(c) Point 9 on the string has the greatest downward acceleration (d) Point 2 on the string has a downward velocity and upward acceleration. 1 2 3 8 9 7 6 4 5 Fig. B.3 5. Given are four arrangements of three fixed electric charges. In each arrangement, a point labeled P is also identified—test charge, +q, is placed at point P. All of the charges are of the same magnitude Q, but they can be either positive or negative as indicated. The charges and point P all lie on a straight line. The distances between adjacent items, either between two charges or between a charge and point P, are all the same. I. + + + P II. + + � P III. + + � P IV. + � � P The correct order of choices in a decreasing order of magnitude of force on P is (a) II > I > III > IV (b) I > II > III > IV (c) II > I > IV > III (d) III > IV > I > II 6. Find the equivalent resistance across AB. A B 2 W 2 W 2 W 2 W 2 W Fig. B.4 PAPER 1 Section A: Single Correct Answer Type 1. An elevator is accelerating upwards with an accelera- tion of 6 m/s 2 . Inside it, a person of mass 50 kg is standing on a weighing machine which is kept on an inclined plane having the angle of inclination 60°. The reading of the weighing machine is Weighing machine a = 6 m/s 2 60° Fig. B.1 (a) 40 kg (b) 160 kg (c) 80 kg (d) 50 kg 2. A spool is pulled horizontally by two equal and opposite forces as shown in Fig. B.2. Which of the following statements is correct? F Rough F Fig. B.2 (a) The center of mass moves towards left. (b) The center of mass moves towards right. (c) The center of mass remains stationary. (d) The net torque about the center of mass of the spool is zero. 3. The rate of dissipation of heat by a black body at a temperature T is Q. The rate of dissipation of heat by another body at temperature 2T and emissivity 0.25 is (a) 16Q (b) 4Q (c) 8Q (d) 8.5Q 4. Figure B.3 shows a sinusoidal wave of period T trav- elling to the right along a string at time t = 0. Which of the following statements is incorrect? (a) Point 3 on the string is moving upward with maximum speed. (b) Point 5 on the string has the greatest upward acceleration. Appendix B Mock Test 1 Cengage Learning India Pvt. Ltd.
Transcript
(c) Point 9 on the string has the greatest downward acceleration
(d) Point 2 on the string has a downward velocity and upward acceleration.
12
3
8
9
7
64
5Fig. B.3
5. Given are four arrangements of three fixed electric charges. In each arrangement, a point labeled P is also identified—test charge, +q, is placed at point P. All of the charges are of the same magnitude Q, but they can be either positive or negative as indicated. The charges and point P all lie on a straight line. The distances between adjacent items, either between two charges or between a charge and point P, are all the same.
I. + + +P
II. + + �P
III. + + �P
IV. + ��P
The correct order of choices in a decreasing order of magnitude of force on P is
(a) II > I > III > IV (b) I > II > III > IV (c) II > I > IV > III (d) III > IV > I > II 6. Find the equivalent resistance across AB.
A
B
2 W
2 W
2 W
2 W2 W
Fig. B.4
PaPer 1Section a: Single Correct answer Type 1. An elevator is accelerating upwards with an accelera-
tion of 6 m/s2. Inside it, a person of mass 50 kg is standing on a weighing machine which is kept on an inclined plane having the angle of inclination 60°. The reading of the weighing machine is
Weighing machine
a = 6 m/s2
60°
Fig. B.1
(a) 40 kg (b) 160 kg (c) 80 kg (d) 50 kg 2. A spool is pulled horizontally by two equal and
opposite forces as shown in Fig. B.2. Which of the following statements is correct?
F
Rough
F
Fig. B.2
(a) The center of mass moves towards left. (b) The center of mass moves towards right. (c) The center of mass remains stationary. (d) The net torque about the center of mass of the
spool is zero. 3. The rate of dissipation of heat by a black body at a
temperature T is Q. The rate of dissipation of heat by another body at temperature 2T and emissivity 0.25 is
(a) 16Q (b) 4Q (c) 8Q (d) 8.5Q 4. Figure B.3 shows a sinusoidal wave of period T trav-
elling to the right along a string at time t = 0. Which of the following statements is incorrect?
(a) Point 3 on the string is moving upward with maximum speed.
(b) Point 5 on the string has the greatest upward acceleration.
Appendix BMock Test 1
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B.2 Physics
(a) 1 W (b) 2 W (c) 3 W (d) 4 W 7. In Fig. B.5, a wire carries current I. What is the value
of the B dl·Ú (as in Ampere’s law) on the helical
loop shown in the figure? The integration in done in the sense shown. The loop has N turns and a part of helical loop on which arrows are drawn is outside the plane of paper.
I
Fig. B.5
(a) –m0(NI) (b) m0(I) (c) m0(NI) (d) Zero 8. A flag is mounted on a car moving due north with
the velocity of 20 km/h. Strong winds are blowing due east with the velocity of 20 km/h. The flag will point in direction
(a) East (b) North-East (c) South-East (d) South-West 9. Water coming out of the mouth of a tap of area
of cross-section 2.5 cm2 is falling vertically in a streamline flow with the speed 3 m/s. The area of cross-section of the water column 80 cm below the tap is
Fig. B.6
(a) 0.5 cm2 (b) 1 cm2
(c) 1.5 cm2 (d) 2 cm2
10. A particle is executing S.H.M. A and B are two extreme positions in which its velocity is zero. It passes through a certain point P at intervals of 0.5 s and 1.5 s with a speed of 3 m/s. What is the maximum speed of the particle?
(a) 6 2 m/s (b) 2 2 m/s
(c) 4 2 m/s (d) 3 2 m/s
Section B: Multiple Correct answers Type
11. For a particle moving in straight line with increas-ing speed, the appropriate sign of acceleration a and velocity v can be
(a) a > 0 and v > 0 (b) a < 0 and v < 0 (c) a > 0 and v < 0 (d) a < 0 and v > 0 12. A number of forces of different magnitudes and direc-
tions which are variable in nature are used to move a particle along a smooth curved horizontal path.
(a) The work done on the particle by the resultant force equals the change in the kinetic energy of particle.
(b) No work done is possible on a particle in cir-cular motion.
(c) Work done is possible on a particle in circular motion but it will not be equal to change in KE.
(d) If the speed of particle changes during circular motion, then some work is being done on the particle.
13. A particle of mass m1 = 4 kg moving at 6i m/s col-lides perfectly elastically with a particle of mass m2 = 2 kg moving at 3i m/s.
(a) Velocity of the center of mass (CM) is 5i m/s. (b) The velocities of the particles relative to the
center of mass have same magnitude. (c) The speed of individual particle before and after
collision remains same. (d) The velocity of particles relative to CM after
collision are v i v i1f / cm 2f / cmm/s m/s= - =, 2
14. The graph between the stopping potential (V0) and wave number (1/l) is as shown in Fig. B.7. If f is the work function, then
q q q
f1 f2 f3
V0 Volts
Metal 1 Metal 2 Metal 3
(1/nm)
0.001 0.002 0.004 1/lFig. B.7
(a) f1 : f2 : f3 = 1 : 2 : 4 (b) f1 : f2 : f3 = 4 : 2 : 1 (c) tan q µ hc / e, where q is the slope.
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Appendix B: Mock Tests B.3
(d) Ultraviolet light can be used to emit photoelec-trons from metal 2 and metal 3 only.
15. A wave is transmitted from medium 1 to medium 2 and the respective velocities in the two media are v1 and v2, respectively,
(a) The phase of the transmitted wave does not change if v2 > v1.
(b) The amplitude of transmitted wave is always less than that of the incident wave.
(c) The frequency of the transmitted wave is always equal to that of the incident wave.
(d) The phase of the transmitted wave does not change if v1 < v2.
Section C: Integer answer Type
16. A cubical vessel with a liquid of density r = 103 kg/m3 is kept at rest on an inclined plane of angle of inc-lination q = 37°. If b = 1 m, then find the pressure difference between A and B.
B
A
q
b
Fig. B.8
17. In two experiments with a continuous flow calorim-eter to determine the specific heat capacity of a liquid,
an input power of 16 W produced a rise of 10 K in the liquid. When the power was doubled, the same temperature rise was achieved by making the rate of flow of liquid three times faster. Find the power lost (in W) to the surrounding in each case.
18. A block of mass m = 1 kg is projected on a smooth horizontal floor with a speed v0 = 3 m/s towards a fixed light spring of stiffness k = 16 N/m. The time of contact of the smooth block with the spring is p*
sec. Find the value of ‘*’.
m
v0k
Fig. B.9
19. A string 120 cm in length sustains a transverse standing wave. The points of the string where the displacement amplitude is each 3.5 mm are separated by the distance of 15 cm. Find the overtone in which the string vibrates.
20. In the circuit shown in Fig. B.10, each battery is 5 V and has an internal resistance of 0.2 W. The reading of the ideal voltmeter is V. Find V in volt.
v
Fig. B.10
PaPer 2Section a: One or More Correct answers Type
1. A metallic square loop PQRS is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in Fig. B.11. If VP, VQ, VR and VS are the potentials of points P, Q, R and S then which of the following is an incorrect statement?
(a) VP = VQ (b) VP > VS (c) VP > VR (d) VS > VR
P
S
Q
R
V
Fig. B.11
2. A bomb of mass 3m is kept inside a closed box of mass 3m and length 4L at its center. It explodes in two parts of mass m and 2m. The two parts move in opposite directions and stick to the opposite sides of the walls of box. The box is kept on a smooth horizontal surface.
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4LFig. B.12
What is the distance moved by the box during this time interval?
(a) 0 (b) L6
(c) L12
(d) L3
3. An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. The average power delivered by the source
(a) must be 1000 W (b) may be greater than 1000 W (c) may be less than 1000 W (d) All of the above three are possible. 4. The deviation for a ray at the interface of two media
from denser (1) to rarer (2) with the angle of incidence 30° is 15°. What maximum deviation a ray of same wavelength can undergo at the interface of two media when entering from medium (2)?
(a) 90° (b) 45° (c) 0° (d) 60° 5. Two identical photocathodes receive the light of
frequencies v1 and v2. If the velocities of the photo-electrons (of mass m) coming out are v1 and v2, resp- ectively, then
(a) v v hm1 2 1 2
1 22- = -È
Î͢˚̇
( )/
n n
(b) v v hm1
222
1 22
- = -( )n n
(c) v v hm1 2 1 2
1 22+ = -È
Î͢˚̇
( )/
n n
(d) v v hm1
222
1 22
+ = -( )n n
6. Pick the correct statements: (a) If a point charge is placed off-center inside
an electrically neutral spherical metal shell, then the induced charge on its inner surface is uniformly distributed.
