–1 × 15 m s–1 = 2250 N = 2.25 × 103 N

12. (d) : The greater the change in the momentum in a given time, the greater is the force that needs to be applied.

13. (b) :

The free body diagram of 6 kg block is as shown in Fig. (b). In equilibrium T2 = 6 g = 6 × 10 = 60 N The free body diagram of the point P is as shown in Fig. (c).In equilibriumT 1sinq = 50 N ...(i) T 1cosq = T 2 = 60 N ...(ii) Dividing (i) by (ii), we get

tanθ = =5060

56

θ =

= °−tan 1 5

640

14. (c) : Here, v = 20 m s–1, a = 2 m s–2, g = 10 m s–2

The coin will fall back into the person’s hand after t s.

\ t va g

=+

=×

+= = =

−

−2 2 20

2 104012

103

3 331

2m sm s

s s s( )

.

15. (a) : The free body diagram of 3 kg block is as shown in the Fig. (a).The equation of motion of 3 kg block isT 2 – 3g = 3a

T 2 = 3(a + g) = 3(2 + 10) = 36 N …(i)The free body diagram of 5 kg is as shown in the Fig. (b).The equation of motion of 5 kg block isT 1 – T 2 – 5g = 5a

T 1 = 5(a + g) + T 2 = 5(2 + 10) + 36 (Using (i)) = 96 N

16. (c) : If F is the force of the finger on the book, F = N, the normal reaction of the wall on the book. The minimum upward frictional force needed to ensure that the book does not fall is Mg. The frictional force = mN.

Thus, minimum value of F Mg=

µ.

17. (a) :

2 N 1 N

P

1 cos45°

2 sin45° 1 sin45°F1

F2

2 cos45°

45°

90°

45°

y

x

Applying equilibrium conditions,SFx = 0⇒ F1 + 1sin45° – 2sin45° = 0or F1 = 2sin45° – 1sin45°

= 22

12

12

− = N

and SFy = 0⇒ 1cos45° + 2sin45° – F2 = 0

F2 = 12

22

32

+ = N

18. (b)


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19. (b) : Here,Mass of the gun, M = 100 kg

Mass of the ball, m = 1 kg

Height of the cliff, h = 500 m

g = 10 m s–2

Time taken by the ball to reach the ground is

t hg

= =×

=−

2 2 50010

102m

m ss

Horizontal distance covered = ut

\ 400 = u × 10where u is the velocity of the ball.

u = 40 m s–1

According to law of conservation of linear momentum, we get

0 = Mv + mu

v muM

= − = − = −−

−( )( ).

1 40100

0 41

1kg m skg

m s

–ve sign shows that the direction of recoil of the gun is opposite to that of the ball.

20. (b) : If a large force F acts for a short time dt the impulse imparted I is

I = Fdt = dpdt

dt

I = dp = change in momentum

21. (d) : Here, mass of the man M = 65 kgAs the man is standing stationary w.r.t. the belt, acceleration of man = acceleration of belt.\ Acceleration of man, a = 1 m s–2

\ Net force on the man F = Ma = (65 kg)(1 m s–2) = 65 N

22. (a) : Since the graph between x and t is a straight line and passing through the origin.\ x = tSince the graph between y and t is a parabola.\ y = t2

\ v dxdt

advdtx x

x= = = =1 0and

and v dydty = = 2t and ay = 2 m s–2

The force acting on the particle isF = may = (0.5 kg)(2 m s–2) = 1 N along y-axis

23. (b) :

30°A B

30°

mgcos30°

mgsin30

°N

W

T

T

mg

As the system is in equilibrium,\ T = W …(i)and T = mgsin30° …(ii)From (i) and (ii), we get

W = mgsin30° = (100 N)12

= 50 N

24. (d) : At t = 4 s, the body has constant velocity

u = 34

m s–1.

After t = 4 s, the body is at rest i.e., v = 0\ Impulse = m(v – u)

= 2 kg (0 – 34

m s–1) = −32

kg m s–1

25. (a) : The net force on the particle is T and is directed towards the centre of the circle. The tension T provides the necessary centripetal force to the particle moving in the circle.

26. (a) : The force on 7th coin is due to weight of the three coins lying above it.\ F = 3mgThis force acts vertically downwards.

27. (b)

28. (a) : At the lowest point, mg acts downwards and T 1 upwards so that net force = mg – T 1.At the highest point, both mg and T2 act downwards so that net force = mg + T 2.Hence, option (a) is correct.

29. (c) : Here R = 300 m, q = 15°, g = 9.8 m s–2, m = 0.2The maximum permissible speed is given by

v Rgmax

( tan )tan

. ( . . ). .

=+

−=

× × +− ×

µ θµ θ1

300 9 8 0 2 0 271 0 2 0 27

= 38.2 m s–1

30. (c)

nn


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Available atleading bookshopsthroughout India.

It is a matter of great satisfaction and encouragement to realise that our efforts in “Excel in Physics, Excel in Chemistry and Excel in Biology” have received overwhelming response. Nearly all the questions asked in the CBSE Board Examination 2012 were available in the books; fully solved.

We feel pleased to present the revised edition of the books. The books give com-prehensive account of the subject according to the current syllabus and pattern of the CBSE Board Examination.

This will impart the students a clear and vivid understanding of the subject matter.


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Page 36

36 physics for you | February’13

1. Why are alloys used for making standard resistance coils? [1]

2. How does focal length of a lens change when red light is replaced by blue light? [1]

3. Name the series of hydrogen spectrum lying in the infrared region. [1]

4. Would sky waves be suitable for transmission of TV signals of 60 MHz frequency? [1]

5. What is the colour code for a resistor of resistance 3.5 kW with 5% tolerance? [1]

6. Which physical quantity has the unit Wb m–2? Is it a scalar or a vector quantity? [1]

7. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. [1]

8. Why is ground wave transmission of signals restricted to a frequency of 1500 kHz? [1]

9. A uniform field

E exists –––––––––––––––

++++++++++++++

A B

D C

between two charged plates as shown in figure. What would be work done in moving a charge q along the closed rectangular path ABCDA?

[2]10. What is an ideal diode? Draw the output

waveform across the load resistor R, if the input waveform is as shown in the figure.

R

– 6 V

0

+ 6 V

[2]11. Find the ratio of intensities of two points P and

Q on a screen in Young’s double slit experiment when waves from sources S1 and S2 have phase

difference of (i) 0° and (ii) π2

respectively. [2]

12. Two conductors are made of the same material and have the same length. Conductor A is solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB. [2]

13. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor 2. Calculate by what factor the voltage sensitivity changes. [2]

14. A coil of 0.01 H inductance and 1 W resistance is connected to 200 V, 50 Hz ac supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current. [2]

15. With the help of an example, explain, how the neutron to proton ratio changes during alpha decay of a nucleus. [2]

16. Two long parallel wires are hanging freely. If they are connected to a battery (i) in series, (ii) in parallel, what would be the effect on their positions? [2]

17. Obtain equivalent capacitance of the following network. For a 300 V supply, determine the charge and voltage across each capacitor. [3]

18. Which of the following waves can be polarised? (i) X-rays (ii) Sound waves. Give reasons. Two polaroids are used to study polarisation.

One of them (the polariser) is kept fixed and the other (the analyser) is initially kept with its axis parallel to the poalriser. The analyser is then rotated through angles of 45°, 90° and 180° in turn. How would the intensity of light coming out of analyser be affected for these angles of rotation,


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Page 37

physics for you | February’13 37

as compared to the initial intensity and why? [3]

19. Illustrate the basic elements required for transmitting and receiving an audio signal with the help of a block diagram. [3]

20. Two cells of voltage R

2 V 5 Ω

10 V 10 Ω

10 V and 2 V and

internal resistances 10 W and 5 W respectively, are connected in parallel with the positive end of 10 V battery connected

to negative pole of 2 V battery as shown in figure. Find the effective voltage and effective resistance of the combination. [3]

21. A source contains two phosphorus radionuclides

1532P (T1/2 = 14.3 days) and 15

33P (T1/2 = 25.3 days). Initially 10% of the decays come from 15

33P. How long one must wait until 90% do so? [3]

22. The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the

capacitance of the capacitor becomes 23

rdof its

original value. What is the new distance between the plates? [3]

OR Define the term electric potential due to a point

charge. Find the electric potential at the centre of a square of side 2 m, having charges 100 mC, –50 mC, 20 mC and – 60 mC at the four corners of the square. [3]

23. How would you establish an instantaneous displacement current of 2.0 A in the space between the two parallel plates of 1 mF capacitor? [3]

24. The figure shows 15 A

25 A

25 cm

10 cm

2 cm

a rectangular current carrying loop, placed 2 cm away from a long straight, current carrying conductor. What is the direction and magnitude of the net force acting on

the loop? [3]

25. Two lenses of powers + 15 D and – 5 D are in contact with each other forming a combination lens.(a) What is the focal length of this combination?(b) An object of size 3 cm is placed at 30 cm from

this combination of lenses. Calculate the position and size of the image formed. [3]

26. In an experiment of photoelectric effect, Neeta plotted graphs for different observation between photoelectric current and collector plate potential but her friend Megha has to help her in plotting the correct graph. Neeta thanked Megha for timely help.(a) What value was displayed by Megha and

Neeta.(b) Draw the correct graph between I and V. [4]

27. What is induced emf? Write Faraday’s law of electromagnetic induction. Express it mathematically. A conducting rod of length l, with one pivoted, is rotated with a uniform angular speed w in a vertical plane, normal to a uniform magnetic field B. Deduce an expression for the emf induced in this rod.

In India, domestic power supply is at 220 V, 50 Hz, while in USA it is 110 V, 50 Hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply. [5]

OR Explain the phenomenon of self induction. Define

coefficient of self inductance. What are its units? Calculate self inductance of a long solenoid. [5]

28. (a) With the help of a circuit diagram explain the working of transistor as oscillator.(b) Draw a circuit diagram for a two inputs OR

gate and explain its working with the help of input, output waveforms. [5]

OR Define the terms potential barrier and depletion

region for a p-n junction. Explain with the help of a circuit diagram, the use of a p-n diode as a full wave rectifier. Draw the input and output wave forms. [5]

29. Define magnifying power of an optical telescope. Draw a ray diagram for an astronomical refracting telescope in normal adjustment showing the paths through the instrument of three rays from a distant object. Derive an expression for its magnifying power. Write the singificance of diameter of the objective lens on the optical performance of a telescope. [5]


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OR State Huygens principle and prove laws of

reflection and refraction on the basis of Huygens principle. [5]

SOLUTIONS1. Alloys are used for making standard resistance

coils because they have low value of temperature coefficient (less temperature sensitivity) of resistance and high resistivity.

2. According to lens maker’s formula,

1 1 1 1

1 2f R R= − −

( )µ

As mb > mr \ fb < fr i.e. focal length of lens decreases.

3. Paschen series, Brackett series and Pfund series.4. No, signals of frequency greater than 30 MHz

will not be reflected by the ionosphere, but will penetrate through the ionosphere.

5. Given, resistance = 3.5 kW ± 5% = 35 × 102 W ± 5%\ Colour code of given resistor is orange, green, red and gold.

6. Magnetic field induction has the unit Wb m–2. It is a vector quantity.

7. The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced emf should resist this increase, which can be achieved by a decrease in current.

8. In ground wave propagation, the loss of energy due to interaction with matter increases with the increase in frequency of wave. Therefore, the waves of frequency above 1500 kHz get heavily damaged in ground wave propagation. Hence the ground wave propagation is restricted to a frequency of 1500 kHz.

9. Work done in moving a charge along the closed rectangular path would be zero, because field in the entire space is uniform and electrostatic forces are conservative forces.

10. A p-n junction diode which offers zero resistance when forward biased and infinite resistance when reverse biased is called an ideal diode.

The output waveform across R is as shown in the figure below.

+ 6 V

0

11. In Young’s double slit experiment, the resultant intensity at any point on the screen is

I I I I I= + +1 2 1 22 cosφ

where f is the phase difference between the waves at that point.

Here, we consider I1 = I2 = I0 Therefore, When f = 0°, the intensity at point P is ∴ = + + ° =I I I I I IP 0 0 0 0 02 0 4cos

When φ π=2

, the resultant intensity at point Q is

I I I I I IQ = + + =0 0 0 0 022

2cos π

∴ =

II

P

Q

21

12. Resistance of conductor A,

RA = ρπ

l( . )0 5 10 3 2× −

Resistance of conductor B,

RB = ρ

πl

[( ) ( . ) ]1 10 0 5 103 2 3 2× − ×− −

\ RR

A

B=

× − ××

= =− −

−( ) ( . )

( . )..

1 10 0 5 100 5 10

0 750 25

31

3 2 3 2

3 2

13. Given, Is′ = I I Is s s+ =20

100120100 , R′ = 2R

Then, initial voltage sensitivity, Vs = IR

s

New voltage sensitivity,

Vs′ = IR

IR

Vss s

′

′=

× =120100

12

35

\ % decrease in voltage sensitivity

= V V

V

V V

Vs s

s

s s

s

− ′× =

−×100

35 100 = 40%

14. Here, R = 1 W, u = 50 Hz, L = 0.01 H The inductive reactance is XL = wL = 2puL = 2 × 3.14 × 50 × 0.01 = 3.14 W The impedance of the circuit is

Z = R XL2 2+ = ( ) ( . ) . .1 3 14 10 86 3 32 2+ = Ω

The phase difference between current and voltage is

tan f = XR

L = 3.14

f = tan–1(3.14) 72° 72

180× π rad

Time lag, Dt = φω

ππ

=×

× ×=

72180 2 50

1250

s


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15. Consider the a-decay of nucleus

ZA

ZAX Y → +−

−24

24He

92238

90234

24U Th He → +

Before decay, neutron to proton ratio, n/p

= 238 9292− = 1.58

After decay, neutron to proton ratio, n/p

= 234 90

90−

= 1.60

Thus ratio increases.

16. (i) When a battery is connected in series to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they move further apart. (ii) When a battery is connected in parallel to two

long parallel wires, the currents in the two wires will be in same direction. Due to which a force of attraction will be acting between them and they come closer to each other.

17.

C1 = 100 pF

C4 = 100 pF

C2 = 200 pF C3 = 200 pF+

–300 V

Here, C2 and C3 are in series.

\ 1 1

2001

2002

2001

100Cs= + = =

Cs = 100 pFNow Cs and C1 are in parallel.\ Cp = Cs + C1 = 100 + 100 = 200 pFAgain, Cp and C4 are in series, therefore the equivalent capacitance of the network is

\ 1 1 1

4C C Cp= + =

1200

1100

3200

+ =

\ C = 2003

pF = 66.7 × 10–12 F

As Cp and C4 are in series,\ Vp + V4 = 300 ...(i)

Charge on C4 is q4 = CV = 200

310 30012× ×−

= 2 × 10–8 C

Potential difference across C4 is

V4 = qC

4

4

8

122 10

100 10=

××

−

− = 200 V

From (i), Vp = 300 – V4 = 300 – 200 = 100 VPotential difference across C1 is V1 = Vp = 100 VCharge on C1 is q1 = C1V1 = 100 × 10–12 × 100 = 10–8 CPotential difference across C2 and C3 in series = 100 VCharge on C2 is q2 = C2 V2 = 200 × 10–12 × 50 = 10–8 CCharge on C3 is q3 = C3 V3 = 200 × 10–12 × 50 = 10–8 C

18. (i) Phenomenon of polarisation is shown by transverse waves only. X-rays are transverse in nature, and hence they can be polarised.

According to Malus law I = I0 cos2q where

I = intensity of light coming from analyserI0 = initial intensity

When q = 45°, then I = I0 cos245° =I0

2 When q = 90°, then I = I0 cos290°= 0 When q = 180°, then I = I0 cos2180° = I0

19.

DetectorTransmitterModulator

Oscillator

Loudspeaker

Transmittingantenna

Receivingantenna

Microphone

A brief description of the various elements is as given below.(i) A microphone converts sound waves into

electrical waves i.e. audio signal.(ii) An oscillator generates carrier waves.(iii) There is a mixing of carrier waves and audio

signal in a modulator.(iv) The modulated waves are fed to transmitter,

these waves are then radiated through transmitting antenna.

(v) The rece iving antenna rece ives the transmitting signal.

(vi) The detector demodulates (separate out) the audio signal from the modulated waves.

(vii) The loudspeaker converts the audio signal back into sound waves.


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20.

I2

I

I1

R

Veff

I

Reff

R

2 V

5 Ω

10 V 10 ΩA

B

F

E

DC

Applying Kirchhoff’s junction rule at junction B, we get I1 = I + I2 ...(i) Applying Kirchhoff’s loop rule for closed loop ABCDEFA gives 10 = IR + 10I1 ...(ii)Applying Kirchhoff’s loop rule for closed loop BCDEB gives 2 = 5I2 – IR = 5(I1 – I) – IR (Using (i))or 4 = 10I1 – 10I – 2IR ...(iii)

Equation (i) – (iii) gives 6 = 3IR + 10I

or 2 = I R +

103

or 2 = (R + Reff)I

Comparing with Veff = (R + Reff)I, we get

Veff = 2 V and Reff = 103

Ω

21. Initially, the source has 90% of 1532P nuclides and

10% of 1533P nuclides. Finally, say after t days, the

source has 10% of 1532P nuclides and 90% of 15

33Pnuclides.\ Initial no. of 15

32P nuclides = 9x;

Initial no. of 1533P nuclides = x

Final no. of 1532P nuclides = y;

Final no. of 1533P nuclides = 9y

As NN

n t Tt T

0

12

12

21 2

1 2=

=

= −/

//

/( )

or N N t T= −0 2 1 2( ) / /

For 1532P isotope,

N0 = 9x; N = y; T1/2 = 14.3 days,

\ y = 9x(2)–t/14.3 ...(i)

For 1533P isotope,

N0 = x; N = 9y; T1/2 = 25.3 days\ 9y = x(2)–t/25.3 ...(ii)

Dividing Eq. (ii) by Eq. (i), we get

9 19

2 81 21

14 31

25 3 11 14 3 25 3= =−

×( ) . . / . .

ttor

∴ =×

log. .

log10 1081 1114 3 25 3

2t

or 1 9085 11 0 301014 3 25 3

. .. .

=×

×t

∴ =× ××

=t 1 9085 14 3 25 311 0 3010

208 5. . ..

. days

22. Here, distance between parallel plates d = 4 mm = 0.004 m, K = 3, thickness, t = 3 mm = 0.003 m

Let the new distance between the plates be d1.

