XII RT-5 [JEE Mains] Code-A Page # 1 PHYSICS XII-ABCD, F1, SMS, F2 REVIEW TEST-5 DATE : 23.09.2012 JEE MAINS PAR T -A [SINGLE CORRECT CHOICE TYPE] Q.1 to Q.33 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1 An infinite V-shaped wire carrying current I is shown in the adjacent figure. Find the magnitude of magnetic field at point P due to the wire, if AP = r (A) tan r 2 I 0 (B*) 2 tan r 2 I 0 (C) 2 cot r 2 I 0 (D) sin r 2 I 0 [Sol. d = rsin B= d i 4 0 (sin1 + sin2 ) = sin r i 4 0 [1 sin(90 )] = sin r i 4 0 (2sin 2/2) = 2 tan r i 4 0 B not = 2B = 2 tan r i 2 0 ] Q.2 A wind farm generator uses a three-bladed propeller mounted on a pylon at a height of 20 m. The length of each propeller blade is 12m. A tip of the propeller breaks off when the propeller is vertical. At that instant, the period of the motion of the propeller is 1.2 s. The fragment flies off horizontally, falls and strikes the ground at P. V 0 P 12m 20m In figure, the distance from the base of the pylon to the point where the fragment strikes the ground is closest to (A) 120 m (B) 130 m (C) 150 m (D*) 160 m
Transcript
XII RT-5 [JEE Mains] Code-A Page # 1
PHYSICSXII-ABCD, F1, SMS, F2 REVIEW TEST-5 DATE : 23.09.2012
JEE MAINS
PART-A[SINGLE CORRECT CHOICE TYPE]
Q.1 to Q.33 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1 An infinite V-shaped wire carrying current I is shown in the adjacent figure. Find the magnitude ofmagnetic field at point P due to the wire, if AP = r
(A)
tan
r2
I0 (B*)
2tan
r2
I0(C)
2cot
r2
I0(D)
sin
r2
I0
[Sol. d = rsin
B = d
i
4
µ0
(sin
1 + sin
2)
= sinr
i
4
µ0 [1 � sin(90 � )]
= sinr
i
4
µ0 (2sin 2/2)
= 2
tanr
i
4
µ0
B not = 2B = 2
tanr
i
2
µ0
]
Q.2 A wind farm generator uses a three-bladed propeller mounted on a pylon at a height of 20 m. The lengthof each propeller blade is 12m. A tip of the propeller breaks off when the propeller is vertical. At thatinstant, the period of the motion of the propeller is 1.2 s. The fragment flies off horizontally, falls andstrikes the ground at P.
V0
P
12m
20m
In figure, the distance from the base of the pylon to the point where the fragment strikes the ground isclosest to(A) 120 m (B) 130 m (C) 150 m (D*) 160 m
XII RT-5 [JEE Mains] Code-A Page # 2
PHYSICS
[Sol. v0 = 12 × 2.1
2 = 20 32 =
2
2
)20(
x1021
10
40064 2 = x2 ; 8 × 20 = 160 m ]
Q.3 Which of the following is an accurate statement ?(A) all points on a rotating disk experience the same radial acceleration.(B*) all points on a rotating disk have the same angular velocity.(C) all point on a rotating disc have same velocity.(D) the vector sum of the tangential acceleration and the centripetal acceleration can be zero for a point on circumference of a rotating disk.
Q.4 In figure, determine the character of the collision. The masses of the blocks, and the velocities before andafter are given. The collision is
(A*) perfectly elastic (B) partially inelastic(C) completely inelastic (D) this collision is not possible
[Sol. e = 2
x]
Q.5 The block is pulled by a small distance and released. The block comes to equilibrium position after time(T is time period of oscillation)
Smooth
(A*) 4
T(B)
2
T(C)
5
T(D) T
Q.6 Block A (0.30 kg) is on a frictionless table. Spring connects block A to a frictionless peg at O. The blockis in uniform circular motion about O, and the spring has length (exended) of 1m. The linear speed ofblock is 2.0 m/s. In figure the amount of stretch of spring is 0.06 m. The force constant of spring is :
AO1m
2m/s
(A) 18 N/m (B*) 20 N/m (C) 22 N/m (D) 24 N/m
[Sol. k × 0.06 = 1
23.0 2
; k = 20 ]
XII RT-5 [JEE Mains] Code-A Page # 3
PHYSICSQ.7 A square coil of side 10 cm consists of 20 turns and carries a current of 12A. The coil is suspended
vertically and the normal to the plane of the coil makes an angle of 30º with
the direction of a uniform horizontal magnetic field of magnitude 0.80 T. Whatis the magnitude of torque experienced by the coil?(A*) = 0.96 N�m(B) = 1.08 N�m(C) = 2.5 N�m (D) None of these
[Sol. Given, side of square coil = 10cm = 0.1 mNumber of turns (n) = 20Current in square coil l = 12AAngle made by coil = 30º
Magnetic field B = 0.80 TThe magnitude of torque experienced by the coil
= NI AB sin = 20 × 12 × (10 × 10�2)2 × 0.80 × sin30º ; = mN96.02
80.024
]
Q.8 A particle of mass m and charge q moving with velocity v entres a region of uniform magenitc field
of induction B
. Then which of the statement is wrong?