(b) If a point charge is placed off-center inside an electrically neutral, isolated spherical metal shell, then the induced charge on its outer sur-face is uniformly distributed.
(c) A non-metal shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charges were concentrated at the center of the shell.
(d) If a charged particle is located inside a non-metal shell of uniform charge, there is no elec-trostatic force on the particle due to the shell.
7. There are two coils A and B as shown in Fig. B.13.
+ � S
A
B
Fig. B.13
(a) When switch S is closed, the direction of the momentary current induced in coil B will be in anticlockwise direction.
(b) When switch S is closed, the direction of the momentary current induced in coil B will be in clockwise direction.
(c) When switch S is opened, the direction of the momentary induced current in coil B will be in clockwise direction.
(d) When switch S is opened, the direction of the momentary induced current in coil B will be in anticlockwise direction.
8. A small ball of mass m suspended from the ceiling at a point O by a thread of length moves along a horizontal circle with a constant angular velocity w. Which of the followings is/are correct?
l
O
Cmw
Fig. B.14
(a) Angular momentum is constant about O. (b) Angular momentum is constant about C. (c) Vertical component of angular momentum
about O is constant. (d) Magnitude of angular momentum about O is
constant.
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Appendix B: Mock Tests B.5
Section B: Comprehension TypeFor Problems 9–11As shown in Fig. B.15, blocks of masses M/2, M and M/2 are connected through a light string as shown; pulleys are light and smooth. Friction is only between block C and floor. System is released from rest.
q
A
B
C
M
M/2
M/2
tan=2
qm
Fig. B.15
9. Find the acceleration of block B. (a) > g sin q (b) < g sin q (c) g sin q (d) Any of the above can be possible 10. Regarding the accelerations of A and C, we can say
that: (a) accelerations of both will be same (b) acceleration of A will be greater than that of C (c) acceleration of C will be greater than that
of A (d) any of the above is possible. 11. Find the tension in the string. (a) 5 Mg sin q/8 (b) Mg sin q/4 (c) Mg sin q/6 (d) Mg sin q/8For Problems 12–14An infinite cylindrical wire of radius R and having current density varying with its radius r as, J = J0[1 – (r/R)]. Then answer the following questions.
R
y
xO
•
•
Fig. B.16
12. The position where magnetic field strength is maxi-mum is
(a) R (b) 3R/4 (c) R/2 (d) R/4 13. The magnitude of maximum magnetic field is
(a) J R006
m (b) 316 0 0J Rm
(c) 516 0 0J Rm (d) 5
6 0 0J Rm
14. Graph between the magnetic field and radius is
(a)
B
R r
(b)
B
R r
(c)
B
R r
(d)
B
R r
For Problems 15–19An inductor of inductance 3 H is given across which a po-tential difference varying with time is shown in Fig. B.17. At t = 0, current in the inductor is zero.
12
02 4
t s( )
V (volt)
Fig. B.17
15. Area under V–t graph represents (a) total charge flown (b) change in current (c) product of inductance and change in current (d) product of inductance and total charge flown 16. Maximum value of current in the inductor will take
place at time t equal to (a) 4 s (b) 2 s (c) 0 (d) 1 s 17. Find the current in the circuit at t = 3 s. (a) 1 A (b) 3 A (c) 5 A (d) 7 A
Section C: Matching Column Type
18. Heat given to process is positive. Match the following columns I and II.
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30
20
10
10 20
E
H
GF
V (m )3
P (atm)
Fig. B.18
Column I Column II (p) EF (a) DW < 0 (q) FG (b) DW > 0 (r) GH (c) DQ < 0 (s) HE (d) DQ > 0 19. Column I and column II contain four entries each.
Entries of column I are to be matched with entries of column II.
A battery of emf E is connected across a conductor as shown in Fig. B.19. As one observes from A to B, match the following.
AB
E
r1r2
Fig. B.19
Column I Column II (p) Current (a) increases (q) Drift velocity (b) decreases of electron (r) Electric field (c) remains same (s) Potential drop (d) cannot be deter- per unit length mine 20. Match quantities in column I and with those quantities
in column II having same dimensions.Column I Column II
(p) Torque (a) Planck’s constant (q) Angular Momentum (b) Work (r) Latent heart (c) Strain (s) Angle (d) Gravitatinal Potential
Mock Test 2
PaPer 1Section a: Single Correct answer Type
1. A projectile is fired vertically upwards from the sur- face of the earth with a velocity Kve, where ve is the escape velocity and k < 1. If R is the radius of the earth, the maximum height to which it will rise measured from the center of earth will be (neglect air resistance)?
(a) 1 2- kR
(b) Rk1 2-
(c) R (1–k)2 (d) Rk1 2+
2. In the measurement of the focal length f of a concave mirror, the object distance u = 40 ± 0.1 cm, the image distance v = 20 ± 0.2 cm. The maximum % error in the measurement of f is
(a) 1.75 (b) 0.75 (c) 0.3 (d) 2.25
3. A small ball is projected horizontally between two large blocks. The ball is given a velocity V m/s and each of the large blocks move uniformly with a veloc-ity of 2V m/s. The ball collides elastically with the blocks. If the velocity of the blocks do not change due to the collision, then find out the velocity of the ball after the second collision.
Assume friction to be absentFig. B.20
(a) 5V (b) 7V (c) 9V (d) None of these 4. Four forces of the same magnitude act on a square as
shown in Fig. B.21. The square can rotate about point
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O, mid point of one of the edges. The force which can produce greatest torque is
(a) F1 (b) F2 (c) F3 (d) F4
F4
F1O
F3 F2
Fig. B.21
5. The voltage of an AC source varies with time accord-ing to the relation: E = 120 sin 100 pt cos 100 pt V. What is the peak voltage of the source?
(a) 60 V (b) 120 V (c) 30 V (d) V 6. A capacitor having initial charge q0 = CE/2 is con-
nected to a cell of emf E as shown in Fig. B.22. The total heat generated in the circuit after the closing of switch is
E
C
-q0+q0
R
Fig. B.22
(a) (1/2)CE2 (b) (1/8)CE2
(c) (1/4)CE2 (d) None of these
7. A uniform magnetic field B i j k= + +3 4 exists in
region of space. A semicircular wire of radius 1 m carrying current 1 A having its center at (2, 2, 0) is placed in x-y plane as shown in Fig. B.23. The force on semicircular wire will be
y
45°
1 m
(2, 2)
xFig. B.23
(a) 2 ( )i j k + + (b) 2 ( )i j k - +
(c) 2 ( )i j k + - (d) 2 ( )- + +i j k
8. The molar heat capacity C for an ideal gas going through a given process is given by C = a/T, where a is a constant. If g = CP /CV, the work done by one mole of gas during heating from T0 to hT0 through the given process will be:
(a) 1a
ln h
(b) a RTln h hg
- --
ÊËÁ
ˆ¯̃
11 0
(c) a ln h – (g – 1) RT0 (d) None of these 9. A container filled with mercury and having a wooden
block floating in it is allowed to fall freely under gravity. During the fall, the upthrust on the wooden block will be
(a) zero (b) equal to the weight of mercury displaced by the
immersed portion of the block (c) equal to the weight of the block in air (d) equal to the loss of weight of the block in water 10. Flux passing through the shaded surface of sphere
when a point charge q is placed at the center is (radius of the sphere is R)
qR/2
Fig. B.24
(a) q/e0 (b) q/2e0 (c) q/4e0 (d) zero
Section B: Multiple Correct answer Type
11. A box of mass m is released from rest at position 1 on the frictionless curved track shown in Fig. B.25. It slides a distance d along the track in time t to
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reach position 2, dropping a vertical height h. Let v and a be the instantaneous speed and instantaneous acceleration, respectively, of the box at position 2. Which of the following equations is/are not valid for this situation?
m
d
1
2h
Fig. B.25
(a) h = vt (b) h = (1/2)gt2
(c) d = (1/2)at2 (d) mgh = (1/2)mv2
12. Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same. Each is com-pressed till their pressure becomes twice the initial pressure. Then:
(a) if the compression is isothermal, then their final volumes will be same.
(b) if the compression is adiabatic, then their final volumes will be different.
(c) if the compression is adiabatic, then the tri-atomic gas will have maximum final volume
(d) if the compression is adiabatic, then the mono-atomic gas will have maximum final volume.
13. Three simple harmonic motions in the same direction having each of amplitude a and the same period are superposed. If each differes in phase from the next by p/4, then
(a) Resultant amplitude is ( )2 1+ a (b) Phase of resultant motion relative to first is 90°. (c) The energy associated with the resulting motion
is ( )3 2 2+ times the energy associated with any single motion.
(d) Maximum speed of resultant SHM will be more than double of the initial SHMs.
14. In Young’s double slit experiment, the phase differ-ence between the waves at a point on screen having intensity less than the average intensity on screen may be
(a) p4
(b) 23p (c) p (d) 7
8p
15. An elastic metal rod will change its length when it (a) falls vertically under its weight (b) is pulled along its length by a force acting at
one end (c) rotates about an axis at one end (d) slides on a rough surface
Section C: Integer answer Type
16. A particle moves with uniform acceleration along a strainght line AB. Its velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes t1 seconds to go from A to M and t2 seconds to go from M to B. Then find the ratio t1/t2.
17. The Sun’s mass is about 3.2 ¥ 105 times the Earth’s mass. The sun is about 400 times as far from the Earth as the Earth is from the Moon. Assume that the Sun-Moon distance is constant and equal to Sun-Earth distance. Find the ratio of the magnitudes of the gravitational pull of the Sun on the Moon (Fms) and of the Earth on the Moon (Fme).
18. A body rolls on two horizontal plates 1 and 2 moving with velocities 3v and –v, respectively. The distance of instantaneous axis of rotation from the lowest plate is equal to R /*. Find the value of ‘*’.
v 2
1
R
3v
Fig. B.26
19. A point source of sound is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners at x = –8 m and x = +2.0 m. A third listener is positioned along the positive y-axis. What is her y-coordinate (in m) if the same wave front reaches her at the same instant as it does the first two listeners?
20. The inductor in an L–C oscillation has a maximum potential difference of 10 V and maximum energy of 100 mJ. Find the value of capacitor in mF in L–C circuit.