∴ = =− −

CA

dC

A

d tK

ε ε01

0

1 1 1and

Since C C1

23

=

(Given)

∴− −

=ε ε0

1

0

1 123

A

d tK

Ad

1

1 12

31d t

Kd− −

=

1

0 003 1 13

23 0 004

1d − −

=×. .

1

0 003 23

10 006

1d − ×=

. .

10 002

10 0061d −

=. .

d1 – 0.002 = 0.006 d1 = 0.006 + 0.002 = 0.008 m = 8 mm

OR Electric potential at a point is the amount of work

done to bring a unit positive charge from infinity to that point against the electrostatic forces.

2 m

AC BD= = ( ) + ( ) =2 2 22 2

m


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Page 41


Focal length, f = 1 110

100P

= × = 10 cm

(b) 1 1 1f v u

= −

1 1 1 110

130v f u

= + = +−

or

1 3 130

230v

=−

=

\ v = 15 cm

m = vu

hh

I

o=

or hI = vu

= 1530

3−

× = – 1.5 cm

26. (a) The values displayed by them is sharing and caring.

(b)

–V0 0Retardingpotential

Collectorplate potential

Stoppingpotential

Phot

ocur

rent

I I I3 2 1> >I3I2I1

27. Refer point 4.1(3, 4, 5, 7 (d)) page no. 224, 225 (MTG Excel in Physics).Advantage: At 220 V supply power loss due to heating effect is lesser.Disadvantage: At 220 V peak value of current is more. Thus, it is more dangerous.

OR Refer point 4.2(1) page no. 226 (MTG Excel in

Physics).28. (a) Refer point 9.4(10) page no. 541, (MTG Excel in

Physics). (b) Refer point 9.5(1) page no. 542, (MTG Excel in

Physics).OR

Refer point 9.3[(2), 6(ii)] page 532, 534 (MTG Excel in Physics).

29. Refer point 6.9(3), page 347 (MTG Excel in Physics).

OR Refer point 6.10(6) and point 6.11 page no. 403,

404 (MTG Excel in Physics).mmm

\ AO = OC = BO = OD = 1 m Potential at the centre of the square O is

V

qAO

qBO

qOC

qOD

A B C D= + + +

14 0πε

V = ××

−×

+×

−×

− − − −9 10 100 10

150 10

120 10

160 10

19

6 6 6 6

= 9 × 109 × 10 × 10–6 = 9 × 104 V

23. Here, ID = 2.0 A, C = 1 mF = 10–6 F

I

ddt

ddt

EA AdEdtD

E= = =εφ

ε ε0 0 0( )

=

= =ε ε0 0A

ddt

Vd

Ad

dVdt

CdVdt

E

Vd

CA

d=

and = ε0

or

dVdt

ICD= = = ×

−−2 0

102 10

66 1. V s

Thus a displacement current of 2.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of 2 × 106 V s–1.

24.

15 AB

25 A

25 cm

10 cm

2 cm

C P

QDA

Force between wires PQ and CD

F1 = µ

π0 1 2

2I I l

r

= 2 10 15 25 0 25

2 10

7

2× × × ×

×

−

−

.= 93.75 × 10–5

= 9.375 × 10–4 N (Repulsive)Force between wires PQ and AB

F2 = 2 10 15 25 0 25

0 12

7× × × ×− ..

= 1.56 × 10–4 N (Attractive)Net force on the rectangular loop F = F1 – F2 = (9.375 – 1.56) × 10–4

= 7.815 × 10–4 N (Repulsive i.e., towards left).

25. (a) Here, P1 = 15 D, P2 = –5 DPower of the combination, P = P1 + P2 = 15 D – 5 D = 10 D


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42 physics for you | february‘13

Page 42

1. The random error in the arithmetic mean of 100 observations is x, then the random error in the arithmetic mean of 400 observations would be

(a) 4x (b) 14

x

(c) 2x (d) 12

x

2. After one second the velocity of a projectile makes an angle of 45° with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are (Take g = 10 m s–2)

(a) 14.62 m s–1, 60° (b) 14.62 m s–1, tan–1(2)(c) 22.36 m s–1, tan–1(2) (d) 22.36 m s–1, 60°

3. A block of mass 1 kg is

2 kg

1 kg

F = 30 N

2 m

placed over a plank of mass 2 kg. The length of the plank is 2 m.

Coefficient of friction between the block and the plank is 0.5 and the ground over which plank is placed is smooth. A constant force F = 30 N is applied on the plank in horizontal direction. The time after which the block will separate from the plank is (Take g = 10 m s–2)(a) 0.73 s (b) 1.2 s (c) 0.62 s (d) 1.6 s

4. A boy of mass 30 kg starts running from rest along a circular path of radius 6 m with constant tangential acceleration of magnitude 2 m s–2. After 2 s from start he feels that his shoes started slipping on ground. The friction between his shoes and ground is

(Take g = 10 m s–2)

(a) 12

(b) 13 (c)

14 (d)

15

5. A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section 0.1 cm2 and the other of brass of cross-section 0.2 cm2. Along the rod at which distance a weight may be hung to produce equal stresses in both the wires?

(a) 43

m from steel wire

(b) 43

m from brass wire

(c) 1 m from steel wire

(d) 14

m from brass wire

6. Two cylinders A and B, fitted with pistons, contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K then the rise in temperature of the gas in B is(a) 30 K (b) 18 K (c) 50 K (d) 42 K

7. A spherical conductor A of radius r is placed concentrically inside a conducting shell B of radius R(R > r). A charge Q is given to A, and then A is joined to B by a metal wire. The charge flowing from A to B will be

(a) QR

R r+

(b) Qr

R r+

(c) Q (d) zero

8. A 1 mF capacitor is connected in the circuit shown below. The emf of the cell is 3 V and internal resistance is 0.5 W. The resistors R1 and R2 have


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Page 43

values 4 W and 1 W respectively. The charge on the capacitor in steady state must be

3 V

R1

R2

0.5

4

1

1 F

(a) 1 mC (b) 2 mC(c) 1.33 mC (d) zero

9. A long cylindrical wire kept along z-axis carries a

current of density

J J r k= 0^, where J0 is a constant

and r is the radial distance from the axis of the cylinder. The magnetic field inside the conductor at a distance d from the axis of the cylinder is

(a) µ0 0J (b) µ0 0

2J d

(c) µ0 0

2

3J d

(d) µ0 0

3

4J d

10. An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with a frequency 25π

Hz. At the position x = 0.04 m, the object has

kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is

(Potential energy is zero at mean position) (a) 6 cm (b) 4 cm (c) 8 cm (d) 2 cm

11. Velocity of sound in an open organ pipe is 330 m s–1. The frequency of wave is 1.1 kHz and the length of tube is 30 cm. To which harmonic does this frequency correspond?(a) 2nd (b) 3rd (c) 4th (d) 5th

12. A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object the image is formed at infinity. The thickness t is(a) 10 cm (b) 5 cm (c) 20 cm (d) 15 cm

13. An ideal massless spring S can be compressed 2 m by a force of 200 N. This spring is placed at the bottom of the frictionless inclined plane which makes an angle q = 30° with the horizontal. A 20 kg mass is released from rest at the top of the inclined plane and is brought to rest momentarily after compressing the spring 4 m. Through what distance does the mass slide before coming to rest?(a) 2.2 m (b) 4 m (c) 8.17 m (d) 1.9 m

14. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m s–2. He reaches the ground with a speed of 3 m s–1. At what height did he bail out?(a) 293 m (b) 111 m (c) 91 m (d) 182 m

15. Two blocks A and B of masses 2m and m,

A 2m

mB

respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration A and B immediately after the string is cut, are respectively

(a) gg,2 (b)

gg

2,

(c) g, g (d) g g2 2

,

16. If the work done in blowing a soap bubble of volume V is W, then the work done in blowing a soap bubble of volume 2V will be(a) W (b) 2W (c) 2W (d) W(4)1/3

17. A hydrogen atom in an excited state emits a photon which has the longest wavelength of the Paschen series. Further emissions from the atom cannot include the(a) longest wavelength of the Lyman series(b) second longest wavelength of the Lyman

series(c) longest wavelength of the Balmer series(d) second longest wavelength of the Balmer

series

18. A horizontal rod rotates about a vertical axis through one end. A ring, which can slide along the rod without friction, is initially close to the axis and then slides to the other end of the rod. In this process, which of the following quantities will be conserved?

[L = angular momentum, KT = total kinetic energy, KR = rotational kinetic energy](a) L only (b) L and KT only(c) L and KR only (d) KT only

19. A photon of energy 10.2 eV corresponds to light of wavelength l0. Due to an electron transition from n = 2 to n = 1 in a hydrogen atom, light of wavelength l is emitted. If we take into account the recoil of the atom when the photon is emitted,(a) l = l0 (b) l < l0(c) l > l0 (d) the data is not sufficient to reach a

conclusion


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20. An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time t1. A proton of mass mp also initially at rest takes time t2 to move through an equal distance in this uniform electric field. Neglecting

the effect of gravity, the ratio tt2

1 is nearly equal to

(a) 1 (b) m

mp

e

1 2/

(c) mm

e

p

1 2/

(d) 1836

21. A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest. After the collision, A and B move with velocities v1 m s–1 and v2 m s–1 respectively making angles of 30° and 60° with respect to the

original direction of motion of A. The ratio vv

1

2will be

(a) 34

(b) 43

(c) 13

(d) 3

22. A bullet of mass 20 g moving with 600 m s–1

collides with a block of mass 4 kg hanging with the string. What is the velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision?

(Take g = 10 m s–2)(a) 200 m s–1 (b) 150 m s–1

(c) 400 m s–1 (d) 300 m s–1

23. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated upwards in order to reduce its period

to T2

? (g is the acceleration due to gravity)

(a) 4g (b) g (c) 2g (d) 3g

24. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres of A and B is

(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

25. A 5 V battery with internal resistance 2 W and 2 V battery with internal resistance 1 W are connected to a 10 W resistor as shown in figure. The current in 10 W resistor is

10 5 V, 2 2 V, 1

P1

P2

(a) 0.27 A, P1 to P2 (b) 0.27 A, P2 to P1(c) 0.03 A, P1 to P2 (d) 0.03 A, P2 to P1

26. A galvanometer of 50 W resistance has 25 divisions. A current of 4 × 10–4 A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of(a) 2500 W as a shunt (b) 2450 W as a shunt(c) 2550 W in series (d) 2450 W in series

27. What is the charge induced in coil of 100 turns of resistance 100 W, if magnetic flux changes from 2 T m2 to – 2 T m2?(a) 4 C (b) 2 C (c) 2.8 C (d) 0.4 C

28. If an electron revolves around a proton, then its time period T is (R = radius of orbit)(a) ∝ R2 (b) ∝ R3/2 (c) ∝ R3 (d) ∝ R

29. A uniform circular disc of mass 12 kg is held by two identical springs as

shown in the figure. When the disc is pressed down slightly and released, it executes SHM with a time period of 2 s. The spring constant of each spring is(a) 236 N m–1 (b) 118.3 N m–1

(c) 59.15 N m–1 (d) 108.3 N m–1

30. Two thin uniform circular rings each of radius 10 cm and mass 0.1 kg are arranged such that they have common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through common centre and perpendicular to the plane of either of the rings in kg m2 is(a) 15 × 10–3 (b) 5 × 10–3 (c) 15 × 10–4 (d) 18 × 10–4

soLuTioNs1. (b) : Since error is measured for 400 observations

instead of 100 observations. So error will reduce

by 14

factor.Hence, the random error in the arithmetic mean

of 400 observations would be x4

.

2. (c) : Total time of flight is T = 4 s and if u is its initial speed and q the angle of projection. Then


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Page 45

Tu

g= =

2 4sinθ

or usinq = 2g ...(i)After 1 s velocity vector makes an angle of 45° with horizontal i.e., vx = vy

ucosq = usinq – gt ucosq = usinq – g ( ) t = 1 s ucosq = 2g – g (Using (i))or ucosq = g ...(ii)Squaring and adding (i) and (ii), we get

u2sin2q + u2cos2q = (2g)2 + (g)2

u2 = 5g2 = 5(10)2 m2 s–2

\ u = 22.36 m s–1

Dividing (i) by (ii), we get,

uu

gg

sincos

θθ

= =2 2

tanq = 2 or q = tan–1(2)3. (a) : Maximum frictional force between the block

and the plank is fmax = mmg = (0.5)(1)(10) = 5 NThe free body diagrams of the block and the plank are as shown in the figure (a) and (b) respectively.

5 N5 N

(a)30 N

(b)

Acceleration of block, a125

15= = −m s

Acceleration of plank, a2230 5

2252

=−

= −m s

\ Relative acceleration of plank a = a2 – a1

a = − = −25

25 7 5 2. m s

∴ = =×

=tSa

2 2 27 5

0 73.

. s

4. (b) : After 2 s speed of boy will be v = 2 × 2 = 4 m s–1

At this moment centripetal force on boy is

F

mvRc = =

×=

2 30 166

80N N

Tangential force on boy is Ft = ma = 30 × 2 N = 60 NTotal force acting on boy is

F F Fc t= + = + =2 2 2 280 60 100( ) ( ) N

At the time of slipping, F = mmg

or 100 = m × 30 × 10 or µ =13

5. (a) :

W

B

x

A

T1 T2

2 m

As stresses are equal, TA

TA

1

1

2

2=

i eTT

AA

T T. . ..

1

2

1

22 1

0 10 2

2= = =or ...(i)

Now for translatory equilibrium of the rod, T1 + T2 = W ...(ii)From (i) and (ii), we get

TW

TW

1 2323

= =;

Now if x is the distance of weight W from steel wire, then for rotational equilibrium of rod,

T1x = T2 (2 – x) or Wx

Wx

323

2= −( )

∴ =x43

m

6. (d) : In cylinder A, heat is supplied at constant pressure while in cylinder B, heat is supplied at constant volume.\ (DQ)A = nCPDTA

and (DQ)B = nCVDTB

Given: (DQ)A = (DQ)B

nCPDTA = nCVDTB

or ∆ ∆TCC

TBP

VA=

For diatomic gas,CC

P

V=

75

∴ = × =∆TB

75

30 42K K

7. (c) : When charge of amount q has flown from A to B, the charge on A is (Q – q).The potentials of A and B are

V

Q qr

qRA =

−+

14

140 0πε πε

V

Q qR

qRB =

−+

14

140 0πε πε

∴ − = − −

>V V Q qr RA B

14

1 1 00πε

( )

\ VA > VB for all values of q. Charge will flow A to B till q = Q.

8. (b) : In steady state current in the branch containing the capacitor is zero and hence emf e is shared between r and R2 in the ratio of their resistances.


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Page 46

Potential difference across R2 is

VR

R rR22

2

3 11 0 5

2=+

=+

=ε ( )( )

.V

VΩ

Ω Ω

Potential difference across capacitor of 1 mF is same across R2.\ Potential difference across 1 mF capacitor is VC = 2 V\ Charge on capacitor, Q = CVC = (1 mF)(2 V) = 2 mC

9. (c) : Current density,

J J r k= 0^

Current within a distance d,

I J dS J rrd dr k kr

d d

= ⋅ = ⋅==∫∫ ∫∫

φ

π πφ

0

2

00

0

2

0( )( )

^ ^

[ ]

^∴ =dS rd dr k

φ = 2

30

3πJ

d

z

yFrom Ampere’s Law,

B dl Jd

C

⋅ =∫ µ π0 0

32

3Here loop C is a circle of radius d, B is the magnetic field at d.

B d Jd( )2 230 0

3π µ π=

B

J d=

µ0 02

3

10. (a) : ω πυ= =2 km

\ k = (2pu)2m ...(i)Total energy of oscillation is E = (0.5 J + 0.4 J) = 0.9 J

E KA=12

2

where A is the amplitude of oscillation

∴ =0 9 12

2. kA

or (Using (i))Ak m

= =1 8 1 8

2 2. .

( )πυ

= =

=1

21 80 2

1

2 251 80 2

350πυ π

π

.

...

m = 6 cm

11. (a) : Fundamental frequency of an open organ pipe is

υ0

1

23302 0 3

550= =×

=−v

lm s

mHz

.The given frequency u = 1.1 kHz = 1100 Hz = 2u0

Therefore, the given frequency corresponds to 2nd

harmonic.12. (d) : Image will be formed at infinity if object is

placed at focus of the lens i.e., at 20 cm from the lens. Hence,

shift = 25 – 20 = 1 1−

µ

t

or 5 1 11 5

= −

.

t

or cmt =×

=5 1 5

0 515.

.

13. (c) : As the spring is compressed by 2 m with the application of a force of 200 N, hence its spring constant k is given by

k

F= = = −

22002

100 1Nm

N m

= 30°

20 kg

h

Suppose l be the distance along the inclined plane which the mass travels before it comes to rest. Applying the conservation of energy,

12 1

2kx mgh mgl= = sinθ

or 12

100 4 20 9 8 12

2× × = × × ×. l

∴ = =l80098

8 17. m

14. (a) : Initial velocity of parachutist after bailing out u2 = 2gh = 2 × 9.8 × 50 = 980 ...(i)When it reaches the ground32 = u2 – 2 × 2 × h1 or 32 = 980 – 2 × 2 × h1 (Using (i))

or orh1980 9

49714

=−

= h1 = 242. 75 m

\ Total height = 242.75 m + 50 m

293 m15. (b) : Just before the string is cut, force on the spring

pulling up = kx = 3mg. After string is cut, free body diagram of block A gives 2maA = 3mg – 2mg

or amg

mA =2 =

g2

A

3mg

2mg

Free body diagram of block B givesmaB = mg or aB = g

mg

B

16. (d) : W = TDA = T(2 × 4pR2)

and V R=43

3π

When volume is doubled new radius becomes R′ = (2)1/3R\ W′ = T × 2 × 4pR′2 = T × 2 × 4p(2)2/3R2

= T × 2 × 4p(4)1/3R2 = (4)1/3W


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Page 47

17. (d) :

DC

B

n = 4n = 3

A

n = 2

n = 1The photon emitted comes from the transition A. Further transitions possible are B (longest l of the Balmer series), C (longest l of the Lyman series) and D (second longest l of the Lyman series).

18. (b) : As no external torque acts on the system, its angular momentum is conserved. Since there is no loss of energy due to friction, the total energy is conserved. However, the ring acquires some translational kinetic energy as it slides outwards, and hence rotational kinetic energy is not conserved.