Paper 2
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Appendix B: Mock Tests B.9
PaPer 2Section a: One or More Correct answer Type
1. A parallel plate capacitor of capacitance 200 mF is charged by a battery of emf 100 V. The battery is now disconnected and temperature of the plate is equal to atmospheric temperature. The plates are now connected by a thin wire of negligible heat capacity. Assume 50% of their stored energy inc- reases their temperature till the capacitor gets com-pletely discharged and energy equally distributes over the plates. If thermal capacity of each plate is 0.5 J K–1 and coefficient of linear expansion is 2 ¥ 10–5 °C–1, percentage increase in the volume of the plates is
(a) 0.001% (b) 0.002% (c) 0.003% (d) 0.004% 2. In the circuit shown in Fig. B.27, X is joined to Y for
a long time, and then X is joined to Z. The total heat produced in R2 is
X
L
E R1
Y
Z
R2
Fig. B.27
(a) LER
2
122
(b) LER
2
222
(c) LER R
2
1 22 (d) LE R
R
22
122
3. A helicopter is moving to the right at a constant horizontal velocity. It experiences three forces F Fgravitational drag, and force on it caused by rotor Frotor . Which of the following diagrams can be cor-rect free body diagram representing forces on the helicopter?
Direction of motion
Fig. B.28
(a) Fnet
Fgravitational
Frotor
(b) Fdrag
Fgravitational
ma
Frotor
(c)
Frotor
Fdrag
Fgravitational
(d)
Frotor
Fdrag
Fgravitational
ma
4. A ray of light strikes a cubical slab as shown in Fig. B.29. Then the geometrical path length traversed by the light in the slab will be:
(a) 2 3 (b) 6
(c) 23
(d) 32
32
+
m � (3/2)
2 2m
2 2mair
2m60°
(3 / 2)m =
Fig. B.29
5. For a silver atom, energy corresponding to Ka transi-tion is 21.75 keV. Also the minimum energy of a strik-ing electron so as to produce L X-rays is 3.56 keV. Now, if an electron with an energy of 23 keV strikes a silver target, the characteristic X-rays spectrum will have
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(a) Only Ka and L X-rays (b) Only L X-rays (c) K and L X-rays (d) Only Ka X-rays 6. Two resistances are expressed as (3 ± 0.3) and (5 ±
0.1) W (a) The combined resistance in series is (8W ±
5%) W. (b) The combined resistance in series is (8W ±
12%) W. (c) The combined resistance in parallel is (15/8 +
7.5%) W. (d) The combined resistance in parallel is (15/8 ±
17%) W. 7. Three equal point charges (Q) are kept at the three
corners of an equilateral triangle ABC of side a. P is a point having equal distance a from A, B and C. If E is the magnitude of electric field and V is the potential at point P, then
(a) E Qa
= 34 0
2p e (b) E Q
a= 6
4 02p e
(c) V Qa
= 34 0p e
(d) E Qa
= 3 64 0
2p e 8. Consider a body undergoing S.H.M. Let, at any in-
stant of time, the net force acting on the body be F ,
the displacement of the body be r , the acceleration
of the body be a and its velocity be
v. Which of the following be always negative (throughout the motion).
(a) F r◊ (b)
a r◊ (c) v r◊ (d)
a F◊
Section B: Comprehension TypeFor Problems 9–11
A
V
B
P0P0
C D
h2
h1
Fig. B.30
A tube of uniform cross-section is used to siphon water from a vessel V as shown in Fig. B.30. The pressure over the open end of water in the vessel is atmospheric pressure (P0). The heights of the tube above and below the water level in the vessel are h1 and h2. 9. Determine the velocity vB of the water issuing out
at B.
(a) v ghB = 2 2
(b) v g h hB = +2 1 2( )
(c) v g h hB = -2 2 1( )
(d) v ghB = 2
10. If h2 = 3.0 m, the maximum value of h1, for which the siphon will work, will be
(a) 3.0 m (b) 6.0 m (c) 7.2 m (d) 4.8 m 11. Given h1 = h2 = 3.0 m, the gauge pressure of water
in the highest level CD of the tube will be (a) 3.0 ¥ 104 N/m2 (b) 5.9 ¥ 104 N/m2
(c) 3.9 ¥ 104 N/m2 (d) 1.5 ¥ 104 N/m2
For Problems 12–14A charged ball of mass 9 kg is suspended from a string in a uniform electric field
E i j= + ¥( )3 5 105 N/C. The ball is in equilibrium with q = 37°. Take g = 10 m/s2.
q
q
y
x
Fig. B.31
12. Find the charge on the ball. (a) 10 mC (b) 100 mC (c) 1 mC (d) 200 mC 13. Find the tension in the string. (a) 50 N (b) 100 N (c) 150 N (d) 200 N 14. If the direction of electric field is reversed, find
the new equilibrium position of the ball. Give your answer in terms of angle made by string with verti-cal.
(a) tan- ÊËÁ
ˆ¯̃
1 34
(b) cot- ÊËÁ
ˆ¯̃
1 314
(c) cot- ÊËÁ
ˆ¯̃
1 34
(d) tan- ÊËÁ
ˆ¯̃
1 314
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Appendix B: Mock Tests B.11
For Problems 15–17In a Young’s double slit experiment set up, the source S of wavelength 6000 Å illuminates two slits S1 and S2 which act as two coherent sources. The source S oscillates about its shown position according to the equation y = 1 + cos pt, where y is in mm and t is in seconds. P is taken to be the origin of the co-ordinate system.
S
P1 mm
1 m2 mm
2 m
S1
S2
C
y
x
Fig. B.32
15. At t = 0, fringe width is b1, and at t = 2 s, fringe width of the figure is b2. Then
(a) b1 > b2 (b) b2 > b1 (c) b2 > b1 (d) Data is insufficient. 16. At t = 2 s, the position of central maxima is (a) 2 mm above C (b) 2 mm below C (c) 4 mm above C (d) 4 mm below C 17. At t = 1 s, a slab of thickness 2 ¥ 10–3 mm and the
refractive index 1.5 is placed just in front of S1. The central maxima is formed at
(a) 1 mm above C (b) 1 mm below C (c) 2 mm above C (d) 2 mm below C
Section C: Matching Column Type 18. A disc of radius R is at rest at a point A on a horizontal
surface. A constant horizontal force starts acting at the center. Between A and B, there is sufficient friction to prevent slipping, but there is no friction to the right of B. The disc takes time T to go from A to B. To the right of B:
A B
F
Fig. B.33
Column I Column II (p) Angular acceleration (a) zero of the disc (q) Angular velocity of (b) constant but not the disc zero
(r) Distance covered in (c) is less than that time T during part AB (s) Angle rotated by (d) is more than that disc in time T during part AB 19. A battery of emf E is connected across a conductor
as shown in Fig. B.34. As one observer from A to B. Match the following:
AB
E
r1r2
Fig. B.34
Column I Column II (p) Current (a) increases (q) Drift velocity of (b) decreases electron (r) Electric field (c) remains same (s) Potential drop across (d) cannot be deter- the length mined 20. Each of the two concentric conducting spherical shells
is given a charge q. Match the columns.
R
2R
q q
Fig. B.35
Column I (p) The total energy of the system if none of the
spheres is earthed. (q) The total energy of the system if the outer
sphere is earthed. (r) Now original charges are restored and the inner
sphere is earthed, then the total energy of the system.
(s) If both are earthed. Column II
(a) qR
2
064pe (b) 5
16
2
0
qRpe
(c) qR
2
016pe (d) None of these
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B.12 Physics
Mock Test 3
PaPer 1Section a: Single Correct answer Type 1. A monkey is climbing up a tree at a speed of 3 m/s.
A dog runs towards the tree with a speed of 4 m/s. What is the relative speed of the dog as seen by the monkey?
(a) >7 m/s (b) Between 5 m/s and 7 m/s (c) 5 m/s (d) <5 m/s 2. With what acceleration a sho-uld the box in Fig. B.36
moving up so that the block of mass M exerts a force 7Mg/4 on the floor of the box?
M
a
Fig. B.36
(a) g/4 (b) g/2 (c) 3g/4 (d) 4g 3. A 1-kg mass is projected down a rough circular track
(radius = 2.0 m) as shown in Fig. B.37. The speed of the mass at point A is 3 m/s and at point B, it is 6.0 m/s. How much work is done on the mass between A and B by the force of friction? (Take g = 9.8 m/s2)
90°
2.0 m
A
B
V
Fig. B.37
(a) –7.3 J (b) –8.1 J (c) –6.1 J (d) –24 J 4. Two identical carts constrained to move on a straight
line, on which sit two twins of same mass, are mov-ing with equal velocity. At some time snow begins to drop uniformly. After the snow fall is stopped, Ram, sitting on one of the carts, picks the snow from cart and throws off the falling snow sideways and in the second cart Shyam is asleep.
(a) Cart carrying Ram will have more speed finally than that carrying Shyam.
(b) Cart carrying Ram will have less speed finally than that carrying Shyam.
(c) Cart carrying Ram will have same speed finally that of carrying Shyam.
(d) Depends on the amount of snow thrown. 5. A conducting sphere of radius R and a concentric
thick spherical shell of inner radius 2R and outer radius 3R is shown in Fig. B.38. A charge +10Q is given to the shell and inner sphere is earthed. Then the charge on inner sphere is
R
Fig. B.38
(a) –4Q (b) –10Q (c) Zero (d) None 6. A current-carrying wire is placed in the grooves of
an insulating semi circular disc of radius R, as shown in Fig. B.39. The current enters at point A and leaves from point B. Determine the magnetic field at point D.
30°30°
B
A
DC
R
i i
Fig. B.39
(a) mp
0
8 3i
R (b) m
p0
4 3i
R
(c) 34
0mp
iR
(d) none of these
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Appendix B: Mock Tests B.13
7. Switch S is closed at t = 0 in the circuit shown in Fig. B.40. The change in flux in the inductor (L = 500 mH) from t = 0 to an instant when it reaches steady state is
St = 0 5 W
500 mH
5 W5 W
50 Fm
20 V
Fig. B.40
(a) 2 Wb (b) 1.5 Wb (c) 0 Wb (d) none 8. A cork suspended from the bottom of a container
filled with water with a string as shown in Fig. B.41. If the container accelerates in a horizontal direction towards right, which one is correct?
Fig. B.41
(a) Inclination of string with vertical is tan–1 (a/g) towards left.
(b) Inclination of the string with vertical is tan–1 (a/g) towards right.
(c) Inclination of the string with vertical is p/2 – tan–1 (a/g) towards left.
(d) Inclination of the string with vertical is p/2 – tan–1 (a/g) towards right.
9. The gravitational potential in a region is given by V = 20 (x + y) J/kg. Find the magnitude of gravi-tational force on a particle of mass 2 kg placed at origin.
(a) 40 2 N (b) 40 N (c) Zero (d) 20 N 10. Two moles of helium, four moles of hydrogen and
one mole of water vapor (for water vapors CV = 3R,
where R is gas constant) form an ideal gas mixture. The molar specific heat at constant pressure of the mixture is (neglecting vibrational degrees of freedom)
(a) 73
R (b) 167
R (c) 103
R (d) 237
R
Section B: Multiple Correct answer Type
11. A particle has a rectilinear motion and Fig. B.42 gives its displacement as a function of time. Which of the following statements is/are true with respect to the motion?
O
A
B D
C
Fig. B.42
(a) In the motion between O and A, the velocity is positive and acceleration is negative.
(b) Between A and B, both velocity and acceleration are positive.
(c) Between B and C, the velocity is negative and acceleration is positive.
(d) Between C and D, the acceleration is positive. 12. Figure B.43 shows three blocks on a rough surface
under the influence of a force P of same magnitude in all the three cases. The coefficient of friction is same between each block and ground. What possible relation holds between magnitudes of normal reac-tion and friction forces. (Assume that blocks do not overturn about edge.) Here, fA, fB and fC are frictional forces and NA, NB and NC are reactions.
q
P(a)
qP
(b)
P
(c)
Fig. B.43
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B.14 Physics
(a) NA > NC > NB (b) fA > fC > fB (c) fC > fA = fB (d) NC > NA = NB 13. Two particles of equal mass m are projected from the
ground with speeds v1 and v2 at angles q1 and q2 as shown in Fig. B.44. Given q2 > q1 and v1 cos q1 = v2 cos q2. Which statement/s is/are correct?
m m
v1 v2
q1 q2(1) (2)
Fig. B.44
(a) The center of mass of particles will move along a vertical line.
(b) The center of mass of particles will move along a line inclined at some angle with vertical.
(c) The particle (1) will be above the center of mass level when both particles are in air.
(d) The particle (2) will be above the center of mass level when both particles are in air.
14. In the circuit shown in Fig. B.45, E1 and E2 are two ideal sources of unknown emfs. Some currents are shown. Potential difference appearing across 6-W resistance is VA – VB = 10 V.
2.00 A
3.00 AB
4.00 W
3.00 W3.00 W 6.00 W4.00 W
R
AE2E1
Fig. B.45
(a) The current in the 4.00-W resistor is 5 A. (b) The unknown emf E1 is 36 V. (c) The unknown emf E2 is 54 V. (d) The resistance R is equal to 9 W. 15. In Young’s double slit experiment, two wavelengths
of light are used simultaneously where l2 = 2l1. In the fringe pattern observed on the screen,
(a) maxima of wavelength l2 can coincide with minima of wavelength l1.
(b) fringe width of l2 will be double that of fringe width of l1 and nth order maxima of l2 will coincide with 2nth order maxima of l1.
(c) nth order minima of l2 will coincide with 2nth order minima of l1
(d) nth order minima of l2 will coincide with (2n – 1)th order maxima of l1
Section C: Integer answer Type 16. In a finite square grid, each link having resistance r
is fitted in a resistance-free conducting circular wire. Calculate x if the equivalent resistance between A and B is 21r/8x.
B
A
Fig. B.46
17. In the figure a conducting rod AD makes contact with the metal rails BA and CD which are 50 cm apart in a uniform magnetic field of induction 0.5 Wbm-2 perpendicular to the plane of the diagram. The total resistance of the circuit ABCD is 0.02 W.
C
B A
D
Fig. B.47
Work done in 10 seconds is found to be 100x J. Calculate x.
18. A concave mirror with its optic axis vertical and mirror facing upward is placed at the bottom of the water tank. The radius of curvature of mirror is 40 cm and refractive index for water m = 4/3. The tank is 20 cm deep and if a bird is flying over the tank at a height 60 cm above the surface of water, the position of image of a bird is at v = 1.25y cm. Calculate y.
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Appendix B: Mock Tests B.15
19. Light of intensity I falls along the axis on a perfectly reflecting right circular cone having semi-vertical angle q and base radius R. If E is the energy of one photon and c is the speed of light, calculate b if the force acting on the cone due to light is found to be
b p qR Ic
2
21 2( cos )- .
20. A wheel A is connected to a second wheel B by means of inextensible string, passing over a pulley C, which rotates about a fixed horizontal axle O, as shown in Fig. B.48. The system is released from rest. The wheel A rolls down the inclined plane OK, thus pull-ing up the wheel B which rolls along the inclined plane ON.
a b
OA B
C
NK
Fig. B.48
Determine the velocity (in m/s) of the axle of the wheel A, when it has travelled a distance s = 3.5 m down the slope. Both wheels and the pulley are ass-umed to be homogeneous disks of identical weight and radius. Neglect the weight of the string. The string does not slip over C.
[Take a = 53° and b = 37°]
PaPer 2Section a: One or More Correct answer Type
1. Suppose there is an, empty box in which a block of mass m is hanging in equilibrium with the help of a vertical spring of stiffness K. Now the box is moved downwards in water (rw = density of water) with a constant acceleration a by applying a vertically downward force F as shown in Fig. B.49. The volume of the box is V. The time period of oscillation of the block in the frame of the box is
M
K
F
Water
Fig. B.49
(a) 2pr
K FM Vgw
( )( )
(b) 2p rK VgM F
w( )( )
(c) 2pr
K F MgM Vg Mgw
( )( )
++
(d) None of these 2. An observer (O) and source (S) move horizontally
with speeds v1 and v2 as shown in Fig. B.50. v3 is the velocity of sound wave. At t = 0, the horizontal separation between O and S is L. The wave pulse from the source S at t = 0 reaches the observer at time [for v1 < v2 < v3]
O SV1 V2
LFig. B.50
(a) t = Lv2
(b) t = Lv1
(c) t > Lv3
(d) t < Lv3
3. A plano convex lens has thickness of 4 cm and radius of curvature of its curved surface is 30 cm. Consider two cases:
(i) the rays parallel to the principal axis are incident on plane surface
(ii) the rays parallel to the principal axis are incident on curved surface
Which of the following is correct? (a) The rays in both the cases will converge at
same distance from the surface on which rays are incident.
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B.16 Physics
(b) To calculate the focal length of lens,
1f
= (m – 1) 1 1
1 2R R-
È
ÎÍ
˘
˚˙ is applicable.
(c) To calculate the position of image 1 1 1f u v
= -
should be used. (d) None of these. 4. In a Coolidge tube, the potential difference used to
accelerate the electrons is increased from 12.4 kV to 24.8 kV. As a result, the difference between the wavelength of Ka-line and minimum wavelength increases two-fold. The wavelength of the Ka line is
(a) 1 Å (b) 0.5 Å (c) 1.5 Å (d) None of these 5. A Vernier caliper with a least count of 0.01 cm was
used to measure diameter of a cylinder as 4 cm and a scale (0–15 cm) with the least count of 1 mm was used to measure a length of 5 cm. The % error in the volume of the cylinder is
(a) 3.0 (b) 4.0 (c) 5.0 (d) 2.5 6. When current (I) in R-L series circuit becomes cons-
tant, where L is a pure inductor, which of the follow-ing given statements is/are correct?
RI L
Supplyvoltage
Fig. B.51
(a) Voltage across R is RI. (b) Some part (not 100 %) of the energy supplied by
the battery will be dissipated in R and remaining will continue to store in L.
(c) Voltage across L is equal to zero.
(d) Magnetic energy stored is 12
2LI .
7. Consider the rays shown in Fig. B.52 as paraxial. The image of the virtual point object O formed by the lens LL is
(a) Virtual (b) Real
(c) Located below the principal axis (d) Located left of the lens
L
L
F
O(Principal axis)
(First principal focus)
Fig. B.52
8. Two strings of lengths L1 and L2 and mass per unit length 4m and m, respectively, are joined and tied horizontally by two fixed vertical walls as shown in Fig. B.53. A standing wave is created and the “junc-tion” is found to be a node. If L1 = L2, then
L1 L2
m2m1
Fig. B.53
(a) The minimum number of loops in L1 is 2. (b) The minimum number of loops in L1 is 1. (c) The wavelength in L1 and L2 are different. (d) The frequency in L1 and L2 are same.
Section B: Comprehension Type
For Problems 9–11Two plane harmonic sound waves are expressed by the equations:
y1(x, t) = A cos (0.5 px – 100 pt) y2(x, t) = A cos (0.46 px – 92 pt)
(All parameters are in MKS): 9. How many times does an observer hear maximum
intensity in one second? (a) 4 (b) 10 (c) 6 (d) 8 10. What is the speed of the sound? (a) 200 m/s (b) 180 m/s (c) 192 m/s (d) 96 m/s 11. At x = 0, how many times the amplitude of y1 + y2 is
zero in one second? (a) 192 (b) 48 (c) 100 (d) 96
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Appendix B: Mock Tests B.17
For Problems 12–14A parallel beam of light falls on a solid transparent sphere.
A
Fig. B.54
12. Which is correct? (a) If the beam is thick, then the whole beam can
be focussed at A. (b) The whole beam can be focussed at A only if
the beam is thin enough (c) If the beam is thin, then the beam cannot be
focussed before A. (d) None of these 13. If however thin beam is focussed at A, then find the
refractive index of the sphere; (a) 1.5 (b) 1.7 (c) 2.0 (d) 2.5 14. For what value of refractive index µ, the thin beam
can be focussed at centre of sphere. (a) 1.5 (b) 2 (c) 2.5 (d) none of theseFor Problems 15–17The plates each of area A of a parallel plate capacitor are given charges Q and –Q. The plates are joined by a non-conducting spring of force constant k. The natural length of the spring is d, the initial separation between the plates. The left plate is connected to a vertical wall through a massless non-conducting rope and the right plate is connected to a block of mass m through similar rope. Assume the pulley to be massless, neglect dielectric effect of the spring and Q2 < 2mAe0g.
m
d
Wall Q – Qk
Smooth table
Fig. B.55
Read the paragraph carefully and answer the following questions. 15. If the block is released from rest, the maximum
elongation of the spring is
(a) 2 2
0kmg Q
A-
Ê
ËÁˆ
¯̃e (b) 2 2 2
0kmg Q
A-
Ê
ËÁˆ
¯̃e
(c) 22
2
0kmg Q
A-
Ê
ËÁˆ
¯̃e (d) None of these
16. If T be the time period of oscillation, then
(a) T > 2p mk
(b) T < 2p mk
(c) T = 2p mk
(d) T will depend on Q.