19. (c) : The total energy available from the transition = 10.2 eV = energy of emitted photon + kinetic energy of recoiling atom.\ Energy of emitted photon < 10.2 eV\ l > l0

20. (b) : The acceleration of the electron is

a

eEme

e=

...(i)

Starting from rest, the distance travelled by the electron in time t1 in a uniform electric field E is

d a t td

aee

= =12

212

1or

...(ii)

The acceleration of the proton is

a

eEmp

p=

...(iii)

Starting from rest, the same distance d travelled by the proton in time t2 in the same electric field is

d a t td

app

= =12

222

2or

...(iv)

Divide (iv) by (iii), we get

∴ = =tt

aa

m

me

p

p

e

2

1 (Using (i) and (iii))

21. (a) :

Applying the law of conservation of linear momentum along a direction perpendicular to the direction of motion (i.e. along y-axis), we get0 + 0 = 4v1sin30° – v2sin60°4v1sin30° = v2sin60°vv

1

2

604 30

34

=°°

=sinsin

22. (a) : Here, Mass of the bullet, m1 = 20 g = 20 × 10–3 kg = 0.02 kg Mass of the block, m2 = 4 kgAccording to law of conservation of linear momentum, we get m1u1 + m2u2 = m1v1 + m2v2where v1 and v2 be the velocities of the bullet and block after the collision. 0.02 × 600 + 4 × 0 = 0.02 × v1 + 4v2or 0.02 × 600 = 0.02v1 + 4v2 ...(i)

Here, v gh212 2 10 0 2 2= = × × = −. m s

Substituting this value of v2 in Eq. (i), we get

12 = 0.02v1 + 8

or 0.02v1 = 12 – 8

∴ = = −v114

0 02200

.m s

23. (d) : Here, Tlg

= 2π ...(i)

When lift is accelerated upwards with acceleration

a, let time period becomes T2

. Then

T lg a2

2=+

π

...(ii)

Dividing Eq. (i) by Eq. (ii), we get

2 1

1 2

=+

= +

g ag

ag

/

Squaring both sides, we get 4 1= +

ag

or a = 3g

24. (b) : Here, rA = 1 mm, rB = 2 mm, d = 5 cmWhen spheres are connected by a conducting wire, charge flows from the sphere at higher potential to the sphere at lower potential, till their potentials become equal.

Now,CC

rr

A

B

A

B= =

12

As potential V is same,


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BRAIN MAP Digital Electronics and Logic Gates

A B A B+=×

1A A+= 0A A×=

A B A B+=×

A A= A A=

A B A B×=+

A A B A B+×=+ ( )A A B A B×+=×

A B A B×=+

Inputs OutputA0

B0

0 11 01 1

Y0111

Inputs OutputA0

B0

0 11 01 1

Y0001

Input OutputA01

Y10

Y A=

Inputs OutputA0

B0

0 11 01 1

Y0110

Inputs OutputA0

B0

0 11 01 1

Y1000

Y A B=+

Y A B A B A B=×+×=Å

Inputs OutputA0

B0

0 11 01 1

Y1001

Y A B A B A B=×+×=Å

Inputs OutputA0

B0

0 11 01 1

Y1110

Y A B=×

A

BY


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Page 50

∴ = = =

qq

C VC V

CC

A

B

A

B

A

B

12

Now,EE

q

rq

r

A

B

A

A

B

B

=4

4

02

02

πε

πε

= ×

qq

rr

A

B

B

A

2

=

=12

21

22

25. (d) :

10 5 V, 2 2 V, 1

P1

P2

I1

B

CD

I

I

I

( – )I I1

( – )I I1

( – )I I1

A

Applying Kirchhoff’s second law for closed loop AP2 P1DA, we get 10I1 + 2 I – 5 = 0 2I + 10I1 = 5 ...(i)

Applying Kirchhoff’s second law for closed loop P2 BCP1P2, we get –2 + 1(I – I1) – 10I1 = 0 or I – 11I1 = 2 ...(ii)or 2I – 22I1 = 4 ...(iii)

Subtracting Eq. (iii) from Eq. (i), we get

32 1 1

320 031 1 2 1I I P P= = ≈or A A from to.

26. (d) : G = 50 W Ig = Current for full scale deflection = Current per division × total no. of divisions = 4 × 10–4 × 25 = 10–2 AGiven V = 25 VHence, required resistance,

R

VI

Gg

= − = −−

2510

502

= 2500 – 50 = 2450 WThis resistance of 2450 W should be connected in series to convert the galvanometer into a voltmeter.

27. (a)

28. (b) : Time period of revolution of electron around a proton is given by

T

Rv

= =Orbit circumference

Electron speed2π

=

=2

4

2 4

0

03π

πε

π πεR

e

mR

mR

e

\ T ∝ R3/2

29 . (c) : Let k be spring constant of each spring. Here, two springs are connected in parallel, the effective spring constant is given by keff = k + k = 2k

T

mk

mk

= =2 22

π πeff

or Tm

k2

242

=π

or km

T=

2 2

2π

∴ =× ( ) ×

= × ( ) = −k2 3 14 12

46 3 14 59 15

22 1.

. . N m

30. (c) : Because both the rings have common centre and their planes are mutually perpendicular, hence, an axis which is passing through the centre of one of the rings and perpendicular to the plane of its plane, will be along the diameter of other ring. Hence, moment of the inertia of the system about the given axis is

I = ICM + Idiameter = + =MR MR MR2 2 212

32

= = × −3

20 1 0 1 15 102 4 2( . )( . )kg m kg m

vvv

Life’s True Beauty“I think your education is imperfect, if you do

not realize my young friends, that life is not

merely a question of getting food, clothes and

shelter. Man does not live by bread alone.

This has been realized from ancient times.

I think that the finest thing in life are

not these, but music, colour, flowers,

beauty, aesthetic sense, the satisfaction

derived from these. It is these finer things in

life that makes life worth living.

- C V Raman


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Page 51

physics for you | February ’13 51

Atoms And nuclei1. In the following reaction, the energy released

is

411H → 4

2He + 2e+ + Energy

Given :

Mass of 11H = 1.007825 u

Mass of 42He = 4.002603 u

Mass of e+ = 0.000548 u(a) 12.33 MeV (b) 24.67 MeV (c) 25.7 MeV (d) 49.34 MeV

2. The total energy of an electron in the excited state corresponding to n = 3 state is E. What is its potential energy with proper sign?(a) –2E (b) 2E (c) –E (d) E

3. The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of wavelength l is emitted for a transition 3 → 1. What will be the wavelength of emission for transition 2 → 1?

(a) λ3

(b) 43λ (c)

34λ

(d) 3l

4. In Bohr model of hydrogen atom, the ratio of period of revolution of an electron in n = 2 and n = 1 orbit is (a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1

5. A radioactive isotope A with a half life of 1.25 × 1010 years decays into B which is stable. A sample of rock from a planet is found to contain both A and B present in the ratio 1 : 15. The age of the rock is(a) 9.6 × 1010 years (b) 4.2 × 1010 years(c) 5 × 1010 years (d) 1.95 × 1010 years

6. A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 3 : 1. The ratio of radii of the fragments is

(a) 1 : 31/3 (b) 31/3 : 4(c) 4 : 1 (d) 2 : 1

7. A hydrogen atom and a Li2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then(a) l l E EH Li H Liand> >

(b) l l E EH Li H Liand= <

(c) l l E EH Li H Liand= >

(d) l l E EH Li H Liand< <8. A radioactive nucleus emits 3a-particles and

5b-particles. The ratio of number of neutrons to that of protons will be

(a)

A ZZ− −

−12

6 (b)

A ZZ

−− 1

(c) A Z

Z− −

−11

6 (d) A ZZ− −

−11

19. A radioactive sample has half life of 5 days. To

decay from 8 microcurie to 1 microcurie, the number of days taken will be(a) 40 (b) 25 (c) 15 (d) 10

10. A free neutron decays spontaneously into(a) a proton, an electron and anti-neutrino(b) a proton, an electron and a neutrino(c) a proton and electron(d) a proton, and electron, a neutrino and an

anti-neutrino.11. The radius of germanium (Ge) nuclide is measured

to be twice the radius of 94Be. The number of

nucleons in Ge are(a) 72 (b) 73 (c) 74 (d) 75

12. The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is

(a) 259 (b)

176 (c)

95 (d) 4

3


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Page 52

52 physics for you | February ’13

13. Fusion reaction takes place at high temperature because(a) nuclei break up at high temperature(b) atoms get ionised at high temperature(c) kinetic energy is high enough to overcome

the coulomb repulsion between nuclei(d) molecules break up at high temperature

14. Which of the following is more effective in inducing nuclear fission?(a) Fast neutron (b) Fast proton(c) Slow proton (d) Slow neutron

15. The electric potential between a proton and an

electron is given by V V r

r = ,0

0ln

where r0 is a

constant. Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number

(a) rn ∝ n (b)

rnn ∝ 1

(c) rn ∝ n2 (d) rn

n ∝ 12

16. To generate power of 3.2 MW, the number of fissions of 235U per minute is

(Energy released per fission = 200 MeV, 1 eV = 1.6 × 10–19 J)(a) 6 × 1018 (b) 6 × 1017

(c) 1017 (d) 6 × 1016

17. A hydrogen atom is excited up to 9th level. The total number of possible spectral lines emitted by the hydrogen atom is(a) 36 (b) 35 (c) 37 (d) 38

18. The 23290Th atom has successive alpha and beta

decays to the end product 20882Pb. The numbers of

alpha and beta particles emitted in the process respectively are(a) 4, 6 (b) 4, 4 (c) 6, 2 (d) 6, 4

19. The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged muon with mass 207 times that of an electron, is am equal to(a) 207a0 (b) a0

207

(c) a0

207 (d) a0 207

20. A nucleus with mass number 220 initially at rest emits an a particle. If the Q value of the reaction is 5.5 MeV, the kinetic energy of the a particle is(a) 4.4 MeV (b) 5.4 MeV

(c) 5.6 MeV (d) 6.5 MeV

21. An element X decays first by positron emission and then two a-particles are emitted in successive radioactive decay. If the product nucleus has mass number 227 and atomic number 89, the mass number and atomic number of element X are(a) (273, 93) (b) (235, 94)(c) (238, 93) (d) (237, 92)

22. The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead,

it is reduced to I8

. The thickness of lead which

will reduce the intensity of I2

will be

(a) 12 mm (b) 18 mm (c) 9 mm (d) 6 mm

23. F pp, F nn and F np are the nuclear forces between proton-proton, neutron-neutron and neutron-proton respectively. Then relation between them is(a) Fpp = Fnn ≠ Fnp (b) Fpp ≠ Fnn = Fnp(c) Fpp = Fnn = Fnp (d) Fpp ≠ Fnn ≠ Fnp

24. Which energy state of doubly ionised lithium has the same energy as that of the ground state of hydrogen ? Given Z for lithium = 3.(a) 4 (b) 3 (c) 2 (d) 1

25. When the electron in hydrogen atom is excited from the 4th stationary orbit to the 5th stationary orbit, the change in the angular momentum of the electron is

(Planck’s constant, h = 6.63 × 10–34 J s) (a) 4.16 × 10–34 J s (b) 3.32 × 10–34 J s (c) 1.05 × 10–34 J s (d) 2.08 × 10–34 J s

semiconductor electronics26. In the following figure, the diodes which are

forward biased, are

(i)

R+ 5 V

+10 V

(ii)

–10 V

R

(iii)

–12 VR

–5 V (iv)

R

+ 5 V


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Page 53


(a) only (iii) (b) (i) and (iii)(c) (ii) and (iv) (d) (i), (ii) and (iv)

27. Truth table for the given circuit is

A

B D

Y

C

(a)

A B Y0 0 10 1 01 0 11 1 0

(b)

A B Y0 0 10 1 01 0 01 1 1

(c)

A B Y0 0 00 1 11 0 01 1 1

(d)

A B Y0 0 00 1 11 0 01 1 0

28. For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct?(a) a ≠ b ≠ c and a = b = g = 90°(b) a = b = c and a ≠ b ≠ g = 90°(c) a = b = c and a = b = g = 90°(d) a ≠ b ≠ c and a ≠ b and g ≠ 90°

29. If an intrinsic semiconductor is heated, the ratio of free electrons to holes is(a) greater than one (b) less than one(c) equal to one (d) decreases and becomes zero

30. While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio is(a) 82 (b) 83 (c) 8.2 (d) 8.3

31. The logic circuit shown below is equivalent to

Z

Y

X

(a)

X

YZ

(b) X

YZ

(c) X Z

(d) X Z

32. Consider the following statements A and B and identify the correct choice of the given answers.(A) The width of the depletion layer in a p-n

junction diode increases in forward bias.(B) In an intrinsic semiconductor the fermi energy

level is exactly in the middle of the forbidden energy gap.

(a) A is true and B is false(b) Both A and B are false(c) A is false and B is true(d) Both A and B are true

33. A common emitter transistor amplifier has a current gain of 50. If the load resistance is 4 kW and input resistance is 500 W, the voltage gain of the amplifier is(a) 100 (b) 200 (c) 300 (d) 400

34. The voltage drop across a forward biased diode is 0.7 V. In the following circuit, the voltages across the 10 W resistance in series with the diode and 20 W resistance are

10

20

10 10 V

(a) 0.70 V, 4.28 V (b) 3.58 V, 4.28 V(c) 5.35 V, 2.14 V (d) 3.58 V, 9.3 V

35. To obtain p-type extrinsic semiconductor, the impurity element to be added to germanium should be of valency(a) 2 (b) 3 (c) 4 (d) 5

36. In the given circuit, the current through the resistor 2 kW is

+– 12 V 2 k

1 k

20 V

(a) 2 mA (b) 4 mA (c) 6 mA (d) 10 mA

37. In a p-n junction photodiode, the value of the photo electromotive force produced by monochromatic light is proportional to(a) the barrier voltage at p-n junction(b) the intensity of light falling on the cell(c) the frequency of light falling on the cell(d) the voltage applied at the p-n junction.

38. Two pieces, one of germanium and the other of the aluminium are cooled from T1 K to T2 K. The resistance of


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Page 54


(a) aluminium increases and that of germanium decreases.

(b) each of them decreases.(c) aluminium decreases and that of germanium

increases.(d) each of them increases.

39. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 1016 m–3. Doping by indium increases nh to 4.5 × 1022 m–3. The doped semiconductor is of(a) n-type with electron concentration ne = 5 × 1022 m–3.(b) p-type with electron concentration ne = 2.5 × 1010 m–3.(c) n-type with electron concentration ne = 2.5 × 1023 m–3.(d) p-type having electron concentration ne = 5 × 109 m–3.

40. A full wave p-n diode rectifier uses a load resistor of 1500 W. No filter is used. The forward bias resistance of the diode is 10 W. The efficiency of the rectifier is(a) 81.2% (b) 40.6% (c) 80.6% (d) 40.2%

communicAtion systems41. If the maximum amplitude of an amplitude

modulated wave is 25 V and the minimum amplitude is 5 V, the modulation index is

(a) 15 (b)

13 (c)

32 (d) 2

342. A signal wave of frequency 12 kHz is modulated

with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively(a) 2512 kHz and 2508 kHz(b) 2522 kHz and 2488 kHz(c) 2502 kHz and 2498 kHz(d) 2522 kHz and 2498 kHz

43. The frequency band used in the downlink of satellite communication is(a) 9.5 to 2.5 GHz (b) 896 to 901 MHz(c) 3.7 to 4.2 GHz (d) 840 to 935 MHz

44. A basic communication system consists of (A) transmitter (B) information source(C) user of information(D) channel (E) receiverChoose the correct sequence in which these are arranged in a basic communication system.(a) ABCDE (b) BADEC(c) BDACE (d) BEADC

45. 1000 kHz carrier wave is amplitude modulated by the signal frequency 200 – 4000 Hz. The channel width of this case is(a) 8 kHz (b) 4 kHz(c) 7.6 kHz (d) 3.8 kHz

46. An example of point to point mode of communication is(a) FM radio (b) standard FM radio(c) television (d) telephony

47. If both the length of an antenna and the wavelength of the signal to be transmitted are doubled, the power radiated by the antenna(a) is doubled (b) is halved(c) remains constant (d) is quadrupled

48. Identify the incorrect statement from the following.(a) AM detection is carried out using a rectifer

and an envelope detector.(b) Pulse position denotes the time of rise or fall

of the pulse amplitude.(c) Modulation index m is kept ≥ 1, to avoid

distortion.(d) Facsimile (FAX) scans the contents of the

document to create electronic signals.

49. Through which mode of propagation, the radio waves can be sent from one place to another(a) ground wave propagation(b) sky wave propagation(c) space wave propagation(d) all of them

50. A television tower of height 140 m can broadcast its signal upto a maxiumum area of (Radius of earth = 6.4 × 106 m)(a) 1.56 × 106 km2 (b) 5.6 × 103 km2

(c) 5.6 × 1010 km2 (d) 1.56 × 109 km2

solutions1. (c) : The given nuclear reaction is

411H → 4

2He + 2e+ + EnergyThe energy released during the process is

Q = [4m(11H) – m(4

2He) – 2(me+)]c2 = [4 × 1.007825 – 4.002603) – 2 × 0.000548]u × c2

= [4.0313 – 4.002603 – 0.001096]u × c2

= (0.027601 u)c2

= (0.027601 u)(931.5 MeV/u) = 25.7 MeV2. (b)3. (d) :

2 E

EE 1

2

3

43


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E3 – E1 = 2E – E = E

hc Eλ

= ....(i)

From figure,E2 – E1 = E2 – E3 + E3 – E1

= − + −43

2 2E E E E

= − + =23

13

E E E

hc hcλ λ′

= 13

(Using (i))

l′ = 3l4. (c) : In Bohr model of hydrogen atom, the period

of revolution of electron in nth orbit is given as

Th n

men =

4 02 3 3

4ε

Tn ∝ n3

\ TT

2

1

321

81

=

=

5. (c) : According to Rutherford and Soddy law for radioactive decay,Number of atoms remained undecayed after time t is N = N0e–lt

eNN

tNN

tλ λ= =0 0or ln

t

NN

= 1 0λ

ln

t

T NN

= 1 2 02

/ln

ln

= ×1 25 102

161

10.ln

ln

= × × ×1 25 10 4 2

2

10. lnln

= 5 × 1010 years6. (a) : As the heavy nucleus at rest breaks, therefore

according to law of conservation of momentum, we get m1v1 + m2v2 = 0

or

vv

mm

1

2

2

1

31

= =

.... (i)

As nuclear density is same,

∴ = =

mm

R

R

R

R1

2

13

23

13

23

4343

ρ π

ρ π

orR

R

mm

13

23

1

2

13

= =

(Using (i))

\ R1 : R2 = 1 : 31/3

7. (b) : In second excited state, n = 3,

So,

l l hH Li= =

3

2πwhile E ∝ Z2 and ZH = 1, ZLi = 3So, E E E ELi H H Lior= <9

8. (d) : During the emission of a-particle, the mass number and atomic number decreases by four and two respectively. During the emission of b-particle the mass number remains the same while the atomic number increases by 1.