17. If Q2 = 2mAe0g, the elongation of the spring is
(a) Zero (b) 22
2
0kQA
mge
-Ê
ËÁˆ
¯̃
(c) 2 22
0kmg Q
A-
Ê
ËÁˆ
¯̃e (d) None of these
Section C: Matching Column Type
18. A uniform rope of mass m and length l hangs vertically from a rigid support. A transverse pulse of wave-length l0 is produced at the lower end. Here the speed of the pulse is v0. At any time t, the pulse speed is v and wavelength of the pulse is l. ThenColumn I Column II
(p) Pulse speed v along (a) Constant the string at any time. (q) Acceleration of (b) Variable pulse at any time along the string. (r) Frequency of the (c) Depends on wave pulse when it travels length along string. (s) Tension along the (d) Independent of length of string. wavelength
Fig. B.56
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B.18 Physics
19. A cyclic process for an ideal gas is shown in Fig. B.57. DQ is the heat supplied to gas and DW is the work done by the gas. Match the columns.
PB C
DAT
Fig. B.57
Column I Column II (p) During AB (a) DQ > 0 (q) During BC (b) DQ < 0 (r) During CD (c) DW > 0 (s) During DA (d) DW < 0 20. The axis of a hollow cone shown in Fig. B.58 is
vertical. Its base radius is R. It is kept in a uniform electric field E parallel to its axis.Column I Column II
(p) Magnitude of flux (a) pR2E through base of cone
(q) Magnitude of flux (b) pR2E/2 through curved part of cone (r) Magnitude of flux (c) Zero through curved part MNQP of cone (s) Net flux through (d) non-zero entire cone
h/4
P
M
Qh
N
h/4 R
E
E
Fig. B.58
Solutions of Mock Test 1
PaPer 1Section a 1. Ans (a): N – mg cos 60° = ma cos 60° N = m cos 60° (a + g) = 400 N
60°a 60°
a cos 60°N
Fig. S.1
Reading will be 40 kg. 2. Ans (b): The spool is having the tendency to rotate
in clockwise direction. The point of contact of spool with ground will move in backward direction; fric-tion will act in forward direction. Hence, the center of mass will move in forward direction and the net torque about the center of mass is in anticlockwise sense.
3. Ans (b): By Stefan’s law, the rate of dissipation of heat by a black body, Q = sT4.
Rate of dissipation of heat by a body of emissivity e, Q¢ = esT ¢4 When T ¢ = 2T and e = 0.25, Q¢ = 0.25 ¥ s ¥ (2T)4
= 0.25 ¥ 16 ¥ sT4 = 4sT4 = 4Q 4. Ans (d): Displacement y (x, t) = A cos (kx – wt) At t = 0, point 1 on the string has maximum displace-
ment Velocity vy (x, t) = wA sin (kx – wt)
1
2
3
4
5
6
7
8
9vy
vy
ayay
ayvy ay
ayay
vy
vy
x
vy = 0
ay = 0
vy = 0
Fig. S.2
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Appendix B: Mock Tests B.19
At t = 0, vy is maximum at kx = p/2.
x = plp
l2 2 4¥
=
i.e., point 3 on the string. Hence, (a) is correct. Acceleration, ay = (x, t) = – w2A cos (kx – wt) At t = 0, ay is the maximum downward acceleration. If cos kx = 1 fi kx = 2 p fi x = l. i.e., point 9 on the string has maximum downward
acceleration. Hence, (c) is correct. At t = 0, kx = –p, x = l/2, i.e., the string has maximum
upward acceleration. Hence, (b) is correct. Velocity vy is positive and acceleration ay is negative
when 0 < kx < p/2. vy is negative and ay is positive p < kx < 3p/2. vy and ay are positive when p/2 < kx < p. vy and ay are negative when 3p/2 < kx < p. Hence, (d) is correct. 5. Ans (c): In (I) and (II), forces on P due to all three
charges are in same directions, but in case (II) dis-tances are minimum.
Hence forces (II > I). In case (III) force on P will be towards left and in
(IV), force will be towards right. But on calculation forces (IV > III).
6. Ans (a): Fig. B.4 can be redrawn as:
A
B
2 W
2 W
2 W
2 W
2 W
Fig. S.3
7. Ans (c): Here B (due to I) and
dl will be in same
direction.
So, B dl◊Ú = B dl cos 0∞Ú
= B dlÚ = B dl BN r=Ú 2p = N(B2pr) = Nm0I
8. Ans (c): v j v if w= =20 20,
Velocity of wind w.r.t. flag:
v v v i jw f w f/ = - = -20 20
-20S
20E
/W FV
Fig. S.4
9. Ans (c): A1 = 2.5 cm2, v1 = 3 m/s, h = 0.8 m, g = 10 m/s2
v22 = v1
2 + 2gh = 25 fi v2 = 5 m/s
A1v1 = A2v2 fi A2 = A vv1 1
2 = 1.5 cm2
10. Ans (d): According to the problem, the particle is at P when t = 0.5 s and once again at P when t = 0.5 s. Let us assume that particle moves from P to A and back from A to P in 0.5 s; from P to B and back from B to P in 1.5 s.
P O BA
1/4
(Mean position)Fig. S.5
This implies that T = 0.5 + 1.5 = 2 s
w p p= =2T
Let OP = x, where O is the mean position. Then to go from O to A or from O to B, time taken
is 12
s. As per the problem, time taken for the particle
to go from P to A is 14
s. Hence time taken to go
from O to P is t sO PÆ = - =12
14
14
.
We have x = a sin wt.
When the particle is at P, x = a acos p ¥ÊËÁ
ˆ¯̃
=14 2
Now, vp = w pa x a a2 2 22
2- = -
i.e., 3 = p a2
a = 3 2p
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B.20 Physics
Now, vmax = aw = 3 2 3 2p
p¥ = m/s
Section B
11. Ans (a, b): For increasing speed, velocity and ac-celeration both should be in same direction, either both towards right or both towards left.
12. Ans (a, d): If any work is done on the particle (work will be done by tangential forces and not by centrip-etal forces), it will go in the form of KE. If work is done on the particle, then its speed will change.
13. Ans (a, d):
m1 = 4 kg6 m/s
m2 = 2 kg3 m/s
m2m1
v1 v2
Fig. S.6
vcm = 4 6 2 34 2
5¥ + ¥+
= m/s
u1/cm = u1 – vcm = 6 – 5 = 1 m/s u2/cm = u2 – vcm = 3 – 5 = 2 m/s Hence (b) is incorrect. After collision: v1 = 4 m/s, v2 = 7 m/s v1/cm = v1 – vcm = 4 – 5 = –1 m/s v2/cm = v2 – vcm = 7 – 5 = 2 m/s 14. Ans (a, c):
f f fl l l l l l1 2 3
1 2 3 1 2 3
1 1 1 1 2 4: : : : : : : := = =hc hc hc
o o o o o o
From Einstein’s photoelectric equation
hc eV V hce es sl
fl
f= + fi = -
Therefore, slope tan q = hce
1 0 0011
1
lo= -. nm
lo1 = 1000 nm = 10000 Å; 1
2lo
= 0.002 nm–1
fi lo2 = 500 nm = 5000 Å; 1
3lo
= 0.004 nm–1
fi lo3 = 250 nm = 2500 Å
Hence, UV light can be used to eject photoelectrons from all the metals 1, 2 and 3.
15. Ans (a, b, c, d): The amplitude of the transmitted
wave is given by a a vv vt i=
+ÊËÁ
ˆ¯̃
2 2
2 1
Note that the transmitted wave is always in phase with the incident wave.
at /ai < 1 only when v1/v2 < 1. Otherwise, at may be more than ai.
The frequency of the incident, reflected and transmit-ted waves are always equal.
Section C
16. Ans (6): The pressures at A and B are PA = P0 + rgy1 PB = P0 + rgy2
B
A
q
q
y2
h1
h1
y y2 1
–
y1q
Fig. S.7
Then, PB – PA = P0 + rgy2 – (P0 + rgy1) = rg (y2 + y1) where y2 – y1 = AB sin q = b sin q Then, we have PB – PA = rgb sin q 17. Ans (8): Let the power lost to surrounding is Q.
Therefore,
16 – Q = dmdt
SÊËÁ
ˆ¯̃
( )10
and 32 – Q = 3 10dmdt
SÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙( )
\ 3216
3--
=QQ
fi Q = 8 W
18. Ans (4): Since time of contact is equal to time of half cycle, therefore
T = p pmk
=4
19. Ans (3): The length of the string is 1.20 m. If V be the velocity of transverse waves in the string, the fun-
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Appendix B: Mock Tests B.21
damental frequency is f = V2 1 20¥ .
. The ends of the
string are nodes and the string may vibrate in any of its higher harmonics. If it vibrates in the nth harmonic, the wave length of this harmonic is ln = 2.40/n, as the string will vibrate with n loops. Figure S.8, shows one of the loops and the data given in the problem. P and Q represent the space point of string where amplitude is 3.5 mm with PQ = 15 cm. It is required to find a, the amplitude at the antinode.
3.5 mm 3.5 mm
P 15 cm A Q
a
N N
Fig. S.8
Since PQ = 15 cm, N1N2 = ln
2 > 15 cm
Hence, ln > 30 cm
Hence, the string vibrates in one of the first four har-monics, which have wavelengths 240 cm, 120 cm, 80 cm and 60 cm, respectively. From symmetry, from the figure, we see that N1P = PA = AQ = QN2 = 7.5 cm so that N1N2 = 30 cm = ln /2. Hence, ln = 60 cm, i.e., the string vibrates in its fourth harmonic. Also if the equation to the loop is y = a cos kx with x = 0 at the antinode A, we see that
Y = 3.5 a cos kx = a a acos cos28 4 2
pl
l p◊ = =
Hence, a = =( ) ( . )2 3 5 5 mm
The string vibrates in its third overtone. 20. Ans (0): There are eight batteries. Let the current in the circuit is I,
Then I = 8 58 0 2
¥¥ .