ZA

ZA

ZAX Y Y3

612 5

112α β → → ′−

−−

−( )

( )( )

( )

∴ = − − −−

No. of neutronsNo. of protons

A ZZ

12 11( )

= − −−

A ZZ

111

9. (c) : Here, Half life, T1/2 = 5 daysInitial activity, R0 = 1 microcurieFinal activity, R = 8 microcurie

As RR

n

0

12

=

where n is the no. of half lives

∴

=

=

18

12

12

12

3n nor

or n = 3

As or days daysn tT

t nT= = = =1 2

1 2 3 5 15/

/ ( )( )

10. (a) : A free neutron is unstable. It decays spontaneously into a proton, an electron and anti-neutrino. n → p + e– + u– neutron proton electron anti-neutrino

11. (a) : Nuclear radii, R = R0(A)1/3 (where R0 = 1.2 fm) or R ∝ (A)1/3

∴ =RR A

Be

Ge

( )( )

/

/9 1 3

1 3

or Be

Be

RR A2

9 1 3

1 3= ( )( )

/

/

\ (A)1/3 = 2 × (9)1/3

or A = 23 × 9 = 72The number of nucleons in Ge is 72.

12. (a) : For Brackett series,1 1

41

2 2λ= −

Rn

where n = 5, 6, 7, 8, ........For longest wavelength, n = 5

∴ = −

1 14

152 2λLongest

R


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= −

=R R116

125

9400 .... (i)

For shortest wavelength, n = ∞

∴ = −∞

=2

1 14

1162λShortest

R R

.... (ii)

Dividing (ii) by (i), we getλ

λLongest

Shortest= × =R

R164009

259

13. (c) : For fusion to take place, high temperature is needed because at high temperature, the kinetic energy becomes large enough to overcome the coulomb repulsion between nuclei.

14. (d)

15. (a) : Given : V V rr

= 00

ln

\ Potential energy, U = eV

or = = 10

00

0

0U eV r

rdUdr

eVrr r

ln

∴

Force, F dUdr

eVr

= − = − 0

or | | = 0FeV

rThis force provides the necessary centripetal force.

∴ mvr

eVr

veVm

20 0 = or = .... (i)

By Bohr’s postulate, mvr nh = 2π

or = 2

v nhmrπ .... (ii)

From equations (i) and (ii), we get

nhmr

eVm2

= 0π

or = 2

0

r nhm

meVπ

×

or = 2

1 0

r hmeV

nπ

×

∴ ∝r nn 16. (a) : Here, Power = 3.2 MW = 3.2 × 106 W

Energy released per fission = 200 MeV = 200 × 106 eV = 200 × 106 × 1.6 × 10–19 JNumber of fissions per minute

= × ×× × × −3 2 10 60

200 10 1 6 10

6

6 19.

. = 6 × 1018

17. (a) : Number of spectral lines emitted is

Nn n= −( )1

2

Here n = 9 \ N = − = × =9 9 12

9 82

36( )

18. (d) : Let number of a particles emitted be x and number of b particles emitted be y.Difference in mass number = 4x = 232 – 208 = 24 or x = 6Difference in charge number = 2x – y = 90 – 82 = 812 – y = 8 or y = 4

19. (b) : a

h

me0

202=

ε

π .... (i)

ah

m eµ

ε

π=

20

2207( ) .... (ii)

Dividing (ii) by (i), we geta

aa

aµµ

0

01207 207

= =or

20. (b) : Here, A = 220, Q = 5.5 MeVThe kinetic energy of the a particle is

KE AA

Qα = − = − ×4 220 4220

5 5. MeV= 5.4 MeV

21. (b) : Let A and Z be mass number and atomic number of element X.

ZA

ZA

ZAX Y Yβ α+

→ → ′− −−

12

58

Since the product nucleus has mass number 227 and atomic number 89.\ A – 8 = 227 or A = 235and Z – 5 = 89 or Z = 94

22. (a) : As I = I0e–mx

∴ = −18

36e µ .... (i)

and

12

= −e xµ

.... (ii)

From Eq. (i), we get

12

336

= −e µ

Using (ii), we getor e–3 mx = e–m36 or x = =36

312 mm

23. (c) : Nuclear force is charge independent.\ Fpp = Fnn = Fnp

24. (b) : The energy of nth state of a hydrogen like atom is given as

E CZn

n =2

2 , where C is a constant.

For ground state of hydrogen atom, Z = 1, n = 1

∴ = =EC

C1

2

21

1( )

( )


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For nth state of Li2+ ion (Z = 3)

EC

nC

nn = =( )3 92

2 2

As En = E1

∴ = = =9 9 322C

nC n nor or

25. (c) : According to Bohr’s quantisation condition

L nhn =

2π∴ = =For n L h4 4

24,π

and for n L h= =5 525,

π \ Change in angular momentum when an electron is excited from n = 4 to n = 5 is

∆L L L h h= − = −5 452

42π π

= = ×

×

−h2

6 63 102 3 14

34

π.

.J s

= 1.05 × 10–34 J s

26. (b) : p-n junction diode is said to be forward biased if p side of junction is at higher potential and n side of junction is at lower potential. Therefore, option (b) is true.

27. (c) :

A

B D

Y

C0 0

0

00

0

0

1

0

00

A

B D

Y

C0 0

0

11

0

1

1

1

11

A

B D

Y

C1 1

0

00

0

0

0

0

00

A

B D

Y

C1 1

1

00

1

1

0

1

11

28. (c) : For cubic crystal, a = b = c and a = b = g = 90°

29. (c)30. (a) : Here, DIE = 8.3 mA

DIC = 8.2 mAAs DIE = DIB + DIC\ DIB = DIE – DIC = 8.3 mA – 8.2 mA = 0.1 mA

Forward current ratio, hIIfeC

B VCE

=

∆∆

is constant

=

=

8 20 1

82..

mAmA

31. (d) :

Z

Y

X

X–

A–C–C

AY–

Output of first OR gate is A = +( )X YInputs of second OR gate are X Aand .Output of second OR gate is C X A= + = + + = + ⋅ = + ⋅X X Y X X Y X X Y( ) ( ) = + =X Y X( )1Final output is Z C X X= = =

32. (c) : In a p-n junction the width of the depletion layer decreases in forward bias.

33. (d) : Here, b = 50, RL = 4 kW = 4 × 103 W Ri = 500 W

Voltage gain, ARRv

L

i= = × × =β 50 4 10

500400

3

34. (b) : Let the currents through the 20 W (parallel) and 10 W (in series with the diode) be I1 and I2 respectively.

10

20

10 10 V

I2

I1( + )I I1 2B

( + )I I1 2F

E

D

A

C

Applying Kirchhoff’s second law for closed loop ABEFA, we get 20I1 + 10(I1 + I2) – 10 = 0 .... (i)Applying Kirchhoff’s second law for closed loop BCDEB, we get0.7 + 10I2 – 20I1 = 0 .... (ii)Solving (i) and (ii), we get I1 = 0.214 A and I2 = 0.358 A


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Thus, voltage across the 10 W resistance in series with the diode = 0.358 A × 10 W = 3.58 VAnd voltage across the 20 W resistance = 0.214 A × 20 W = 4.28 V

35. (b) : To obtain p-type extrinsic semiconductor a trivalent impurity should be added to germanium.

36. (c) : In the circuit, Zener diode is used as a voltage regulating device. Therefore, the current in the

circuit, I = = × =−122

6 10 63Vk

A mAΩ

37. (b) : In photodiode, the photoelectromotive force is produced by photo-voltaic action i.e. a potential difference is created between two points whose magnitude depends upon the intensity of incident light.

38. (c) : The temperature coefficient of resistance of aluminium is positive and of germanium is negative. Hence the resistance of aluminium decreases and that of germanium increases with decreases in temperature.

39. (d) : As n n ni e h2 = ×

(1.5 × 1016 m–3)2 = ne × (4.5 × 1022 m–3)

∴ = ××

= ×−

−−ne

( . )( . )1 5 104 5 10

5 1016 2

229m

mm

3

33

As nh >> ne, so semiconductor is p-type and ne = 5 × 109 m–3.

40. (c) : Here, ri = 10 W, RL = 1500 WEfficiency of the full wave rectifier is

ηπ

= =+

PP

I R

I r Rm L

m f L

dc

ac

( / )

( / ) ( )

2

2

2

2

=

+0 812. Rr R

L

f L

=

×+

= =0 812 1500

10 15000 806 80 6

.. . %

41. (d) : Maximum amplitude, Amax = Ac + Am ...(i)Minimum amplitude, Amin = Ac – Am ...(ii)Solving (i) and (ii), we get

AA A

c =+max min2

, AA A

m =−max min2

Modulation index, µ = =−+

AA

A AA A

m

c

max min

max min

µ =

−= =

25 5 2030

23

V V25 V + 5 V

42. (d) : Here, us = 12 kHz, uc = 2.51 MHz = 2510 kHz The upperside band frequency = uc + us = (2510 + 12) kHz = 2522 kHzThe lower side band frequency = uc – us = (2510 – 12) kHz = 2498 kHz

43. (c) : The frequency band used in the downlink of satellite communication is 3.7 to 4.2 GHz.

44. (b) : The block diagram of a communication system is shown in the figure below.

InformationSource Transmitter Channel Receiver User of

Information

45. (a) : Bandwidth is equal to twice the frequency of modulating signals.Therefore, Band width = 2um = 2 × 4000 Hz = 8 × 103 Hz = 8 kHz

46. (d) : In point-to-point communication mode, communication takes place over a link between a single transmitter and a receiver. Telephony is an example of such a mode of communication.

47. (c) : Power radiated by the antenna is proportional

to lλ

2. When both the length of the antenna l and

wavelength of the signal l are doubled, the power radiated by the antenna remains constant.

48. (c) : Modulation index m is kept ≤ 1 to avoid distortion.

49. (d) 50. (b) : Here, h = 140 m

R = 6.4 × 106 mCoverage range, d Rh= 2Area covered = pd2 = p2Rh

= × × × ×227

2 6 4 10 1406.

= 5632 × 106 m2 = 5.6 × 109 m2

= 5.6 × 103 km2

nn


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1. The dimensions of ab

in the equation P a tbx

=− 2

,where P is pressure, x is distance and t is time are(a) [M2LT–3] (b) [ML0T–2](c) [ML3T–1] (d) [MLT–3]

2. A balloon is going upwards with velocity 12 m s–1. It releases a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground?

[Take g = 10 m s–2](a) 5 s (b) 6 s (c) 7 s (d) 8 s

3. A block of mass 10 kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100 N is applied on it, then acceleration of block will be

[Take g = 10 m s–2](a) 10 m s–2 (b) 5 m s–2

(c) 15 m s–2 (d) 0.5 m s–2

4. A ball with charge – 50e is placed at the centre of a hollow spherical shell has a net charge of –50e. What is the charge on the shell’s outer surface?(a) –50e (b) Zero (c) –100e (d) +100e

5. A voltmeter of resistance 998 W is connected across a cell of emf 2 V and internal resistance 2 W. The potential difference across the voltmeter is(a) 1.99 V (b) 3.5 V(c) 5 V (d) 6 V

6. The magnetic moment produced in a substance of 1 g is 6 × 10–7 A m2. If its density is 5 g cm–3, then the intensity of magnetisation will be(a) 8.3 × 106 A m–1 (b) 3.0 A m–1

(c) 1.2 × 10–7 A m–1 (d) 3 × 10–6 A m–1

7. A planet is revolving around a star in an elliptic orbit. The ratio of the farthest distance to the closest distance of the planet from the star is 4.

The ratio of kinetic energies of the planet at the farthest to the closest position is(a) 1 : 16 (b) 16 : 1 (c) 1 : 4 (d) 4 : 1

8. A homogeneous disc with a radius 0.2 m and mass 5 kg rotates around an axis passing through its centre. The angular velocity of the rotation of the disc as a function of time is given by the formula w = 2 + 6t. The tangential force applied to the rim of the disc is(a) 1 N (b) 2 N (c) 3 N (d) 4 N

9. A body of mass 0.01 kg executes simple harmonic motion (SHM) about x = 0 under the influence of a force as shown in the adjacent figure. The period of the SHM is (a) 1.05 s (b) 0.52 s (c) 0.25 s (d) 0.03 s

10. A transverse wave in a medium is described by the equation y = A sin 2(wt – kx). The magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity, if the value of A is

(a) λπ2 (b)

λπ4 (c)

λπ

(d) 2λπ

11. The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled. If the pressure on the surface is equal to h metre of mercury column and relative density of mercury is r, then the depth of lake in metre is(a) 2rh (b) 4rh (c) 8rh (d) 7rh

12. A stone of mass 2 kg is projected upwards with kinetic energy of 98 J. The height at which the kinetic energy of the body becomes half its original value is

F(N)

80

0.20–0.2

–80x(m)

Questions for Practice


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(Take g = 10 m s–2)(a) 5 m (b) 2.5 m (c) 1.5 m (d) 0.5 m

13. A particle of mass m collides with another stationary particle of mass M. If the particle m stops just after collision, the coefficient of restitution of collision is equal to

(a) 1 (b) mM

(c) M mM m

−+

(d) mM m+

14. A body cools down from 60°C to 55°C in 30 s. Using Newton’s law of cooling, calculate the approximate time taken by same body to cool down from 55°C to 50°C. Assume that the temperature of surroundings is 45°C.(a) 40 s (b) 55 s(c) 50 s (d) 60 s

15. A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s–1. The induced emf when the radius is 2 cm, is (a) 2p mV (b) p mV

(c) πµ

2V (d) 2 mV

16. In the series LCR circuit as shown in the figure, the voltmeter V and ammeter A readings are

]9

V

AR = 50 L C

400 V 400 V

100 V 50 Hz,(a) V = 100 V, I = 2 A (b) V = 100 V, I = 5 A(c) V = 400 V, I = 2 A (d) V = 300 V, I = 2 A

17. The electric field part of an electromagnetic wave in a medium is represented by

Ex = 0,

E t xy = ×

− ×

−2 5 2 10 106 2. cos ,NC

rads

radm

π π

Ez = 0. The wave is

(a) moving along the + x direction with frequency 106 Hz and wavelength 100 m.

(b) moving along + x direction with frequency 106 Hz and wavelength 200 m.

(c) moving along – x direction with frequency 106 Hz and wavelength 200 m.

(d) moving along + y direction with frequency 2p × 106 Hz and wavelength 200 m.

18. A point source of light is placed at a depth of h below the surface of water of refractive index m. A floating opaque disc is placed on the surface of water so that light from the source is not visible from the surface. The minimum diameter of the disc is

(a) 212 1 2h

( ) /µ − (b) 2h(m2 – 1)1/2

(c) h

[ ( ) ]/2 12 1 2µ − (d) h(m2 – 1)1/2

19. An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is

(a) 12 (b) 2

1 (c) 5

4 (d) 3

420. One milliwatt of light of wavelength 4560 Å is

incident on a cesium surface of work function 1.9 eV. Given that quantum of efficiency of photoelectric emission is 0.5%, Planck’s constant, h = 6.62 × 10–34 J s, velocity of light, c = 3 × 108 m s–1, the photoelectric current liberated is (a) 1.836 × 10–6 A (b) 1.836 × 10–7 A(c) 1.836 × 10–5 A (d) 1.836 × 10–4 A

21. The distance between poles of horse shoe magnet is 10 cm and its pole strength is 10–4 A m. The magnetic field at a point P midway between the poles is(a) zero (b) 8 × 10–9 T(c) 2 × 10–7 T (d) 8 × 10–7 T

22. Find the potential difference between the points E and F in the figure given below. Assume E and F are the midpoints of AB and DC respectively.