= 25 A
PD across voltmeter = potential difference across rightmost battery = E – Ir = 5 – 25 ¥ 0.2= 0
PaPer 2Section a
1. Ans (d): Since the entire loop is in magnetic field, so no net emf will be induced in the loop. Hence, no current will flow in the loop.
P
S R
Q
e e
Fig. S.9
So VP = VQ, VS = VR, VP > VS, VP > VR, VQ > VR. 2. Ans (d): Let the box moves towards right by a dis-
tance x, then the displacement of 2m is x + 2L and displacement of m is x – 2L towards right.
4L
m 3m x
Fig. S.10
Dxcm = m x m x m xm m m
1 1 2 2 3 3
1 2 3
D D D+ ++ +
fi 0 = 3mx + 2m(x + 2L) + m(x – 2L)
fi x = - L3
3. Ans (c): Pav = EvIv cos f = 100 ¥ 10 cos f = 1000 cos f Watt So depending upon the value of f, Pav can be equal
to or less than 1000 W. 4. Ans (b): Angle of refraction = Angle of incidence +
Angle of deviation fi r = 30° + 15° = 45° m1 sin 30° = m2 sin 45°
fi mm
1
2
1 21 2
2= =//
; mm
2
1
12
=
When ray enters from medium (2) for maximum deviation, angle of incidence ª 90°.
\ m2 sin 90 = m1 sin q fi q = 45° Therefore, maximum deviation = 45°
5. Ans (b): h h mvn n m1 0 121
2= +
h h mvn n2 0 212
= +
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B.22 Physics
fi h m v v( ) ( )n n1 2 12
221
2- = -
fi v v hm1
222
1 22- = -( )n n
6. Ans (b, c, d): Concept based. 7. Ans (a, c): According to Lenz’s law, current in coil
A grows in clockwise direction (when switch S is closed). This causes increase in magnetic flux linking coil B in clockwise direction. Therefore, current is in-duced in coil B in anticlockwise direction. Similarly, when switch is opened the magnetic flux linked with coil B decreases. The direction of induced current in coil B will be clockwise.
8. Ans (b, c, d): Angular momentum about O: L0 = lmv, as l keeps changing direction, so L0 also
changes in direction but its magnitude remains con-stant. Its vertical component also remains same.
l
O
Cm
Lc
Lo
Fig. S.11
LC = rmv, always perpendicular to r, i.e., always upwards and hence constant.
Section B
9. Ans (c) 10. Ans (a) 11. Ans (d): Here the acceleration of A and C will be
same; let it be a. The acceleration of B will be inde-pendent of A and C; let it be b.
For B: T + Mg sin q – T = Mb fi b = g sin q
A
B
M
TT
TT
M/2
M/2
a
b
aC
q
Fig. S.12
For A: M g T M a2 2
sin q - =
For C: T M g M g M a+ - =2 2 2
sin cosq m q
Solve to get T = Mg sin q/8 12. Ans (b) 13. Ans (b)
14. Ans (d): Current within radius r: I J dAr
= Ú0
fi I J rR
r dr J r rR
r= -È
Î͢˚̇
= -È
ÎÍÍ
˘
˚˙˙Ú 0
00
2 31 2 2
2 3p p
Apply ampere’s law: B2pr = m0I
B r J r rR
B J r rR
2 22 3 2 30 0
2 3
0 0
2p m p m= -
Ê
ËÁˆ
¯̃fi = -
È
ÎÍÍ
˘
˚˙˙
To find max B: dBdr
r R= fi =0 3 4/
B J R RR
R Jmax/ ( / )= -
È
ÎÍÍ
˘
˚˙˙
=m m0 0
2
0 03 4
23 4
3316
15. Ans (c): Area under V–t graph.
t dt
dA
t
V
Fig. S.13
dA = Vdt
dA L dIdt
dt V L dIdtÚ Ú= Ê
ËÁˆ¯̃
=ÊËÁ
ˆ¯̃
Q
A L dI L I II
I
= = -Ú1
2
2 1( )
16. Ans (a): Given I1 = 0
0 2 4t
V
Fig. S.14
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Appendix B: Mock Tests B.23
I2 is maximum if the area is maximum. Maximum area will be at 4 s.
17. Ans (d): A1 = 21
0 2 4 t
V
A1
3
12
Fig. S.15
I AL2
1 213
7= = = A
Section C 18. (p – c) (q – b, d) (r – d) (s – c, a) W is positive during expansion and negative during
compression. 19. (p – c) (q – b) (r – b) (s – b) Current through any cross-section should remain
same as all the cross-sections are in series. I = neAvd. As area increases, so drift velocity dec-
reases.
E is directly proportional to vd, so E also decreases. If E decreases, then PD across a segment of fixed
length also decreases. It can also be done like this
dxx
dR
I
Fig. S.16
dR dxA
= r ; A increases with x.
Potential across dx: dV I dR I dxA
= = r
E dVdx
IA A
= = µr 1
So E and dV/dx both decrease as A increases. 20. (p – b) (q – a) (r – d) (s – c) Theory based.
Solutions of Mock Test 2
PaPer 1Section a
1. Ans (b): If a body is projected from the surface of the earth with a velocity v and reaches a height h, then according to the law of conservation of energy,
12 1
2mv mghh R
=+ /
Here v = kve = k gR2
12
31
2mk gR mg r Rr R
R
. ( )( )= -
+ -
k R r RR
r R2 1+ -ÈÎÍ
˘˚̇
= - or k2r = r – R
or r = Rk1 2-
2. Ans (a): f uvu v
uv u v=+
= + -( ) 1
df v u v uu v
du= + -+
È
ÎÍ
˘
˚˙-( )
( )1
2
+ + -+
È
ÎÍ
˘
˚˙-u u v v
u vdv( )
( )1
2
dff
duu
duu v
dvv
dvu v
= -+
+ -+( ) ( )
Since all errors add up with positive sign, hence
dff
= 0 140
0 160
0 220
0 260
. . . .+ + +
= 0 140
0 320
0 740
. . .+ =
or dff
ÊËÁ
ˆ¯̃
= ¥ =100 0 740
100 1 75. . %
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B.24 Physics
3. Ans (c): Use the following result:
u1
O
v1
O
v0 v0
Before collision After collision
Fig. S.17
e v vu u
=--
2 1
1 2
fi 1 20 1
1 01 1 0=
- - -- -
fi = +v v
u vv u v( )
( )
Using the above result: Now velocity after first collision, V¢ = V + 2(2V) = 5V Velocity after second collision, V≤ = V¢ + 2(2V) = 5V + 4V = 9V 4. Ans (c): Perpendicular distance of F3 is greatest from
O, hence torque of F3 is greatest. 5. Ans (a): E = 120 sin 100 pt cos 100 pt = 60 sin (200 pt) So, peak voltage E0 = 60 V 6. Ans (b): Final charge on capacitor qf = CE Charge flown through battery:
q q q CEf= - =0 2
Wb = DU + Heat
qE qC
qC
= -Ê
ËÁˆ
¯̃+
202
2 2Heat
fi Heat =18
2CE
7. Ans (b): Force on semi-circular wire will be same as the force on straight wire AC.
A
y
C
2, 2 R = 1 m
xFig. S.18
l AC R i j i j= = ∞ + ∞ = +2 45 45 2(cos sin ) ( )
F Il B= ¥
= ¥ + ¥ + +1 2 3 4( ) ( )i j i j k
= - +2 [ ]i j k
8. Ans (b): Q CdT a TT
aT
T
= = =Ú0
0
0
0
hh
hln ln
DU = CVDT = Rr
T-
-1
1 0( )h
W Q U ar
RT= - = ---
È
ÎÍ
˘
˚˙D ln h
h 11 0
9. Ans (a): Upthrust = Weight of fluid displaced. As g = 0, upthrust = 0. 10. Ans (c): a = 60° , Solid angle subtended by BCD: w = 2p (1 – cos a) = p
A E
B D
C
a2R
Fig. S.19
Solid angle subtended by ABDE: w(ABCDE) – w(BCD) = 2p – p = p
Hence, flux through ABDE: fe
pp e
= =q q
0 04 4
Section B
11. Ans (a, b, c): Here, the acceleration of block will be changing both in magnitude and direction, so we cannot apply the kinematical formulae of constant acceleration.
We can apply the conservation of energy, because there is no dissipating force (such as friction) which decreases the energy of system.
12. Ans (a, b, d): Pi = P Pf = 2P Vi = V Vf = ?
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Appendix B: Mock Tests B.25
For adiabatic process PV3 = (2P) v3f
fi vf = 12
1 3ÊËÁ
ˆ¯̃
/
for v is maximum, vf is minimum
For isothermal process, PV = (2P)vf fi vf = v/2 13. Ans (a, c, d): A = a + 2a cos 45° = ( )2 1+ a Energy is proportional to the square of amplitude Maximum velocity is proportional to amplitude.
p/4p/4
a
a
aA
Fig. S.20
14. Ans (b, c, d): Let maximum intensity be I0
I = I0 wt Df2
Iav = I0
2
I < Iav fi cos2 Df2
12
ÊËÁ
ˆ¯̃
<
fi cos Df2
12
ÊËÁ
ˆ¯̃
<
fi pf
p2
32
< <D
15. Ans (b, c): In options (a) and (d), there will be no tension in the rod.
But in options (b) and (c), roll will be under tension. Hence, length will increase.
Section C
16. Ans (2): v as2 22 22
= +
2 m/s v 14 m/sA B
MS/2t1
S/2t2
Fig. S.21
fi v2 – 4 = as (1)
14 22
2 2- =v as
fi 196 – v2 = as (2) From (1) and (2), we get v2 – 4 = 196 – v2 fi v = 10 m/s (3)
Now t va a a1
2 10 2 8=
-=
-=
t va a a
tt21
2
14 14 10 4 12
=-
=-
= fi =
17. Ans (2): Let m = Moon’s mass, Ms = Sun’s mass Me = Earth’s mass rms = the center-to-center distance from the Sun to
the Moon rme = the center-to-center distance from the Earth to
the Moon Fms = gravitational force exerted on the moon by the Sun Fme = gravitational force exerted on the moon by the
Earth
Then FF
M rM r
ms
me
S me
e ms=
2
2
Putting the values,
FF
MM
rr
ms
me
S
e
me
ms=
ÊËÁ
ˆ¯̃
ÊËÁ
ˆ¯̃
2
= ¥ ÈÎÍ
˘˚̇
= =3 2 10 1400
3216
252
.
18. Ans (2): Let the required distance is x. Then, w = v/x = 3v/(2R – x). Hence, x = R/2.
19. Ans (4): t = 8 2+=
-xc
xc
2 – x
x 2– 8
8 + x
Fig. S.22
fi –6 = 2x fi x = –3m fi t = 5c
y
� 3
Fig. S.23
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B.26 Physics
5 32 2+ y fi y = 4
20. Ans (2): QC
2
2 = 100 (1)
QC
= 16 (2)
From (1) and (2),
Q2
100 1010
6
=¥ -
fi Q = 20 ¥ 10–6 C
\ C = Q10
= 2010
mF = 2 mF
PaPer 2Section a
1. Ans (c): 2 ¥ S ¥ DT = 12
12
2CVÊËÁ
ˆ¯̃
,
where S = 0.5 J K–1 (for each plate) \ DT = 0.5 ºC
2. Ans (a): When X is joined to Y, current in L:I = E/R1
Energy stored in L: U LI E LR
= =12 2
22
12
This whole energy will be dissipated as heat after X is connected to Z.
3. Ans (c): Fgravitational acts downwards.
Fdrag acts opposite to the direction of motion.
Now let us see how Frotor acts.
rotorF
(a) (b)
Fig. S.24
Wings of helicopter will push the air perpendicular to their plane of rotation as shown in Fig. S.24(a). From third law, air will apply force
Frotor on the wings as
shown in Fig. S.24(b).
There is no force ma. Basically, ma is the resultant of all the forces acting on a body. Moreover, here a is zero because the velocity of helicopter is con-stant.
4. Ans (b): L = 2 m + 2 m + 2 m = 6 m
2 m
60°
45°
45° 45°
45°
2 m
2 m
TIR
TIR
60°
Fig. S.25
5. Ans (b): Energy required to remove an electron from K shell = energy required to remove an electron from
L shell + Energy difference of K and L shells (i.e., the energy of Ka transition)
= 21.75 + 3.56 = 25.31 keV The incident electron does not have this energy, hence
only L X-rays will be produced. 6. Ans (a, d): In series,
Rse = + ± ++
¥ÊËÁ
ˆ¯̃
= ±( ) . . ( %)3 5 0 3 0 13 5
100 8 5 W
R r rr r
rr
rr
r rr rP = ¥
+ÊËÁ
ˆ¯̃
± + ++
ÊËÁ
ˆ¯̃
¥1 2
1 2
1
1
2
2
1 2
1 2100D D D
= ¥+
ÊËÁ
ˆ¯̃
± + + ++
ÊËÁ
ˆ¯̃
¥5 35 3
0 33
0 15
0 3 0 13 5
100. . . .
= ± + + ¥ = ±ÊËÁ
ˆ¯̃
158
0 1 0 02 0 05 100 158
17( . . . ) % W
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Appendix B: Mock Tests B.27
7. Ans (b, c): Let the points charges have the coordinate
P ∫ (–a/2, 0, 0); Q ∫ (a/2, 0, 0) and R ∫ 0 32
0, ,aÊ
ËÁˆ
¯̃
Therefore, coordinate of point at which E has to be
calcualted is 0 02 3
23
∫Ê
ËÁˆ
¯̃, , .a a
Also, we have E Q
rr= 1
4 03pe
\ E E E Enet = + +1 2 3
= + +14 0
3 0 0 0peQa
P Q R( )
=Ê
ËÁˆ
¯̃=Q
aa k Q
ak
43 3
26
403
02pe pe
RO
P
Q
x
y
Fig. S.26
8. Ans (a, b): Refer Fig. S.27.
O
Meanposition
r
r
F F
a
a
Fig. S.27
When r is positive,
F is negative and vice-versa.
Section B
9. Ans (a): As long as water fills the tube (as shown in Fig. S.28) and the points A and B are open to the atmosphere, the velocity at B will be given by Tor-ricelli’s theorem.
Hence, v ghB = 2 2 , where h2 is the difference in levels A and B.
A
V
B
P0P0
C D
h2
h1
P P0 = Aatm. pressure
P0
Fig. S.28
10. Ans (c): We apply Bernoulli’s theorem for level A and for the highest level, CD, (labeled by subscript 1)
to get P P v ghA = + +1 12
112
r r ...(1)
Since the tube has uniform cross section and water is incompressible, v v ghB1 22= = ...(2)
From (1) and (2),
P P gh ghA1 22
112
2= - ÈÎ
˘˚ -r r
= Patm – rg (h1 + h2) ...(3) The minimum value of P1 = 0 (P1 cannot be negative
because then not water will reach the level). Hence putting P1 = 0,
( ) ..
.maxh Pg
h1 2
5
31 0 10
1 10 9 83 0= - = ¥
¥ ¥-atm
r = 10.2 m – 3.0 m = 7.2 m 11. Ans (b): Putting h1 = h2 = 3.0 m in equation (3) above,
we get P1 = Pressure at level CD = Patm – (rg)(3 + 3) = 1.0 ¥ 105 – 6rg The gauge pressure at level CD = 6rg = 6 ¥ 103 ¥
9.8 N/m2 = 5.9 ¥ 104 N/m2. 12. Ans (b) 13. Ans (a) 14. Ans (d) T sin q = 3q ¥ 105 (1) T cos q = mg – 5q ¥ 105 (2) Solve to get: q = 100 mC T = 50 N After the reversal of the direction of electric field T¢ sin a = 3q ¥ 105 ⇒ T¢ cos a = mg + 5q ¥ 105
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B.28 Physics
T
q
mg
q
T cos qq (5 10 )¥ 5
q (3 10 )¥ 5
T sin q
Fig. S.29
tan a = 3 105 10
5
5q
mg q¥
+ ¥
= 3 10 109 10 5 10 10
314
4 5
4 5¥ ¥
¥ + ¥ ¥=
-
-
fi a = ÊËÁ
ˆ¯̃
-tan 1 314
mg
a
3 10q ¥5
5 10q ¥5
T¢ sin a
T ¢ cos aa
T ¢
Fig. S.30
15. Ans (c): Fringe width will be independent of the position of S.
16. Ans (d): Path difference between the waves reaching A,
y1
S1
S2
D2 = 2 mD1 = 1 m
SA
y
Fig. S.31
Dx dyD
dyD
= +1
2 1 For central maxima: Dx = 0 fi y1 = –2y = –2(1 + cos pt) At t = 2 s, y1 = –4 mm 17. Ans (a): If slab is placed in front of S1, then
Dx dyD
dyD
t= + - -1
1
211( )m
At t = 1 s, y = 0 and for central maxima, Dx = 0.
fi y Dd
t12
11= -( )m
= 2 1 5 1 2 102 10
10 16
33¥ - ¥ ¥
¥= =
-
--( . ) m mm
Section C 18. (p – a, c) (q – b, d) (r – d) (s – d) To the right of B, there is no friction, so the torque
about center becomes zero. Hence, angular accel-eration becomes zero. So, angular velocity becomes constant. To the right of B, it will cover more distance in same time T because now there is no opposing friction. Also angle rotated is also more.
19. (p – c) (q – b) (r – b) (s – b) Current through any cross-section should remain
same. I = neAvd, as area increases so drift velocity decr-
eases. E is directly proportional to vd, so E also decreases. If E decreases, then PD across a segment of fixed
length also decreases. 20. (p – b) (q – c) (r – d) (s – d)
(p) C R RR R
R10
04 2
28= ¥
-=pe pe( )
C2 = 4pe0 (R) = 8pe0R
Rq
-q2q
C2
C1
2R
Fig. S.32
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Appendix B: Mock Tests B.29
U = qC
qC
qR
2
1
2
2
2
0222
516
+ =( )pe
(q) 2q charge will flow in the earth.
U = qC
qR
2
1
2
02 16=
pe
q
-q
R
2RC1
Fig. S.33
(r) PD of inner sphere is zero.
V k xR
qR
= +ÊËÁ
ˆ¯̃
=2
0 fi x = – q/2
R2R X
–X q x+
Fig. S.34
U = xC
q xC
qR
2
1
2
2
2
02 2 32+ + =( )
pe (s) The charge on both will become zero. Here the
net energy of system will be zero.
Solutions of Mock Test 3
PaPer 1Section-a
1. Ans (c): V V V i jD M D M1
4 3= - = -
\
VD M116 9 5= + = m/s
2. Ans (c): FBD of M: If M exerts force F = 7Mg/4 on floor, then from the third law, the floor also exerts force F on box in upward direction.
F – Mg = Ma
fi 74Mg Mg Ma- =
fi a g= 34
3. Ans (c): Conservation of energy principle
KE at A = ¥ ¥ =12
1 0 3 0 4 52. . . J
Loss in PE between A and B = 1.0 ¥ 9.8 ¥ 2 = 19.6 J Gain in KE from A to B if there had been no friction
= 19.6 J Total KE at B if there had been no friction
= 4.5 + 19.6 = 24.1 J
But actual KE at B = ¥ ¥ =12
1 0 6 182. J
Work done by friction = Change in energy = 18 – 24.1 = – 6.1 J 4. Ans (c): mv0 = (m + m¢)v, where m¢ is the mass of
snow that is added.
v¢ = mvm m
0
+ ¢ Since Ram throws off show sideways, no impulse
acts in the direction of motion. fi v¢ for Ram and Shyam will have equal velocities. 5. Ans (a): Potential of sphere should be zero.
-qq
10 +Q q
Fig. S.35
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B.30 Physics
V kqR
kqR
k Q qR
q Q= - + + = fi = -2
103
0 4( )
6. Ans (b): r1 = 2R sin 30° = R, r2 = 2R sin 60° = R 3
CR
D
r2
r1i
i
60° 30°
B
A
Fig. S.36
BD = BAC + BCB
= mp
f mp
f0
11
0
224 4
ir
ir
sin sinÈ
ÎÍ
˘
˚˙ +
È
ÎÍ
˘
˚˙
= mp
mp
0
1 2
0
460 30
4 3i
r ri
Rsin sin∞ - ∞È
ÎÍ
˘
˚˙ =
7. Ans (b): Current through the inductor before closing the switch = 10/(5 + 5) = 1 A
Current through the inductor after closing the switch
(in steady state) I = =205
4A
\ Df = L DI = 1.5 Wb 8. Ans (b): As shown in
Fig. S.37, the water surface will make an angle q with horizon-tal, where tan q = a/g.