6 m(+ )q

(+ )q

+ /2q

A E B

CFD4 m

+ /2q

(a) (1.2 × 109q) volt (b) (1.8 × 109q) volt(c) (1.5 × 109q) volt (d) (3 × 109q) volt

23. As the switch S is closed in the circuit shown in

figure, current passed through it is(a) zero

10 V 4 5 V2

2

S(b) 1 A(c) 2 A(d) 1.6 A


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24. When a spring is stretched by a distance x, it exerts a force, given by F = (– 5x – 16x3) N. The work done, when the spring is stretched from 0.1 m to 0.2 m is(a) 8.1 × 10–2 J (b) 12.2 × 10–2 J(c) 8.1 × 10–1 J (d) 12.2 × 10–1 J

25. A satellite is orbiting around the earth with total energy E. What will happen if the satellite’s kinetic energy is made 2E?(a) Radius of the orbit is doubled(b) Radius of the orbit is halved(c) Period of revolution is doubled(d) Satellite escapes away

26. Sand drops vertically at the rate of 2 kg s–1 on to a conveyor belt moving horizontally with a velocity of 0.2 m s–1. Then the extra force required to keep the belt moving is(a) 0.4 N (b) 0.08 N (c) 0.04 N (d) 0.02 N

27. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v m s–1. If it is to climb the inclined surface then v should be

(a) ≥107

gh (b) ≥ 2gh

(c) 2gh (d) 107

gh28. The average translational kinetic energy of

O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is(a) 0.0015 (b) 0.003(c) 0.048 (d) 0.768

29. Two simple harmonic motions are represented

by the equations y t1 0 1 1003

= +

. sin π

π and

y2 = 0.1 cospt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

(a) π6

(b) −π3

(c) π3

(d) −π6

30. A pipe 30.0 cm long is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1.1 kHz source? Take the speed of sound in air as 330 m s–1.(a) First harmonic (b) Second harmonic(c) Third harmonic (d) Fourth harmonic

31. There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 N m–1 and density

103 kg m–3. The bubble is at a depth of 10.0 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure?(a) 1030 N m–2 (b) 1230 N m–2

(c) 1130 N m–2 (d) 1330 N m–2

32. When a system is taken from state a to state b along the path acb as shown in figure, 60 J of heat flows into the system and 30 J of work is done by the system. Along the path adb, if the work done by the system is 10 J, heat flow into the system is(a) 100 J

(b) 20 J

(c) 80 J

(d) 40 J

33. In the circuit shown, the cell is ideal, with emf = 15 V. Each resistance R is of 3 W. The potential difference across the capacitor of capacitance 3 mF is

R

R R

RR

15 V

C = 3 F

(a) zero (b) 9 V (c) 12 V (d) 15 V

34. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. The magnitude of B (in tesla) is

(Take g = 10 m s–2)(a) 2 (b) 1.5 (c) 0.55 (d) 0.66

35. A 50 Hz ac of crest value 1 A flows through primary of a transformer. If mutual inductance between primary and secondary be 1.5 H, the crest voltage in secondary is(a) 75 V (b) 150 V (c) 471 V (d) 300 V

36. When 100 V dc is applied across a coil, a current of 1 A flows through it. When 100 V 50 Hz ac is applied to the same coil, only 0.5 A flows. The inductance of the coil is(a) 5.5 mH (b) 0.55 mH(c) 55 mH (d) 0.55 H

37. Young’s experiment is performed with light of wavelength 6000 Å wherein 16 fringes occupy a certain region on the screen. If 24 fringes occupy


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Page 62

the same region with another light of wavelength l, then l is(a) 6000 Å (b) 4500 Å (c) 5000 Å (d) 4000 Å

38. A plane electromagnetic wave propagating in the x-direction has wavelength of 6.0 mm. The electric field is in the y-direction and its maximum magnitude of 33 V m–1. The equation for the electric field as function of x and t is

(a) 11 sinp(t – xc

) (b) 33 sinp × 1011(t – xc

)

(c) 33 sinp (t – xc

) (d) 11 sinp × 1011 (t – xc

)

39. A convex lens of focal length 1.0 m and a concave lens of focal length 0.25 m are 0.75 m apart. A parallel beam of light is incident in the convex lens. The beam emerging after refraction from both lenses is(a) parallel to principal axis(b) convergent(c) divergent (d) none of these

40. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic numer Z of hydrogen like ion is(a) 4 (b) 1 (c) 2 (d) 3

41. A particle moving with uniform acceleration has velocity 6 m s–1 at a distance of 5 m from the initial position. After moving another 7 m, the velocity becomes 8 m s–1. The initial velocity and acceleration of the particle are(a) 2 m s–1, 4 m s–2 (b) 4 m s–1, 2 m s–2

(c) 4 m s–1, 4 m s–2 (d) 6 m s–1, 1 m s–2

42. A particle is moving along a circular path of radius 5 m with uniform speed of 5 m s–1. What will be average acceleration when the particle completes half revolution?(a) zero (b) 10 m s–2

(c) 10p m s–2 (d) 10π

m s–2

43. The system is pushed by a force F as shown in figure. All surfaces are smooth except between B and C. Friction coefficient between B and C is m. Minimum value of F to prevent block B from downward slipping is

A B C

2m 2mmF

(a) 32µ

mg (b) 52µ

mg

(c) 52

mmg (d) 3

2

mmg

44. A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is 4 m s–1. There is a constant frictional force of 500 N. The work done by the lifting mechanism is(a) 196 × 103 J (b) 204 × 103 J(c) 214 × 103 J (d) 203 × 105 J

45. A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half of the escape velocity for that planet. The maximum height attained by the body is

(a) R3

(b) R2

(c) R4

(d) R5

SOLUTIONS

1. (b) : P a tbx

=− 2

[a] = [T2], as t2 is subtracted from a.

From, P a tbx

tbx

=−

=2 2

[ ]b =

=− −

tPx

2 2

1 2[T

ML T L]

[ ][ ] = [M–1L0T4]

∴

=

−ab

[T[M L T0

2

1 4]

] = [ML0T–2]

2. (a) : Here, u = – 12 m s–1, as packet is to fall to the ground and it is released initially with the velocity of balloon, so

a = + g = 10 m s–2, S = 65 m, t = ?

S = ut + 12

at2

65 = – 12t + 5t2 or 5t2 – 12t – 65 = 0

t =± +

=±

= −12 144 1300

1012 38

105 2 6s .or s

Time cannot be negative. Therefore, t = 5 s.

3. (b) :

F

N

m

mgf

Here, m = 10 kg, g = 10 m s–2, m = 0.5 F = 100 N Force of friction, f = mN = mmg = 0.5 × 10 kg × 10 m s–2 = 50 N Force that produces acceleration F′ = F – f = 100 N – 50 N = 50 N

a Fm

= ′ = = −5010

5 2Nkg

m s


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Page 63

8. (c) : Given : w = 2 + 6t

α

ω= =

ddt

6

t= FR = Ia where R is the radius of the disc.

∴ = =F I

R

MR

Rα α1

22

F MR=12

α =12

× 5 × 0.2 × 6 = 3.0 N

9. (d) : The slope of F-x gives the spring constant k. From the given graph,

k = Slope of F-x graph =−−

= −( )( . )

80 00 2 0

400 1Nm

N m

Time period of SHM is

T mk

= 2π = =−

20 01

4000 031π

..

kgN m

s

10. (b) : The given equation of the transverse wave is y = A sin2(wt – kx)

Velocity of the particle =dydt

= 2Awcos2(wt – kx)

Maximum velocity = 2Aw

Velocity of the wave = Coefficient ofCoefficient of

tx k k

= =22ω ω

As per question

2 2 12 4

Ak

Ak

Aωω λ

πλπ

= = = ⇒ =or

11. (d) : From Boyle’s law (T = constant) P1V1 = P2V2

\ (Hrwater + hrmercury) g ( 43pr3)

= hrmercury g (43p(2r)3)

or Hrwater = 8hrmercury – hrmercury

or H = 7hρ

ρmercury

water \ H = 7hr12. (b) : At the time of projection kinetic energy of the

stone,

K mu=12

2

where m is the mass of the stone and u is the velocity of the projection

or u

Km

2 2 2 982

98= =×

=

Using, v2 = u2 – 2gh

4. (d) : The net charge on the outer surface is – (– 50e – 50e) = 100e.5. (a) :

2 V

2

998 V

I I

From Ohm’s law

IR r

=+

=+

ε 2998 2

VΩ Ω

= 2 × 10–3 A

Potential difference across the voltmeter is V = IR = (2 × 10–3 A) × 998 W = 1.996 V6. (b) : Intensity of magnetisation

I MV

M= =

MassDensity

Given: Mass = 1 g = 10–3 kg

and density = 5 g cm–3 = 5 1010

3

2 3 3× −

−kg

( ) m= 5 × 103 kg m–3

Hence, I =× × ×

=− −

−−6 10 5 10

103

7 2 3 3

31A m kg m

kgA m

7. (a) : Angular momentum remains conserved during the revolution of planet. Because gravitational force is a central force.

rmin rmax

vmin

vmax

PlanetStar

According to law of conservation of angular momentum, we get

mvmaxrmin = mvminrmax where vmin is the speed of the planet when it is

farthest from the star and vmax is the speed of the planet when it is closest to the star.

∴ =vv

rr

min

max

min

max ...(i)

As per question

rrmax

min= 4

∴ = =

KEKE

Farthest

Closest

1212

2

2

2mv

mv

rr

min

max

min

max

=

=14

116

2

(Using (i))


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Page 64

∴ = =h

ug

v2

20( )∵

...(i)

h =×

=98

2 9 85

.m

Also (Using (i)), ( )K m gh=

12

2

and ′ = ′ = × ′K mv m gh

12

12

22 ( )

∴ ′ = ′K

Khh

According to the problem

′ =KK2 or

KK

hh2

= ′

′ = = =hh2

52

2 5m m.

13. (b) : According to law of conservation of linear momentum, we get

mu + 0 = 0 + Mv2 ⇒ v2 = muM

...(i) As per definition,

ev vu u

vu

vu

muMu

mM

= −−−

= −−

−= = =

( )( )

2 1

2 1

2 200

(Using(i))

14. (c) : According to Newton’s law of cooling

T Tt

KT T

Ts1 2 1 2

2−

=+

−

where Ts is the temperature of the surroundings. For 1st case

60 5530

60 552

45−=

+−

K ...(i)

Similarly, for 2nd case

55 50 55 502

45−=

+−

t

K ...(ii)

Dividing Eq. (i) by Eq.(ii), we get

5305

12 57 5

t

=..

or or st t30

12 57 5

12 57 5

30 50= = × =..

..

15. (b) : Here, Magnetic field, B = 0.025 T Radius of the loop, r = 2 cm = 2 × 10–2 m Constant rate at which radius of the loop

shrinks, drdt

= × − −1 10 3 1m s

Magnetic flux linked with the loop is f = BAcosq = B(pr2)cos0° = Bpr2

The magnitude of the induced emf is

ε φ π π= = ( ) =d

dtddt

B r B rdrdt

2 2

=0.025 × p × 2 × 2 × 10–2 × 1 × 10–3

= p × 10–6 V = p mV16. (a) : As VL = VC \ XL = XC

The circuit is resonance circuit.

Current, =100 V50

AIΩ

= 2

As V V V VR L C= + −2 2( )

∴ = + = =100 0 1002V V VR R Ror V

Hence, the reading of the voltmeter V is 100 V and the reading of ammeter A is 2 A.

17. (b) : Given, Ex = 0,

Ey = 2.5 NC

cos ,2 10 106 2π π×

− ×

−rads

radm

t x

Ez = 0.This shows that the wave is propagating along + x direction. Comparing the given equation with Ey = E0cos(wt – kx), we get

ω π πυ π υ= × = × =2 10 2 10 106 6 6or 2 or Hz

and k k= × = = ×− −ππ

λπ10 2 102 2or

or λπ

π=

×=

−210

2002 m

18. (a) : The figure shows incidence from water at critical angle qc for the limiting case.

rDisc Air

c

Lightsource

hWater

c

Now, sin θµc =1 so that tan

( ) /θµc =

−1

12 1 2

From figure, tan θcrh

=

where r is the radius of the disc. Therefore, diameter of the disc is

2 2 212 1 2r h

hc= =

−tan

( ) /θµ

19. (d) : Number of spectral lines obtained due to transition of an electron from nth orbit to lower orbits is

Nn n

=−( )1

2In the first case, N = 6

∴ =−

⇒ =61

24

n nn

( )


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Page 65

In the second case, N = 3

∴ =−

31

2n n( ) ⇒ n = 3

Velocity of an electron in hydrogen atom in nth orbit is

venhn =

24

2

0

ππε

; v

nn ∝1

∴ =

vv

4

3

34

20. (a) : Energy of photon =hcλ Power of lamp = P

No. of photons emitted per second, NP

hc=

/ λ \ No. of photoelectrons emitted per second

= = ×

0 5100

1200

.N

Phcλ

\ Photoelectric current = ×

1200

Phc

eλ

= ×

× × × ×

× × ×

− − −

−1

20010 4560 10 1 6 10

6 62 10 3 10

3 10 19

34 8( ) ( ) ( . )

( . ) ( ) = 1.836 × 10–6 A21. (b) : At a point P, midway between the poles,

magnetic field due to each pole is in the same direction.

\ B = B1 + B2 = µ

π0 1

12

2

224

mr

mr

+

As m1 = m2 and r1 = r2

\ B mr

= =× ××

− −

−µ

π0

2

7 4

2 242 10 2 10

5 10( )

( )= 8 × 10–9 T

22. (a) :

6 m(+ )q

(+ )q

+ /2q

A E B

CFD4 m

+ /2q

DE CE AE AD= = + = + =( ) ( ) ( ) ( )2 2 2 23 4 5 m

Potential at E,

V

qAE

qBE

qDE

qCEE = + + +

14

2 2

0πε/ /

= + + +

14 3 3

25

250πε

q q q q/ / = +

14

23 50πεq q

AF BF AD DF= = + = + =( ) ( ) ( ) ( )2 2 2 24 3 5 m Potential at F,

V

qAF

qBF

qDF

qCFF = + + +

14

2 2

0πε/ /

= + + +

14 5 5

23

230πε

q q q q/ /

= +

14

25 30πεq q

∴ − = + − −

V Vq q q q

E F1

423 5

25 30πε

= −

=×

q q4

13

15

215 40 0πε πε

= × × × = ×

215

9 10 1 2 109 9q q. volt.

23. (c) : The currents through various arms will be as shown in figure.

S

10 V4

5 V2

2

CI3I1 I2 BA

Let V be the potential at C.Applying Kirchhoff’s first law at C, we get I1 + I2 = I3

104

52

02

−+

−=

−V V V

10 – V + 10 – 2V = 2V or V = 4 V

∴ = =I342

2VΩ

A

24. (a) : Given: F = – 5x – 16x3 N

W Fdx x x dx

x

x

i

f

= = − −∫ ∫ ( ).

.5 16 3

0 1

0 2

= − −

52

42 4

0 1

0 2x x

.

.

= − − + +

52

0 2 4 0 2 52

0 1 4 0 12 4 2 4( . ) ( . ) ( . ) ( . )

= –0.1 – 0.0064 + 0.025 + 0.0004 = – 0.081 = – 8.1 ×10–2 J 25. (d) : Kinetic energy of the satellite orbiting the

earth is

E mv=12

2, where v is the orbital velocity.

\ 2 12

22

E m v= ( ) 2v is the escape velocity of the satellite from the

earth. \ When the kinetic energy of satellite is made

2E, satellite escapes away.

26. (a) : Force required to keep the belt moving = F

F v dmdt

= = 0.2 m s–1 × 2 kg s–1 = 0.4 N

27. (a) : From conservation of energy Potential energy = Translational + Rotational kinetic energy kinetic energy


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mgh mv I= +

12

12

2 2ω

mgh mv mR vR

= +

12

12

25

2 22

2

or mgh mv mv= +

12

15

2 2

or ormgh mv v gh= ≥

710

107

2 2

28. (c) : The average translational kinetic energy of a

molecule of a gas =32

k TB which is independent

of mass of molecule but depends upon the temperature of the gas.

29. (d) : y1 = 0.1sin(100pt + π3 )

Velocity, vdydt1

1=

= 0.1 × 100pcos(100pt + π3

) y2 = 0.1cospt

or = 0.1sin(pt + π2

)

Velocity, vdydt2

2= = 0.1 × pcos(pt + π2 )

\ Phase difference of the velocity of particle 1 w.r.t. particle 2 is

= (100pt + π3

) – (pt + π2 )

= 99pt + π3

– π2

At t = 0, phase difference = π3

– π2

= – π6

30. (b) : Here, l = 30 cm = 30 ×10–2 m, v = 330 m s–1

In case of an open organ pipe, fundamental frequency = frequency of 1st harmonic

υ1 22330

2 30 10= =

× × −vl

= 550 Hz

The frequncies of 2nd harmonic, 3rd harmonic, 4th harmonic,.... are

2 × 550 Hz = 1100 Hz; 3 × 550 Hz = 1650 Hz; 4 × 550 Hz = 2200 Hz and so on. Hence, the source of frequency 1.1 kHz i.e.,

1100 Hz will resonantly excite second harmonic.31. (c) : Here, r = 1.0 mm = 10–3 m, S = 0.075 N m–1, r = 103 kg m–3; h = 10 cm = 10 ×10–2 m The pressure inside the bubble which is greater

than the atmospheric pressure is

= +2Sr

h gρ

=×

+ × × ×−

−2 0 07510

10 10 10 9 832 3. .

= 150 + 980 = 1130 N m–2

32. (d) : According to first law of thermodynamics, For the path acb,

Qacb = DUacb + Wacb c b

a dP

V

\ DUacb = Qacb – Wacb = 60 J – 30 J = 30 J For the path adb, Qadb = DUadb + Wadb DUacb = DUadb, change in internal energy is path

independent. \ Qadb = 30 J + 10 J = 40 J

33. (c) : When capacitor is fully charged, it draws no current. Hence potential difference across capacitor = Potential difference across C and F.

I1I1

I2 II I

B

A

EC

F

GH

3

3 3

15 V

3 3 D

3 F

Refer figure. Effective resistance of the network between A and

D is R1

3 3 33 3 3

2=+ ×+ +

=( )( )

Ω

Total resistance of the circuit = 2 W + 3 W = 5 W

Current, I =155

VΩ

= 3 A

∴ =

+=I1

3 33 6

1( )( )A

AΩ

Ω Ω

and

AAI2

3 63 6

2=+

=( )( )Ω

Ω Ω Potential difference across A and D = 3 W × 2 A = 6 V Potential difference across D and F, VD – VF = 3 W × 3 A = 9 V Potential difference across C and D, VC – VD = 3 W× 1 A= 3 V Potential difference across C and F = VC – VF

= (VC – VD) + (VD – VF) = 3 V + 9 V = 12 V \ Potential difference across capacitor = VC – VF = 12 V

34. (d) : Magnetic force on a straight wire F = BIl sin90° = BIl Weight of the wire, W = mg Since the wire remains suspended in mid-air so,


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Page 67

BIl = mg

or B mgI l

= =× ×

×=

−( ).

.200 10 102 1 5

0 663

T

35. (c) : If the alternating current in the primary is I = I0 sinwt, the emf induced in the secondary is

ε ωsPM

dIdt

M ddt

I t= = [ sin ]0

ε ω ω πυ ωs M I t MI t= =0 02cos cos [as w = 2pu]

\ (es)max = 2puMI0 [as (coswt)max = 1]

Substituting the given data,

(es)max = 2p ×50 ×1.5 × 1 = 150p = 471 V

Note:

(es)max = = ××

=MdIdt

1 5 114

150

300. V is wrong as it

will give the average value (over a quarter cycle) and not the required peak value.

36. (d) : R VI

= = =100

1100

VA

Ω

ZVI

= = =rms

rms

VA

1000 5

200.

Ω

As R X ZL2 2 2+ =

X Z RL = − = − =2 2 2 2200 100 100 3 Ω

As XL = wL = 2puL = 2p × 50 × L

\ LXL= = =

100100 3100

0 55π π

. H

37. (d) : l1 = 6000 Å, n1 = 16 fringes \ n2 = 24 fringes As n1l1 = n2l2

⇒ λλ

1

2

2

1=

nn

⇒ 6000 24162λ

=

⇒ λ26000 16

2496000

24=

×= = 4000 Å

38. (b) : Angular frequency,

ω πυπλ

π= = =

× ×× −

2 2 2 3 106 10

8

3c

= p × 1011 rad s–1

The equation for the electric field, along y-axis in the electromagnetic wave is

E E txcy = −

0 sinω

= 33 sinp × 1011 (t – xc

)

39. (a) : Power of system = 1 1

1 2 1 2f fd

f f+ −

= +−

−

−11

10 25

0 751 0 25.

.( )( . )

= 1 – 4 + 3 = – 3 + 3 = 0 Since, power of the system is zero therefore, the

incident parallel beam of light will remain parallel after emerging from the system.

40. (c) : The wavelength of the first line of lyman series for hydrogen atom is

1 1

11

234

432 2λ

λ= −

= =R R

Ror

The wavelength of the second line of Balmer series for hydrogen like ion is

1 1

21

43

1616

32

2 22

2′= −

= ′ =

λλZ R Z R

Z Ror

As per question, l = l′

or

43

163 2=

Z or Z2 = 4 or Z = 2

41. (b) : Let u be the initial velocity and a be uniform acceleration of the particle. Using v2 – u2 = 2aS 62 – u2 = 2a × 5 ...(i)

and 82 – u2 = 2a(5 + 7) = 2a × 12 ...(ii) Solving (i) and (ii), we get a = 2 m s–2 and u = 4 m s–1

42. (d) : Change in velocity when the particle completes half revolution is

Dv = 5 m s–1 – (– 5 m s–1) = 10 m s–1

Time taken to complete the half revolution is

t rv

= =×

=−

π ππ

55 1

mm s

s

Average acceleration = ∆vt

=−

−10 1012m s

sπ π= m s

43. (b) : Horizontal acceleration of the system is

a Fm m m

Fm

=+ +

=2 2 5

...(i)

Let N be the normal reaction between B and C. Free body diagram of C gives

N ma F= =2 25

(Using (i))

C

Na


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Page 68

Now B will not slide downward if mN ≥ mBg

or µ25

F mg

≥ or F mg≥

52µ

So F mgmin =5

2µ

44. (c) : Work done against gravitational force = mgh = 1000 × 9.8 × 20 = 196 × 103 J

Work done to impart velocity to the body = 12

2mv

= 12

× 103 × 16 = 8 × 103 J

Work done against frictional force = 500 × 20 = 10 × 103 J Total work done = 214 × 103 J

45. (a) : Escape velocity, v GM

Re =2

where M and R are the mass and radius of the planet respectively.

Velocity of projection, v

v GMR

e= =2

12

2

The total energy of the body when it is projected is

Ei = KE + PE

= − = −

12

12

24

2mv GMmR

m GMR

GMmR

= −

34

GMmR

Let the maximum height attained by the body be h. At this height, the total energy of the body is

E GMm

R hf = + = −+

KE PE 0

From the principle of conservation of energy, Ei = Ef

− = −+

34

GMmR

GMmR h

or 3(R + h) = 4R

or 3 3 43

R h R h R+ = ⇒ =

vvv

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19 years AIIMS Chapterwise Solutions

300`325`

Index

19 yrs.Chapterwise

19 years ( 1994-2012) Solved Papers with Detailed Solutions

10 Model Test Papers

500+ General Knowledge Questions

1400+ Assertion and Reason Questions

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49. The electric field at a distance 32R from the centre

of a charged conducting spherical shell of radius

R is E. The electric field at a distance R2

from the centre of the sphere is

(a) zero (b) E (c) E2

(d) E3

50.

A and

B are two vectors of equal magnitude and q is the angle between them. The angle between

A or

B with their resultant is

(a) θ4 (b) θ

2 (c) 2q (d) q

SOLUTIONS1. (c) : By covering the aperture of diameter d/2,

focal length of lens is not affected. Area reduces by (1/4)th. So does the intensity.\ New focal length = f, and

New intensity = I I I− =

434

2. (b) : As no external torque acts on the system, angular momentum will remain conserved. According to law of conservation of angular momentum, I1w1 = I2w2

or m r m rr r r f f f2 2ω ω=

where the subscripts r and f denote the rear and front.

m(2rf)2wr = mrf2 wf [rr = 2rf (Given)]

\ 4wr = wf or ωω

rf=

4Hence, the angular velocity of rear wheel will be smaller compared to front wheel.

3. (b) : The frequency of kinetic energy is twice the frequency of simple harmonic motion.

4. (c) : Let the block and bullet meet after time t, at a distance x vertically below the top of the cliff.Taking vertical downwards motion of block for time t

x gt= +0 12

2 ( ) ... u = 0 (i)

Taking vertical upwards motion of bullet for time t

100 100 12

2− = −x t gt ...(ii)

Adding Eqs. (i) and (ii), we get 100 = 100t or t = 1s

5. (c) : Here,

Voltage gain = 50Input resistance, Ri = 100 WOutput resistance, Ro = 200 W

Resistance gain = = =RR

o

i

200100

2ΩΩ

Power gainVoltage gain

Resistance gain=

( )2

= × =50 502

1250

6. (d) : According to Kepler’s third law,

T RTT

RR

∝ ∴ =

=3 2 2

1

3 21 021 03/

/..

Percentage difference =

−×

T TT

2 1

1100

= −

×

= − ×= × =

TT

2

11 100

1 03 1 1000 03 100 3( . )

. %

7. (c) : As v Rgmax tan= θ

For the same banking angle

v Rmax ∝

∴′

= ′ ′ =′

vv

RR

RR

vv

max

max maxor max

2

As per question

′ = + = =v v v v vmax max max max max.10100

1110

1 1

∴ ′ =

=

RR

vv

1 11 21

2.

.max

max

R′ = 1.21 R = 1.21 × 20 m = 24.2 m

8. (d) : Here, Vp = 220 V, Is = 2 A, Vs = 440 V h = 80%, Ip = ?

η =V IV I

s s

p por I

V IVps s

p=

η

Substituting the given values, we get

Ip =

=( )( )

( )

440 280

100220

5V A

VA

9. (a) : As long as the block of mass m remains stationary, the block of mass M released from

rest comes down by 2Mgk

(before coming it rest

momentarily again). Thus the maximum extension in the spring is

Cont. from Page No. 28


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x

Mgk

=2

...(i)

For block of mass m to just move up the incline kx = mgsin37° + mmgcos 37° ...(ii)

2 35

34

45

Mg mg mg= × + ×

(Using (i))

or M m=35

10. (d) : When the frequency of incident light is halved, its frequency becomes 0.75u0 which is less than threshold frequency (u0). Hence, no photoelectric emission will take place.

11. (b) : Cross - sectional area, A

D=

π 2

4where D is the diameter of the wire.

∴ =∆ ∆AA

DD

2

or ∆ ∆DD

AA

= = × =12

12

2 1% % ...(i)

Poisson’s ratio, σ =∆∆D Dl l//

or ∆ ∆ll

D D= = =

/.

. %σ

10 4

2 5 (Using (i))

12. (a) : According to definition of centre of mass, we can imagine one particle of mass (1 + 2 + 3) kg at (2, 2, 2). Let the second particle of mass 4 kg be put at (x2, y2, z2).Given (xCM, yCM, zCM) = (0, 0 , 0)

Using x

m x m xm mCM =

++

1 1 2 2

1 2

06 2 4

5 42=

× + ×+

x

x2 = –3Similarly, we get y2 = –3, z2 = –3

13. (a) 14. (a) 15. (b) : Let e be the emf and r be the internal resistance

of the battery.As per question

12 16

16=

×+

εr

...(i)

and 11 1010

=×

+ε

r ...(ii)

Dividing (i) by (ii), we get

1211

8 105 16

311

25

1016

=++

=++

( )( )

( )( )

rr

rr

or

15(16 + r) = 22(10 + r)240 + 15r = 220 + 22r

7r = 20 or r =207

Ω

16. (a) : The velocity vs of the satellite is given byGMm

r

m vr

vGM

r

s

s

s s

s

ss

2

2=

∴ =

…(i)

Kinetic energy of the satellite is

K = 12

msvs2 = 1

2 2m

GMr

GMmrs

s

s

s

= (Using (i))

Potential energy of the satellite is

UGMm

rs

s= −

Total energy of the satellite is

E K UGMm

rGMm

rGMm

rs

s

s

s

s

s= + = − = −

2 2 ...(ii)

The angular momentum L of the satellite is given by

L = msvsrs = msGM

rr

ss

1 2/ (Using(i))

= (GMms 2 rs)1/2 …(iii)

From Eqs. (ii) and (iii), we get L = (2Emsrs 2)1/2

17. (a) : Given : x = 36tand 2y = 96t – 9.8t2

or y = 48t – 4.9t2

Let the initial velocity of projectile be u and angle of projection is q. Then,Initial horizontal component of velocity,

ux = ucosq = dxdt t

=0

= 36

or ucosq = 36 …(i)Initial vertical component of velocity,

uy = usinq = dydt t

=0

= 48

or usinq = 48 …(ii)Dividing (ii) by (i), we get

tanq = 4836

43

=

\ sinq = 45

or q = sin–1 45

18. (d) : Here, body is acting as an observer. According


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Page 72

to the given case, let u be the velocity of the body.

Then, υ υ υ υ1 2=−

=+

v uv

v uv

and

\ Du = u2 – u1 = u 2uv

So, ∆υυ

× = × =100 2 100 2uv

⇒ uv

= = =100

300100

3 m s–1

19. (a) : According to lens maker’s formula

1 1 1 1

1 2f R R= − −

( )µ

Here, m = 53 , R1 = ∞, R2 = 0.3

\ 1 5

31 1 1

0 3f= −

∞

−

. = – 2

0 9.

or f = –0 92.

= – 0.45 m

20. (a) : FItav = = =

−

∆25

0 05500

1N ss

N.

21. (c) : Latent heat = Heat energyMass

= =

−−[ ]

[ ][ ]ML T

MM L T

2 20 2 2

22. (d) : In case of electric dipole, F ∝1

r3

′ =

= ′ =FF

r

r

FF

121

18 8

3

3 or

23. (d) : Here, B = 1 Wb m–2 , H = 150 A m–1

m0 = 4p × 10–7 H m–1

As B = m0mrH

µµ π πr = =

× ×=

−

− − −BH0

2

7 1 1

514 10 150

106

Wb m

H m A m

24. (b) : The equivalent circuit is as shown in figure a and b.

\ Effective resistance between A and D = 5 W + 10 W + 5 W = 20 W

25. (a) : Here, h = 18 × 10–5 poise, r = 0.3 mm = 0.03 cm v = 1 m s–1 = 100 cm s–1

According to Stoke law, Viscous force, F = 6phrv F = 6p × 18 × 10–5 poise × 0.03 cm × 100 cm s–1 = 101.73 × 10–4 dyne

26. (c) : At maximum height h, v = 0 \ Total mechanical energy, E1 = mgh + 0

Let v′ be velocity of stone at height ′ =h h45

.

\ Total mechanical energy, E mg h mv2

245

12

= + ′

According to the law conservation of mechanical energy,

E2 = E1

∴ + ′ =

45

12

2mgh mv mgh

or ′ = −

v g h h2 2 4

5

or

′ =v gh

25

...(i)

Required ratio of kinetic energy to its potential

energy at ′ =h h45

is

K EP E.

. ..

=′

=

1245

1545

2mv

mgh

mgh

mgh

(Using (i))

=14

27. (b) : AB

Y

0 0 00

00

01

1

0

AB

Y

1 1 00

10

00

0

0

0

1

28. (b) :

M2k 2k

k

2k


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Springs on the left of the block are in series, hence their equivalent spring constant is

kk kk k

k12 22 2

=+

=( )( )

Springs on the right of the block are in parallel, hence their equivalent spring constant is

k2 = k + 2k = 3kNow again both k1 and k2 are in parallel.\ keq = k1 + k2 = k + 3k = 4kHence, frequency of the system is

υπ π

= =1

21

24k

Mk

Meq

29. (c) : The oscillator frequency should be same as proton’s cyclotron frquency.

Cyclotron frequency, υπ

=Bq

m2or Bq = 2pmu ...(i)Kinetic energy of proton is

K q B rm

m rm

= =2 2 2 2 2

22

2( )π υ

(Using (i))

= 2p2mu2r2

Substituting the given values, we get

K =× × × × ×

×

−

−2 3 14 1 67 10 10 0 5

1 6 10

2 27 7 2 2

13( . ) ( . ) ( ) ( . )

.MeV

= 5.1 MeV30. (a) : Initial fundamental frequency of a stretched

string is

υµ

=1

2LT

…(i)

where the symbols have their usual meanings.When the length of a stretched string is shortened by 40% and the tension is increased by 44%, then its length and tension becomes

′ = − =L L L L40100

35

T T T T′ = + =44

1003625

Then, final fundamental frequency is

υµ

′ =′

′12L

T

=

=1

2 35

3625

22L

TL

Tµ µ

…(ii)

Dividing (ii) by (i), we get

υυ′

=21

31. (a) : The yellow light is scattered less by the fog particles.

32. (b) : Here, l = 6000 Å = 6000 × 10–10 m = 6 × 10–7 mFor nth dark fringe,

Path difference, S1P – S2P = (2n – 1) λ2

For 3rd dark fringe n = 3

\ S1P – S2P = 52

5 6 102

7λ =

× × −m = 15 × 10–7 m

= 1.5 × 10–6 m = 1.5 mm33. (a)

34. (b) :

µ

δ

=

+

sin

sin( / )

A

A

m2

2

where m is the refractive index, A is the angle of prism and dm is the angle of minimum deviation.Given : A = 60°, m = 1.414

2

602

30=

° +

°

sin

sin

δm

12

602

4560

2=

° +

° =

° +

sin sin sin

δ δm mor

4560

2° =

° + δm or dm = 90° – 60° = 30°

35. (c) : According to Stefan’s law

dTdt

Ams

T Ts= −σ [ ]4 4

or s A

m dTdt

T Ts= −4σ [ ]4

Here, A = 19.2 cm2 = 19.2 × 10–4 m2, m = 34.38 g = 34.38 × 10–3 kg,

s = 5.73 × 10–8 Wm–2 K–4, dTdt

= 0.04°C s–1

T = 400 K, Ts = 300 K

∴ =× × −

× ×

− −

−s ( . )( . )[( ) ( ) ]

. .19 2 10 5 73 10 400 300

34 38 10 0 04

4 8 4 4

3

=× × −

× ×

− −

−( . )( . ) [( ) ( ) ]

. .19 2 10 5 73 10 10 4 3

34 38 10 0 04

4 8 8 4 4

3

= 1400 K36. (d) : From N = N0 e–lt

Ne0 = N0e–l×5


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Page 74

\ 5l = 1 or l = 15

The activity reduces to half its initial value in half life period,

T1 22 2

1 55 2/

ln ln/

ln= = =λ

37. (b) : The magnetic field at the centre O due to current in the inner coil is

B10

410 2 0 2

0 2=

× ×µπ

..

0.3 A0.2A

O

0.4 m

0.2 m

The magnetic field at the centre O due to current in the outer coil is

B20

410 2 0 3

0 4=

× ×⊗

µπ

..

Net magnetic field at the centre O isB = B1 – B2 [Since B1 and B2 are in opposite directions]

= 204

1 34

54

00

µπ

µ−

= T

38. (d) : Heat required to melt 1 kg ice at 0°C into water at 0°C isQ = mice Lice = (1 kg) (80 cal g–1) = (1000 g) (80 cal g–1) = 8 × 104 cal

Change in entropy, ∆S QT

= =×8 10273

4calK

= 293 cal K–1

39. (b) : Here, N = 1000, A = 500 cm2 = 500 × 10–4 m2

B = 2 × 10–5 Wb m–2

Initial flux through the coilfBInitial = BAcosq

= 2 × 10–5 × 500 × 10–4 cos0° = 10–6 WbFinal flux after rotation

fBFinal = BAcos180°

= 2 × 10–5 × 500 × 10–4 × cos180° = – 10–6 WbThe average induced emf in the coil is

εφ φ φ

= − = −− N

t

N

tB B B∆

∆ ∆Final Initial

= −− − − −1000 10 10

0 2

6 6

.V

= 10 × 10–3 V = 10 mV40. (b) : Number of neutrons in 12

6C = 12 – 6 = 6 Number of neutrons in 14

6 C = 14 – 6 = 8

41. (b) : P AT BTV

=− 2

or V AT BTP

=− 2

Since P is constant

\ dV AdT BTdTP

=− 2

or PdV = (A – 2BT) dT

Workdone, W PdV A BT dTT

T

= = −∫ ∫ ( )21

2

= A(T2 – T1) – B(T2 2 – T1 2)

42. (d) : 1530

1430P Si

positron→ + +e

43. (c) :

A

B

D E

F G

C C

C C

C C

The equivalent circuit diagrams are as shown in the figure below.

AC C

C

C C

C

B

D

E

G

F

A

A

2C C

2 /3C

CB

B

2C

2 /3C

The equivalent capacitance between A and B is

C C C C

AB = + =23

23

43


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175`175` 200`


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Page 76

44. (b) : y = bx2

Differentiating w.r.t to t on both sides, we get

dydt

b x dxdt

= 2

vy = 2bxvx

Again, differentiating w.r.t. to t on both sides, we get

dv

dtbv dx

dtbx

dvdt

yx

x= +2 2 = +2 02bvx

[dvdt

x = 0, because the particle has constant

acceleration along y-direction]As per question

dv

dta bvy

x= = 2 2 or v abx

22

= or v abx =

245. (c) : Here, G = 50 W, Ig = 0.05 A, I = 5 A,

A = 2.97 × 10–2 cm2 = 2.97 × 10–2 × 10–4 m2

= 2.97 × 10–6 m2

r = 5 × 10–7 W m

SI G

I I=

−=

×−

=g

g

0 05 505 0 05

5099

..

Ω

Now, SA

SA= =

ρρ

l lor

\ l = ××

×=

−

−5099

2 97 105 10

3 06

7( . )

( ). m

46. (c) :

According to steady flow, A1v1 = A2v2 + A3v3or A3v3 = A1v1 – A2v2

or vA

A v A v33

1 1 2 21

= −[ ]

= × − × = −10 4

0 2 4 0 2 2 1 1.