The string will orient itself perpendicular to the water surface. Hence, inclination of the string with vertical is tan–1 (a/g) towards right.
Alternately: T cos q + mg = B cos q fi (B – T) cos q = mg ...(1)
B sin q – T sin q = ma fi (B – T) sin q = ma ...(2) From (1) and (2), tan q = a/g
9. Ans (a): I dVdx
I dVdyx y= - = - = - = -20 20,
Resultant field: I I Ix y= + =2 2 20 2
Force = mI = 2 20 2 40 2( ) = N 10. Ans (d): For helium, the molar specific heat at con-
stant volume, Cv = 32
R; for hydrogen, Cv = 52
R.
For water vapour, Cv = 3R Specific heat Cv of the mixture
= 2 3
24 5
21 3
2 4 1167
¥ + ¥ + ¥
+ +=
R R RR
CP = Cv + R = 167
R R+ = 237
R
Section B 11. Ans (a, c, d): From O to B, acceleration is negative
and from B to D, acceleration is positive, because from O to B, graph is opening downwards and from B to D graph is opening upwards. From O to A and from C to D, slope is positive, so velocity is positive. From A to C, slope is negative, so velocity is negative.
12. Ans (a, b, c)
fA NAmg
P cos q
P sin qfB
NB mg
P cos q
P sin q
(a) (b)
fANC
mg
P
(c)
Fig. S.38
NA = mg + P sin q NB = mg – P sin q NC = mg Clearly NA > NC > NB Now to find relation betwen frictions: (i) Let sliding does not occur: fA = P cos q, fB = P cos q, fC = P fi fC > fA = fB (ii) Let sliding occurs at all contact surfaces: fA = mNA, fB = mNB, fC = mNC fi fA > fC = fB
q
qT
mg
B a
Fig. S.37
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Appendix B: Mock Tests B.31
13. Ans (a, d): v mv m vm mxcm- = + -
+=1 1 2 2 0cos ( cos )q q
So the horizontal velocity of center of mass is zero. Hence, center of mass will move in vertical direction.
Now v2 cos q2 = v1 cos q1
fi vv
2
1
1
21= >cos
cosqq
(Q q2 > q1)
fi v2 > v1 fi v2 sin q2 > v1 sin q1 At any time, particle (2) will be at greater height than
particle (1). So particle (2) will be above the center of mass level.
14. Ans (a, b, c, d)
+ +E1 E2
2A R
3.00 A10 V4.00 W 3.00 W 3.00 W 6.00 W
4.00 WFig. S.39
After redrawing the circuit, we get
+ +E1
E2
2 A R
3.00 A
10 V4.00 W 3.00 W 6.00 W
4.00 W
1 A 7 A
(2)
5 A
5 A + 3 A
(1)
5 A
(3)
+
–
Fig. S.40
(a) Current through 4 W = 5 A (b) From loop (1): – 8(3) + E1 – 4(3) = 0 fi E1 = 36 V From loop (2): + 4(5) + 5(2) – E2 + 8(3) = 0 fi E2 = 54 V (c) from loop (3): – 2R – E1 + E2 = 0
R = E E2 1
254 36
29- = - = W
15. Ans (b, d) (a) Let nth order maxima of l2 coincides with mth
order minima of l1, then
n Dd
m Dd
n ml l2 12 12
2 2 12
= -ÊËÁ
ˆ¯̃
fi = -
fi 2m = 4n + 1
fi m = 2n + 12
Not possible because m and n are integers
(b) b l l b22 1
12 2= = =D
dD
a Let nth order maxima of l2 coincides with mth
order maxima of l1. Then
n Dd
m Dd
m nl l2 1 2= fi =
Hence (b) is correct (c) Let nth order minima of l2 coincides with mth
order minima of l1. Then
2 12
2 12
2 1n Dd
m Dd
- ÊËÁ
ˆ¯̃
= -ÊËÁ
ˆ¯̃
l l
fi (2n – 1) 2 = 2m – 1
fi m n= -2 12
not possible because m and n are integers (d) Let nth order minima of l2 coincides with mth
order maxima of l1. Then
2 12
2 1n DD
m Dd
-ÊËÁ
ˆ¯̃
=l l
fi m = 2 n – 1 Hence, (d) is correct
Section C 16. Ans (5): Since A, P, Q and R are at same potential.
So, the following circuit can be redrawn as:
A
R Q
P
B
A P
B
xxx
x xx
x x x x x
x
xx/2
x/2x/2A
B
x/2
Fig. S.41
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B.32 Physics
So, equivalent resistance between A and B
= 2110
2140
x r=
17. Ans (5): Rate of mechanical work done = Fv where F is the force and v is the velocity.
F = Bil = B BlvR
l◊ ÊËÁ
ˆ¯̃
= 0.5 ¥ 0.5 ¥ 0.5 = 12.5 Therefore, power = 12.5 ¥ 4 = 50 W Work done in 10 s = 50 ¥ 10 = 500 J
18. Ans (3): 4 3 160
3 4 1 0/ (( / ) )v
+ = -•
=
M
O
I
I¢
I¢¢
A
Fig. S.42
v = – 80 cm I ¢ seen as object for the mirror
fi 1 1100
120
25v
v+-
= - fi = - cm
I ≤ will serve as a object for plane surface
1 4 35
1 4 3v
- = -•
( / ) ( ( / ))
v = +3.75 cm
19. Ans (2): Power of light received by the cone = l (p R2). Let the number of photons hitting the cone per second is n. Then
q
q
q
RFig. S.43
nE = IR2 fi n = pR2I/E By symmetry, the net force on the cone will be verti-
cally downward.
Force due to one photon: f = 2 hl
qsin
This force is perpendicular to the surface of cone. Hence, net force on cone will be
F = nf sin q = n h2l
q qsin sinÊËÁ
ˆ¯̃
= p qR Ic
21 2( cos )-
20. Ans (2): Applying COE, mgs (sin a – sin b)
= 12
2 12
12
2 2 2I mv Iw w+ +ÊËÁ
ˆ¯̃
where w = v/r and I = mr2/2 Putting values and solving, we get v = 2 m/s.
PaPer 2Section a
1. Ans (d): T MK
= 2p in any case
2. Ans (d): As the observe is moving, so relative veloc-ity between pulse and observer = v1 + v3.
So time taken t Lv v
Lv
=+
<1 3 3
.
3. Ans (d): Because of the thickness of the lens, we cannot use the usual formulae of lens and a unique focal length cannot be derived.
4. Ans (c): lmin.= ¥ =1 24 10 1
4
VÅ Å for 12.4 kV and
0.5 Å for 24.8 kV. Now, 2 (la – lmin, 1) = (la – lmin, 2) fi 2la – 2 = la – 0.5 fi la = 1.5 Å
5. Ans (d): V d l= p 2
4
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fi D D DVV
dd
ll
= + = ¥ + =2 2 0 014
0 15
140
. .
% error in volume = DVV
¥ =100 2 5. %
6. Ans (a, c, d): When the current becomes constant, PD across inductor will become zero. Hence, the entire PD will be across R.
Also energy stored in the inductor becomes constant if current through it becomes constant.
7. Ans (a, c, d): The incident rays parallel to the prin-cipal axis pass through (or appears to come from) second focus after refraction.
L
L
F2
F1
I
Fig. S.44
F1 Æ first focus, F2 Æ second focus I Æ Image. Since the second focus is to the left of
lens, so it should be concave lens.
8. Ans (a, c, d): vv
1
2
2
1
1
2
1
2
12
= = fi =mm
ll
ll
L m L n11
22
2 2= =l land
where m and n are the number of loops in L1 and L2, respectively.
Now L L m n1 21 2
2 2= fi =l l fi m = 2n
For minimum, n = 1 fi m = 2
Section B
9. Ans (a): w1 = 100 p
f1 = wp2
= 50 Hz, f2 = 922
pp
= 46 Hz
Beat frequency = f1 = f2 = 4 10. Ans (a): v = w/K = 100 p /(0.5p) = 200 11. Ans (c): At x = 0 y = y1 + y2 = 2A cos 96pt cos 4 pt
For y = 0, cos 96pt = 0 or 4pt = 0 fi 96pt = (2n + 1)(p/2) and 4pt = (2m + 1)p/2 For 0 < t < 1,
- 12
< n < 95.5 and - 12
< m < 3.5
Here n and m are integers, therefore, the net amplitude becomes zero 100 times.
12. Ans (b) 13. Ans (c) 14. Ans (d)
m m mmy R
y R-•
= - fi =-
1 11
If beam is focussed at A, than y = 2R
fi 21
2R R=-
fi =mm
m
The above is valid for paraxial rays. Hence, beam should be thin.
If m > 2, then y < 2R. Hence, beam can be focussed before A.
For thin beam to be focussed at A, y = R.
fi R R=-
fi = •mm
m1
(not possible)
15. Ans (c): Use conservation of energy. 16. Ans (c): Time period is not affected by a constant
force acting along the line of SHM. 17. Ans (a): mg is balanced by the electrostatic force.
Sectiion C 18. (p – b, d) (q – a, d) (r – a, d) (s – b, d) Velocity at a point at distance x from the lower end:
v T xg xg= = =m
mm
a dvdt
g x dxdt
= =-1 2
2
/
ag x
gx g= =-1 2
22
/
/
Velocity, tension, frequency etc, do not depend upon wavelength. But in turn, wavelength depends upon them.
19. (p – b, d) (q – a, c) (r – a, c) (s – b, d) (p) During AB: Temperature is constant (DU = 0)
and pressure is increasing. So from PV = nRT, volume decreases. Hence, DW < 0. So also DQ < 0
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(q) During BC: Temperature increases (DU > 0) and pressure remains constant, so volume increases. Hence DW > 0. So also DQ > 0.
(r) During CD: Opposite to AB (s) During DA: Opposite to BC 20. (p – a, d) (q – a, d) (r – b, d) (s – c) (p) Flux entering through base: f = EA = EpR2
(q) Flux outgoing through curved part is same as the flux entering through base.
(r) Flux through MNQP will be same as the flux through base from radius R/4 to 3R/4. So
f pp
= ÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
È
ÎÍÍ
˘
˚˙˙
=E R R E R34 4 2
2 2 2
(s) Net flux through the entire cone will be zero as in uniform electric field, flux through any closed surface is zero.