[ . . ] m s

47. (c) : The given wave equation is

y t x= +1

1060 2sin( )

Compare it with the standard wave equation

y = Asin(wt + kx)

We get,

(i) Amplitude m cm, A = =1

1010

(ii) Angular frequency, w = 60 rad s–1

(iii) Angular wave number, k = 2 rad m–1

\ Velocity of the wave,

vk

= = =−

−−ω 60

301

11rad s

2 rad mm s

\ Frequency of the wave,

υωπ π π

= = =2

602

30 Hz

\ Wavelength of the wave, λπ π

π= = =2 2

2km

As there is positive sign between t and x terms, the given wave is moving in the negative x direction.

48. (d) : Here, n1 = 4, T1 = 400 K n2 = 2, T2 = 700 KThe temperature of the mixture is

Tmixture = n T n T

n n1 1 2 2

1 2

++

=× + ×

+4 400 2 700

4 2K K

=+

=1600 1400

6500

K KK

49. (a) : Electric field at a point inside the charged conducting spherical shell is zero as charge only resides on the outer surface of the conducting spherical shell.

50. (b) : Here,

A B=Let the resultant

R makes an angle a with the vector

A . Then

tan sincos

sincos

αθ

θθ

θ=

+=

+=( )B

A BA

A AA B

Given

=+

= =sin

cos

sin cos

costanθ

θ

θ θ

θθ

1

22 2

22

22

αθ

=2

mmm


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Page 77

units & measurements

Physical quantity = Numerical value × unit

Homogeneity Principle Dimensions of [LHS] = Dimensions of [RHS]Mean absolute error

∆∆ ∆ ∆

aa a a

nn

mean =+ + +| | | | ... | |1 2

∆ ∆an

aii

n

mean = ×=∑1

1| |

Arithmetic mean amean =+ + +a a a

nn1 2 ....

an

aii

n

mean ==∑1

1Relative error or fractional error

= =

mean absolute errormean value

mean

mean

∆aa

Percentage error mean

meanδa

aa

= ×∆

100%

If in a vernier callipers n VSD coincide with (n – 1) MSD, then vernier constant or its least count is

VCn

n= −

−

1 1 (value of 1 MSD) or 1n

(value of

MSD).Least count of screw gauge or spherometer

=

PitchNumber of divisions on circular scale

and

Pitch = Number of divisions moved on linear scaleNumber of rotaations given

= Linear distance moved in one rotation.

Random error = n , where n = number of events or n = number of quantities.Radius of curvature using spherometer

Rlh

h= +

2

6 2

kinematics

Speed =total path length

time taken

Average speed =total distance travelled

total time taken

i e vs s st t tav. . = 1

1

+ ++ +

2 3

2 3

............

Instantaneous speed

= =

→lim

∆

∆∆t

st

dsdt0

Velocity displacementtime taken

=

Average velocity total displacementtotal time taken

=

Acceleration a =Change in velocity

time takenAverage acceleration

a

vtav =

∆∆

Instantaneous acceleration

avtt

=

→

lim .∆

∆∆0

Equation of motion for a uniform accelerated motion

v = u + at

s ut at= +12

2

v 2 – u2 = 2as

s ua

nn = + −2

2 1( ) where u is initial velocity, v is final velocity, a is

uniform acceleration, s is distance travelled in time t, sn is distance covered in nth second. These equations are not valid if the acceleration is non-uniform.Equation of motion for a body under gravity

v = u + gt


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Page 78

h ut gt= +12

2

v 2 – u2 = 2gh

h u g nn = + −12

2 1( )

Relative velocity If two bodies are moving along the same line in the same direction with velocities vA and vB relative to earth, the velocity of B relative to A will be given by vBA = vB – vA.If the bodies are moving towards or away from each other as directions of vA and vB are opposite velocity of B relative to A will be vBA = vB – (–vA) = vB + vA.

Relative velocity of rain

tanα =vvm

rWhere, vm = velocity of man vr = velocity of rain and a is the angle with the vertical direction at which man should hold umbrella to save himself from the rain.

Unit vector, aaa

^| |

=

where, a is the unit vector drawn in the direction of

a aand| |is the magnitude of the vector.Dot or scalar product

a b ab⋅ = cos ,θ0 ≤ q ≤ pProperties of dot product

a b b a⋅ = ⋅

a b c a b a c⋅ + = ⋅ + ⋅( )

m a b ma b a mb a b m( ) ( ) ( )

⋅ = ⋅ = ⋅ = ⋅ where m is a scalar

i i j j k k^ ^ ^ ^ ^ ^ ,⋅ = ⋅ = ⋅ = 1 i j j k k j^ ^ ^ ^ ^ ^⋅ = ⋅ = ⋅ = 0

If

a a i a j a k b b i b j b k= + + = + +1 2 3 1 2 3^ ^ ^ ^ ^ ^and

a b a b a b a b⋅ = + +1 1 2 2 3 3

a a a a a a⋅ = = + +212

22

33.

b b b b b b⋅ = = + +212

22

33.

If

a b a b⋅ = 0 and and are not null vectors, then

a band are perpendicular. Cross or vector product

a b ab n× = sin .^θ 0 ≤ q ≤ pProperties of vector product

a b b a× = − ×

a b c a b a c× + = × + ×( )

m a b ma b a mb a b m( ) ( ) ( ) ( ) ,

× = × = × = × where m is a scalar.

I f

a a i a j a k

b b i b j b k

= + +

= + +1 2 3

1 2 3

^ ^ ^

^ ^ ^,and then

a bi j k

a a ab b b

× =

^ ^ ^

1 2 3

1 2 3

a b× = the area of a parallelogram with sides

a and

b .If

a b× = 0 and

a and

b are not null vectors, then

a and

b are parallel.Parallelogram law of vector addition

R a b= + , then

R a b ab= + +2 2 2 cosθ

and tan sincos

βθ

θ=

+b

a b

If

R a b a b= − = + −( )

then R a b ab= + −2 2 2 cosθ

and tan sin( )

cos( )sin

cosβ

θθ

θθ

=° −

+ ° −=

−b

a bb

a b180

180

Where, q is the angle between

a band .

Equation of trajectory

y xgx

u= −tan

cosθ

θ

2

2 22

Where u is initial velocity makes an angle q with the horizontal.Time of flight

T

ug

=2 sinθ

Horizontal range

R

ug

=2 2sin θ

Range will be maximum. If q = 45°

R

ugmax =2

If angle of projection is changed from q to q = (90° – q) then range

′ = =° −

Ru

gu

g

2 22 2 90sin sin[ ( )]θ θ= =

ug

R2 2sin θ


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Maximum Height

H

ug

=2 2

2sin θ

Height attained by projectile is maximum if q = 90°

H

ug

Rmax

max= =2

2 2Here, range of projectile

R

ug

=× °

=2 2 90 0sin

When the range is maximum, ( q = 45°)

H

ug

ug

R=

°= =

2 2 2452 4 4

sin max

Projectile on an inclined plane (Motion up the Plane)

Time of flight Tu

g=

−2 sin( )cos

θ ββ

Range

Ru

g=

− −2

22[sin( ) sin ]

cosθ β β

βR will be maximum when sin(2q –b) is maximum.i.e. sin(2q – b) = 1

Ru

gmax ( sin )=

+

2

1 βup the plane

Motion down the plane

Time of flight Tu

g=

+2 sin( )cos

θ ββ

Range, Rug

=+ +

−

2

221

sin( ) sinsin

θ β ββ

R will be maximum, if sin(2q + b) = 1

Rmax =+

−

=−

ug

ug

2

2

211 1

sinsin ( sin )

ββ β

down the plane

At the highest point of a projectile motion given angular projection, the angular momentum of projectile.

L muu

g= ×cos sin

θθ2 2

2

In case of angular projection, the angle between velocity and acceleration varies from 0° < q < 180°.Angular acceleration

αω θ

= =ddt

d

dt

2

2

When a body moves in a circular path with increasing angular velocity, it has two linear acceleration.

Centripetal acceleration

a

vr

r v vc = = = =2

2 22ω ω πυ( )

Tangential acceleration at = ra Resultant acceleration

a a ac t= +2 2

tan .β =

aa

t

c

Centripetal force

F

mvr

=2

.

Laws of motion

Linear momentum

p mv=

Where, m is mass of a body moving with velocity

v.Newton’s second law Force,

F = rate of change of linear momentum

= =

dpdt

ma

.

Where

a is acceleration produced in the body.Impulse = Change in linear momentum = F × t = m(v – u)Equilibrium of concurrent forces :

F F F Fn1 2 3 0+ + + + =....Lami’s theorem :

F F F1 2 3

sin sin sinα β γ= =

where, α = angle between and

F F2 3

β = angle between and

F F3 1

γ = angle between and

F F1 2

Ap parent weight of a man in a lift : When the lift is at rest or moving with constant velocity, the apparent weight = mg. Thus apparent weight = true weight. When the lift is accelerating upwards with acceleration a, then apparent weight = m(g + a). Thus apparent weight is more than the true weight.When the lift is accelerating downwards with acceleration a, then apparent weight = m(g – a).

Thus apparent weight is less than the true weight of man.


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Page 80

In the cable supporting the lift breaks, the lift falls freely with a = g, then apparent weight = m(g – g) = 0.When a person of mass m climbs up a rope with acceleration a, the tension in the rope is T = m(g + a). When the person climbs down the rope with acceleration a, the tension in the rope is T = m(g – a).When the person climbs up or down with uniform speed, the tension in the rope is T = mg.

Thrust on the rocket F udmdt

= −

Where, dmdt

is mass of burnt gases escaping per

second and u = exhaust speed of the burnt gases.Velocity of rocket at any time t.

v umme=

log 0

Acceleration of rocket at any instant

a = upthrust weightmass

−

Laws of friction : The magnitude of the force of static friction between any two surfaces in contact can have the values

fs ≤ ms R ...(i)where the dimensionless constant ms is called the coefficient of static friction, R is the magnitude of normal reaction force. The equality in equation

holds when the surfaces are on the verge of slipping i.e., fs = (fs)max = (fl) ≡ ms R.The magnitude of the force of kinetic friction acting between two surfaces is

fk = mk R where mk is coefficient of kinetic friction.

Acceleration of a body down a rough inclined plane, a = g(sinq – mcosq)

where, q is the angle of inclination and m is the coefficient of friction. Angle of repose m = tanaWhere, a is angle of reposeWork done in moving a body over a rough horizontal surface. W = mR × s = mmg × sWhere, R is normal reaction and s is distance moved by body.Work done in moving a body up a rough inclined plane.

W = (mgsinq + mR)s.Bending cyclist,

angle of bending tanθ=vrg

2

Circular turning of roads The velocity with which a car can take a circular path of radius r without slipping is given by v rgsmax = µThe maximum permissible speed to avoid slipping,

v

rg s

smax

/( tan )tan

=+

−

µ θµ θ1

1 2

v 02 = rg tanq or tanθ =

vrg02

tanθ =−

=h

b h

vrg2 202

Motion in a vertical circle Tension at any position of angular displacement, (q) along a vertical circle is given by

Tmv

rmg= +

2cosθ

At the lowest point of vertical circle, q = 0° Tension at the lowest point is given by

Tmv

rmgL

L= +2

At the highest point of the vertical circle, q = 180°. Tension at the highest point is given by

T

mvr

mgHH= −2

Minimum velocity at the highest point,

v grH =Minimum velocity at the lowest point for

looping the loop, v grL = 5 . When the string is horizontal, q = 90°, minimum velocity, v gr= 3 .Height through which a body should fall for looping the vertical loop h = 5r/2.

work, enerGy anD power

W F S FS= ⋅ =

cosθ Where q is angle between

F Sand

Work done by a variable force, W F x dxx

x

i

f

= ∫ ( )

Kinetic energy : K mv=12

2.

Relation between kinetic energy ( K) and linear momentum (p)


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Kpm

p mK= =2

22or

Work done by a spring force

W F ds= ⋅∫ spring

Work energy theorem : W = Kf – Ki

Elastic potential energy : U kx=12

2

Gravitational potential energy : U = mgh

Power, totalPW

t=

Instantaneous power, PdWdt

Fd sdt

F v= = ⋅ = ⋅

Elastic collision in one dimension

B A

vm m um m

m um m1

1 2 1

1 2

2 2

1 2

2=

−+

++

( )( )

v

m um m

m mm m

u21 1

1 2

2 1

1 22

2=

++

−+

( )

Perfectly inelastic collision in one dimension

B A

v

m u m um m

=++

1 1 2 2

1 2( )

Loss in kinetic energy in elastic collision is

∆K

m mm m

u u=+

−12

1 2

1 21 2

2( )

( )

Coefficient of restitution

e

v vu u

=−−

2 1

1 2

Kinetic energy lost in inelastic collision is

∆ =

+− −K

m mm m

u u e12

11 2

1 21 2

2 2( )

( ) ( )

A ball falls from a height h, it strikes the ground with a velocity u gh= 2 . Let it rebound with a velocity v and rise to a height h1.

e

vu

gh

gh

hh

= = =2

21 1

or orh e h h e h1 12= = .

A ball dropped from a height h and rebounding. The time taken by the ball in rising to height h1

and coming back is 2

22 21h

ge

hg

= .

rotationaL motion

The coordinates of centre of mass are given by

X

m x

m

m x

MCM

i ii

N

ii

N

i ii

N

= ==

=

=∑

∑

∑1

1

1

Y

m y

m

m y

MCM

i ii

N

ii

N

i ii

N

= ==

=

=∑

∑

∑1

1

1

Z

m z

m

m z

MCM

i ii

N

ii

N

i ii

N

= ==

=

=∑

∑

∑1

1

1

where M = m1 + m2 + m3 ..... mN (total mass of system)For a continuous distribution of mass, the coordinates of centre of mass are given by

XM

x dm YM

y dm ZM

zdmCM CM CM= = =∫ ∫ ∫1 1 1; ;

Velocity of centre of mass is given by

vCM = ==

=

=∑

∑

∑m v

m

m v

M

i ii

N

ii

N

i ii

N

1

1

1

Acceleration of centre of mass is given by

aCM = ==

=

=∑

∑

∑m a

m

m a

M

i ii

N

ii

N

i ii

N

1

1

1

Angular velocity : ωθ

=ddt

Angular acceleration : αω

=ddt

Equations of rotational motion w = w0 + at

θ ω α= +021

2t t

w2 – w02 = 2aq


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Torque

τ = ×r F In magnitude t = rF sinq

Angular momentum

L r p= × In magnitude, L = rp sin q

Relationship between torque and angular momentum

i edLdtext. .,

τ =

Moment of inertia : I m ri ii

N

==∑ 2

1

Theorem of perpendicular axis : Iz = Ix + Iy

where, x and y are two perpendicular to axes in the plane and z axis is perpendicular to its plane.

Theorem of parallel axes : I = ICM + Md2

where, ICM is the moment of inertia of the body about an axis passing through the centre of mass and d is the perpendicular distance between two parallel axis.

S.No. Body Axis of rotation Moment of inertia (I)

Radius of gyration (K)

1.Uniform circular ring of mass M and radius R

(i) about an axis passing through its centre and perpendicular to its plane MR2 R

(ii) about a diameter12

2MRR

2

(iii) about a tangent in its own plane32

2MR 32

R

(iv) about a tangent perpendicular to its plane 2MR2 R 2

2.Uniform circular disc of mass M and radius R

(i) about an axis passing through its centre and perpendicular to its plane

12

2MRR

2

(ii) about a diameter14

2MR R2

(iii) about a tangent in its own plane54

2MR 52R

(iv) about a tangent perpendicular to its own plane

32

2MR 32

R

3.Solid sphere of radius R and mass M

(i) about its diameter25

2MR 25

R

(ii) about a tangential axis75

2MR 75

R

4.Hollow sphere of radius R and mass M

(i) about its diameter23

2MR 23

R

(ii) about a tangential axis 53

2MR 53

R


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Relation between torque and moment of inertia

Torque t = Ia

where a is the angular acceleration.Relation between angular momentum and moment of inertia, L = Iw

Kinetic energy of rotational motion, K IR =12

2ω .

Kinetic energy of a rolling body = translational kinetic energy (KT) + rotational kinetic energy (KR)

= + = +

12

12

12

12 2 22

2Mv I MvK

Rω

When a body rolls down an inclined plane of inclination q without slipping its velocity at the

bottom of incline is given by vgh

K

R

=+

2

12

2

where h is the height of the incline.When a body rolls down on an inclined plane without slipping, its acceleration down the

inclined plane is given by ag

K

R

=+

sin .θ

12

2

When a body rolls down on an inclined plane without slipping, time taken by the body to reach

the bottom is given by t

lK

Rg

=

+

2 1

2

2

sinθ

where l is the length of the inclined plane.

GravitationNewton’s universal law of gravitation

F

Gm m

r= 1 2

2

Where, r is the separation between masses of objects m1 and m2.Acceleration due to gravity

g

GM

R= 2 .

Where M and R are the mass and radius of Earth respectively.Relationship between g and G

g

GM

R

G R

RGRe

e

e

ee= = =2

3

2

43 4

3

π ρπ ρ

where Me is the mass of the earth, Re is the radius of the earth and r is the uniform density of the material of the earth.The acceleration due to gravity at height h above the surface of earth is given by

g

GM

R hg

hR

gGM

Rh

e

e e

e

e

=+

= +

=

−

( )2

2

21

For h << Re

\ g gh

Rhe

= −

1 2

The acceleration due to gravity at a depth d below the surface of earth is given by

g

GM

RR d g

R dR

gd

Rde

ee

e

e e= − =

−

= −

3 1( )

5.Solid cylinder of length l, radius R and mass M

(i) about its own axis12

2MRR

2

(ii) about an axis passing through its centre and perpendicular to its own axis

Ml R2 2

12 4+

l R2 2

12 4+

(iii) about the diameter of one of the faces of cylinder

Ml R2 2

3 4+

l R2 2

3 4+

6. Thin rod of length L

(i) about an axis through its centre and perpendicular to the rod

ML2

12L

12

(ii) about an axis through one end and perpendicular to the rod

ML2

3L

3


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At the centre, d = Re \ gd = 0.Gravitational field intensity

EGm

r= − 2

Where, m is test mass.The gravitational field intensity due to spherical shell of radius R and mass M at a point distant r from the centre of the shell is given as follows:

At a point outside the shell i.e. r > R

EGM

r= − 2

At a point on the surface of the shell i.e. r = R

E

GM

R= − 2

At a point inside the shell i.e. r < R, E = 0For solid sphere gravitational field intensity change only at a point inside the sphere i.e., r < R.

EGMr

R= − 3

Gravitational potential : VGM

r= −

The gravitational potential due to a spherical shell of radius R and mass M at a point distant r from the centre of the shell is given as follows:

At a point outside the shell i.e. r > R

VGM

r= −

At a point on the surface of the shell i.e. r = R

VGM

R= −

At a point inside the shell i.e. r < R

VGM

R= −

The gravitational potential due to a solid sphere at a point inside the sphere i.e. r < R

V

GM R r

R= −

−( )32

2 2

3 Relation between gravitational field intensity and gravitational potential

E

dVdr

= −

Gravitational potential energy : UGMm

r= −

Gravitational potential energy of a body of mass m at height h above the surface of the earth is given by

U

GM mR hh

e

e=

−+( )

Gravitational potential energy of a body of mass m on the surface of the earth is given by

UGM m

Rse

e=

−

Orbital speed of satellite, when it is revolving around earth at a height h is given by

v

GMR h

Rg

R hg

GM

Ro

e

ee

e

e

e

=+

=+

=

As 2

Time period of a satellite :

T

R hGM R

R hg

e

e e

e=+

=+

2 23 3π

π( ) ( )

Height of satellite above the earth’s surface

hT R g

Ree=

−2 2

2

1 3

4π

/

Angular momentum of a satellite

L mv r mr

GMr

m GMro= = = [ ] /2 1 2

Kinetic energy of a satellite,

K mvGM mR h

Uo

e

e= =

+=

12

12 2

2( )

| |.

Potential energy of a satellite, UGM mR h

e

e=

−+

.

Total energy (mechanical) of a satellite

E K U

GM mR h

e

e= + = −

+2( )

Escape speed : vGMRe

e

e=

2

properties of soLiDs

Stress restoringarea

=

Longitudinal Stress =FAN

Volumetric Stress =FAV

Tangential Stress =FAT

Longitudinal strain = change in lengthoriginal length

=∆LL

Volumetric strain = change in volumeoriginal volume

=∆VV


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Hooke’s law : Stress = E × Strain or StressStrain

= E

Young’s modulus, Y =normal stress

longitudinal strain

= = =

F AL L

FLA L

FL

r L

//∆ ∆ ∆π 2

Bulk modulus, BF AV V

PVV

= =−

= −normal stress

volumetric strain//

.∆ ∆

BF AV V

PVV

= =−

= −normal stress

volumetric strain//

.∆ ∆

–ve sign shows that volume is decreasing when force is applied.

Modulus of rigidity (h) = tangential stressshearing strain

= =

F A FA

/ .θ θ

In case of a solids and liquids bulk modulus is almost constant. In case of a gas, it is process dependent

In isothermal process, K = Ki = PIn adiabatic process K = Ka = gP

CompressibilityBulk modulus ( )

=1

B

when pressure is applied on a substance, its volume decreases while mass remains constant.

Hence, its density will increases,

′ =−

′ ≈ +

<<ρ

ρρ ρ

11 1

∆∆ ∆

P BP

BP

B/or if

Poisson’s ratio ( P) : lateral strainlongitudinal strain

=−∆∆

r rL L

//

Relations among elastic constants (Y, B, h and s) Y = 3B(1 – 2s), Y = 2h(1 + s),

ση

η=

−+

3 22 6B

B ,

9 1 3Y B

= +η

Breaking force = Breaking stress × Area of cross section of the wire.Every wire is like a spring whose force constant is equal to

K

YAL

KL

= ∝or 1

Work done in a stretched wire,

W = × × ×

12

stress strain volume

= × × = ×12

12

FA

LL

AL F L∆

∆ = × ×12

load elongation

Elastic potential energy stored per unit volume of a stretched wire,

u Y= × × = × ×12

12

2stress strain strain( ) In case of elongation by its own weight,

F (= Mg) will act at centre of gravity of the wire, so that length of wire which is stretched is (L/2).

∴ = = = = ∆L

Mg LAY

MgLAY

gLY

M AL( / )2

2 2

2ρρ

In case of twisting of a cylinder (or wire) of length L and radius r, elastic restoring couple per unit twist is given by

CrL

=πη 4

2 where h is modulus of rigidity of the material of wire.

Interatomic force constant ( k)

k =interatomic force

change in interatomic distance = =F

rYr0

0∆Depression of a beam loaded at the middle by a load W and supported at the ends

δ =

WLYIg

3

48Depression of a cantilever at a free end

δ =WLYIg

3

3properties of fLuiDs

Density (mass)(volume)

, ρ =m

V

Relative density density of a substancedensity of water at

= 4 C°

Pressure PF

AFA

= =thrustarea

( )( )

For a point at a depth h below the surface of a liquid of density r, hydrostatic pressure P is given by P = P0 + hrg

where P0 represents the atmospheric pressure.When a body of density r

B (which may be different

from the density of material of body) and volume V is completely immersed in a liquid of density s, following two forces act on the body :

weight of body W = VrBg acting vertically

downwards through the centre of gravity.


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Buoyant force or upward thrust W′ = Vsg equal to weight of the liquid displaced, acting vertically upwards through the centre of buoyancy.

Depending upon relative magnitudes of above two forces, following three cases are possible:

The density of body is greater than that of liquid (i.e., rB > s). In this situation as weight will be more than upthrust, the body will sink.The density of body is equal to the density of liquid (i.e., r

B = s). In this situation

W = W′ so the body will float fully submerged anywhere in the liquid.The density of body is lesser than that of liquid (i.e. r

B < s). In this situation

W < W′ so the body will move upwards and in equilibrium will float partially immersed in the liquid such that

W = Vinsg or VrBg = Vinsg or Vr

B = Vins

Equation of continuity A1v1 = A2v2

Sur face tension, SFL

= =Force

LengthWork done in forming a liquid drop of radius r, surface tension S is, W = 4pr2S.Work done in forming a soap bubble of radius r, surface tension S is,

W = 2 × 4pr2S = 8pr2S.Work done in increasing the radius of a liquid drop from r1 to r2 is

W S r r= −( )4 22

12π .

Work done in increasing the radius of a soap bubble from r1 to r2 is

W S r r= −( )8 22

12π

When n number of smaller drops of a liquid, each of radius r, surface tension S are combined to form a bigger drop of radius R, then

R = n1/3rThe surface area of bigger drop = 4 pR2 = 4pn2/3r2. It is less than the area of n smaller drops.Work done in breaking a liquid drop of radius R into n equal small drops

W = 4pR2 (n1/3 – 1) S where S is the surface tension.

Excess pressure inside a liquid drop is given by

P

Sr

=2 .

Excess pressure inside a soap bubble is given by

PSr

=4 .

Excess pressure in side an air bubble in a liquid is given by

PSr

=2 .

When an air bubble of radius r is at depth h below the free surface of liquid of density r and surface tension S, then the excess pressure inside the bubble,

PSr

h g= +2

ρ .If r1 and r2 are the radii of curved liquid surface, then excess pressure inside the liquid surface is given by

P S

r r= +

1 1

1 2.

When two soap bubbles of radii r1 and r2 coalesce to form a new soap bubble of radius r, under

isothermal conditions then r r r= +12

22 .

When two soap bubbles of radii r1 and r2 are in

contact with each other and r is the radius of the

interface, then rr r

r r=

−1 2

2 1.

The total pressure inside an air bubble of radius r at a depth h below the surface of liquid of density r is

P P h gSr

= + +02

ρThe rise or fall in a capill ary tube is given by

hSr g

SR g

rR

= = =

2 2cos cosθρ ρ

θ

where q is the angle of contact. According to Newton viscous force ( F) of a liquid between two layers is given by

F Advdx

= − η

where h = coefficient of viscosity of the liquid

Poiseuille’s equation : QPr

lPR

= =π

η

4

8

Rl

r=

84

ηπ

is called liquid resistance.

Stoke’s law : F = 6phrv.

Terminal velocity : vr g

T =−2

9

2( ) .ρ ση

Critical velocity : vK

rc =η

ρ

Reynold number : vN

DN

Dvc

RR

c= =η

ρρ

ηor

Bernoulli’s theorem :


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P gh v+ + =ρ ρ12

2 constant

or P gh v P gh v1 11 1

22 2 2

212

12

+ + = + +ρ ρ ρ ρ

Velocity of efflux v gh= 2Time after which liquid strikes the horizontal surface

tH h

g=

−2( )

Range = = = × −R vt gh h H h2 2 ( )

R H hH

max = =at2

If the hole is at the bottom of the tank, time t taken by the tank to emptied.

t

Aa

H g= 2 /where a is the area of the hole.

thermaL, properties of matter

Relationship between different temperature scales:

T T T T TC F R Ra K−=

−=

−=

−=

−0100

32180

080

460212

273 15100

.

Coefficient of linear expansion of a solid,

α =×

=increase in length

original length rise in temperature∆∆∆L

L TCoefficient of area expansion of a solid,

β =×

=increase in area

original area rise in temperature∆∆A

A TTCoefficient of volume expansion of a solid,

γ =×

=increase in volume

original volume rise in temperature∆∆∆V

V TRelation between a, b and g

αβ γ

= =2 3

.

The specific heat of a substance is given by

sm

QT

=1 ∆

∆The molar specific heat of a substance is given by

C

QT

=1µ

∆∆

Thermal capacity, S = s × m

The latent heat of a substance’s given by LQm

=Principal of calorimetry :

Heat lost by one body = Heat gained by the other.When a bar of length L and uniform area of cross section A with its ends maintained at temperatures T1 and T2, the rate of flow of heat (or heat current) H is given by

HKA T T

L=

−( )1 2

Wiedemann-Franz law : KT

Kσ

= a constant, where = thermal conductivity

where, K = thermal conductivity

and s = electrical conductivity

Thermal resistance of the bar, RL

KAH = .Stefan Boltzmann law : E = sT4

If the body is not a perfectly black body, then

E = esT4

The energy radiated per second by a body of area A = eAsT4

Newton’s law of cooling : dQdt

k T TS= − −( )

Wien’s displacement law : λmT = constant .

Temperature of sun is given by TR S

Rs

=

2

2

1 4

σ

/

.

thermoDynamics

The work done by a gas is W dW PdVV

V

i

f

= = ∫∫ where Vi and Vf are the initial and final volume of

the gas.First law of thermodynamics : DQ = DU + DWEquation of isothermal process PV = constant.

Work done during isothermal process,

W RT

V

Vf

i=

µ ln

; W RT

PP

i

f=

µ ln

Equation of adiabatic process, PV g = constantwhere g = CP/CV.

Work done during adiabatic process,

W

PV P Vi i f f=−

−

( )( )γ 1

; W

R T Ti f=−

−

µ

γ

( )1

Equation of isobaric process VT

= constant.

Work done during isobaric process, W = P(Vf – Vi) = mR(Tf – Ti).

Efficiency of a heat engine,

η = = =−

= −work done

heat absorbedWQ

Q Q

QQQ1

1 2

1

2

11

The coefficient of performance of a refrigerator,

β =heat extracted from the reservoir at low temperature

wT2

oork done to transfer the heat


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= =

−QW

QQ Q

2 2

1 2The efficiency of a Carnot engine is given,

η = −1 2

1

TT

kinetic theory of Gases

Equation of an ideal gas : PV = mRT = kBNT

Boltzmann constant

kR

NBA

=

NA is the Avogadro’s number.

Here, µ = =mM

NNA

where, m is the mass of the gas containing N molecules, M is the molar massEquation of a real gas :

Pa

VV b RT+

− =

µµ µ

2

2 ( )

where, a and b are Van der waals constants

Critical temperature : TaRbC =

827

Critical pressure : Pa

bC =

27 2

Critical volume : VC = 3bAccording to kinetic theory of an ideal gas pressure exerted by an ideal gas is given by

P mn v=13

2

Root mean square speed,

vRTM

k TmB

rms = =3 3

Average speed , vRTM

k TmB= =

8 8π π

.

Most probable speed, vRTM

k TmB

mp = =2 2

.v v vrms mp> >

Average translational kinetic energy of a gas

molecule is E k TB=32

The molar specific heats are given by

C RV ( )rigid diatomic =52

C RP ( )rigid diatomic =72

γ ( )rigid diatomic =75

The mean free path, λπ

=1

2 2n d

osciLLations

Angular frequency w = 2pu =2πT

Veloc ity of a particle in S.H.M. is given by

v A x= −ω 2 2

Acceleration of a particle in S.H.M. is given by

a = –w2x

The kinetic energy of a particle in S.H.M. is given byK m A x= −

12

2 2 2ω ( )

The potential energy of a particle in S.H.M. is

given by, = +12

2 2 2m A tω ω φsin ( )

Total energy of a particle in S.H.M. is given by

E m A=12

2 2ω

Spring pendulum

Tmk

= 2π

The time period of a simple pendulum is given by

T L g= 2π / .

If the length of a simple pendulum is comparable with the radius of earth (Re), then time period T is given by

Tg

L Re

=+

2 11 1

π

If a simple p endulum is suspended in a lift and lift is accelerating downwards with an acceleration a, then its time period is given by

TL

g a=

−2π

For upwards motion, TL

g a=

+2π

For upwards or downwards with constant velocity

v, TLg

= 2π

If a simple pendulum is suspended in a lift and lift is freely falling with acceleration g, then its

time period is given by TL

g g=

−= ∞2π


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If a simple pendulum is suspended in a carriage which is accelerating horizontally with an acceleration a, then its time period is given by

TL

g a=

+2

2 2π

( )

If a simple pendulum is suspended from the roof of a trolley which is moving down an inclined plane of inclination q, then the time period is given by

T

Lg

= 2πθcos

If a simple pendu lum whose bob is of density r oscillates in a non-viscous liquid of density s(s < r), then its time period is given by

TL

g=

−

21

πσρ

Torsional pendulum :

TIC

= 2π

where I is the moment of inertia of the disc about the suspension wire as axis of rotation and C is the restoring torque per unit twist.

CrL

=πη 4

2where r is the radius, L is the length and h is the modulus of rigidity of a wire respectively.The time period of oscillation of a liquid in U-tube is given by

TLg

hg

= =22

2π π

where L = total length of liquid column in a U-tube,h = height of liquid column in each limb of U-tubeAlso h = L/2The time period of oscillation of floating cylinder

in a liquid is given by Tm

A g= 2π

σwhere m is the mass of a cylinder, A is the area of cross section of a cylinder, s is the density of a liquid

or Th

ghg

= = ′2 2πρ

σπ

where h is the height of cylinder of density r and s is the density of a liquid in which cylinder is floating, h′ is the height of the cylinder inside the liquid.Time period of LC oscillations of a circuit

containing capacitance C and inductance L is given by

T LC= 2πIf a wire of length L, area of cross-section A, Young’s modulus Y is stretched by suspending a mass m, then the mass can oscillate with time period

TmLYA

= 2π

If a gas is enclosed in a cylinder of volume V fitted with piston of cross section area A and mass M and the piston is slightly depressed and released, the piston can oscillate with a time period

TMVBA

= 2 2π

waves

Speed, frequency and wavelength relation v = lu

Intensity of a wave : I = 2p2u2A2rv

where u is the frequency, A is the amplitude, v is the velocity of the wave, r is the density of the medium.Energy density of wave

u = 2p2A2u2r where r is the density of the medium.

Wave velocity, vk

=ω

Particle velocity ,

vparticle = dydt

A kx t= − +ω ω φcos( ) = −

ωk

dydx

Particle acceleration, ad y

dty= = −

2

22ω

Relationship between phase difference, path difference and time difference

Phase difference path difference= ×2πλ

Phase difference time difference= ×2πT

Speed of a transverse waves on a stretched string

is given by vT

=µ

where T is the tension in the string, m is the mass per unit length of the string called linear density.Speed of a transverse wave in a solid is given by

v =ηρ

where h is the modulus of rigidity, r is the density of a solid.


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Speed of a longitudinal wave in a medium is

given by vE

=ρ

where E is the modulus of elasticity and r is the density of the medium.

Speed of a longitudinal wave in a metallic bar is

given by vY

=ρ

where Y is the Young’s modulus and r is the density of material of a metallic bar.Spee d of a longitudinal wave in a fluid is given

by vB

=ρ

where B is the bulk modulus and r is density of a fluid.

Newton’s formula : vP

=ρ

Speed of sound in a gas, v v=γ3 rms .

Effect o f temperature : v vt

t = +

0 1

546

where v0 is the speed of sound in the gas at 0°C.Effect of pressure : The speed of sound in a gas is given by

vP RT

M= =

γρ

γ

Speed of sound in gas is independent of the pressure of the gas, provided temperature remains constant.Effect of humidity : With increase in humidity, density of air decreases

vP

=γρ

Vib rations in a stretched string of length L fixed at both ends.

Fundamental frequency

υ

λ µ11 2

12

= = =v v

L LT

For the nth mode, λ n = 2L/n Frequency of nth mode

υλ

υµn

n

v nvL

nnL

T= = = =

2 21 where n = 1, 2, 3, ....

υµp

pL

T=

2,

where p = number of loops.Vibrations of a closed organ pipe

For nth mode, λnL

n=

−4

2( )

Frequency, υλn

n

v v nL

n v= =−

= −( )

( )2 14

2 1 1

υλ1

1 4= =

v vL

Due to the end correction the fundamental frequency of a closed organ pipe is given by

υCv

L ev

L r=

+=

+4 4 0 6[ ] [ . ]

Due to the end correction, the fundamental frequency of an open pipe is given byυO

vL e

=+2 2[ ]

=+v

L r2 1 2[ . ]

Speed of sound in air at room temperature using resonance tube is given by

v = 2u(L2 – L1)

B eat frequency = no. of beats/sec = (u1 – u2) = difference in frequencies.Tuning fork is a source of sound of single frequency and frequency of a tuning fork of arm length L and thickness d in the direction of vibration is given by

υρ ρ

=

= =

d

Lv

d

L

Yv

Y2 2 since

According to Doppler’s effect the apparent frequency heard by the observer is given by

′ =±

υ υ

v vv v

o

s

where vs, vo and v are the speed of source, observer and sound relative to air.

The upper sign on vs (or vo) is used when source (observer) moves towards the observer (source) while lower sign is used when it moves away.If the wind blows with speed vw in the direction of sound, v is replaced by v + vw in the above equation. If the wind blows with speed vw in a direction opposite to that of sound, v is replaced by v – vw in the above equation.

A prac tical and small unit of loudness of sound is decibel (dB). 1 decibel = 1/10 bel.In decibel the loudness of a sound of intensity I is

given by LII

=

10 100

log .

mmm


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Physics for Youfebruaty 2013

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