(A) Its path in the region of the field is circular if 0B.v
(B) Its path in the region of the field is a straight line if 0Bv
(C) The angular velocity of particle m
qB is if 0B.v
(D*) None of these
[Sol. If 0B.v
, means Bv
or the path is circular..
If 0Bv
,means v is either parallel or antiparallel to B
, i.e., path is straight line.
Futher, Bq
m2T
or
m
Bq
T
2
]
Q.9 In figure, a solid uniform cylinder of 5 cm radius is positioned on a frictionless plane inclined at 30° above
horizontal. A force F is exerted by a string wrapped around the cylinder. When F has a certain criticalvalue the center of mass of the spool does not move. When this is the case, what is the angular accelerationof the spool ? F
PHYSICSQ.10 A vented tank of large cross-sectional area has a horizontal pipe 0.12m in diameter at the bottom. This
holds a liquid whose density is 1500 kg/m3 to a height of 4.0 m. Assume the liquid is an ideal fluid inlaminar flow. In figure, the velocity with which fluid flows out is
4.0m
(A) 52 m/s (B) 5 m/s (C*) 54 m/s (D) 10 m/s
[Sol. gh2v = 4102 = 522 ; 54v ]
For F1 & F2Q.10 Two particles 1 and 2 start from origin and move along same straight line in same direction. The acceleration
versus time graph of both particles is shown in figure. Choose correct statement about the motion ofparticles 1 and 2.
(A) During motion from O to P, somewhere one particle will overtake the other.(B) During motion from O to P, average speed of both the particles is same.(C*) During motion from O to P, average speed of both the particles are different.(D) Displacement covered by the two particles from O to T are same.
[Sol. Since, accelerationof 1 is always greater or equal to 2, 2 will never overtake 1. Also, average speed of1 is greater than that of 2. The dispalacement curve of the two will be different as they have differentacceleration.]
Q.11 A heat engine performs the reversible cycle abca with 9.0 moles of an ideal gas. Path ca is an adiabaticprocess. The temperatures at point a and b are 300 K and 500 K, respectively. The adiabatic constantof the gas is 1.60.In figure, the heat absorbed by the gas in path ab, in kJ, is :
Pa b
c
300K 500K
V
(A) zero (B) + 25 (C*) + 40 (D) � 25
[Sol. Q = nCp T = n T1
R
= 9 × 6.1
25R6.1 × 200 = 40000 ]
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PHYSICSFor F1 & F2Q.11 The length of a potentiometer wire is l. A cell of emf E is balanced at a length l/3 from the positive end of
the wire. If the length of the wire is increased by l/2. At what distance (in cm) form postiive end will thesame cell give a balance point, if l = 10 cm.(A*) 5 cm (B) 10 cm (C) 20 cm (D) none of these
[Sol. Let x be the desired lengthPotential gradient in the first case
l0E
3
EE
3E 00
l
l........ (i)
Potential gradient in second case
ll 3
E2
2/3
E 00
l3
E2)x(E 0 ........(ii)
From Eqs. (i) and (ii)
2x
l ]
Q.12 Three charged capacitors, C1 = 17µF, C2 = 34µF and C3=41µF and two open switches, S1 and S2 areassembled into a network with initial voltages and polarities, as shown. Final status of the network isattained when the two switches, S1 and S2 are closed. The figure, the final charge on capacitor C3 inmC, is closest to :
+
+
+
+
�
�
�
�
60VC1
50VC3
S2
S1
90VC2
� �
+ +
(A*) zero (B) 410 (C) 1200 (D) 3300[Sol. After closing both the switches S1 and S2, capacitor C3 gets shorted so potential difference across C3
and hence charge on plates of C3 becomes zero. ]Q.13 An electric dipole is placed well inside the capacitor having capacitance 3F in the orientation shown in
the figure. The plates of capacitor are separated by 1cm. The electric field dipole moment of dipole is10�12 C-m. The torque acting on dipole at the instant shown is ______.
PHYSICSQ.14 Two parallel plate capacitor of equal size are charged by a power supply of voltage V
supply. What is true
of these capacitor when their stored charge is plotted as a function of supply voltage ?
A
B
Q
0
0Vsupply
(A) Capacitor B has a higher dielectric constant between the plates; capacitor A has a higher breakdown voltage(B) Capacitor A has a higher dielectric constant between the plates; Capacitor B has a higher break down voltage.(C) Capacitor B has higher dielectric constant between the plates; Capacitor A has a lower breakdown voltage.(D*) Capacitor A has a higher dielectric constant between the plates; Capacitor B has a lower break down voltage.
Sol. (i) At same voltage A stores more charge than B, but they same size so capacitance of A is more than Bdue to higher dielectric constant between the plates of A.(ii) Clear from graph that breakdown voltage of A is more than that of B.
Q.15 Which of the following electric field configurations describes the lines of force in an electric field betweentwo parallel plates of opposite charge?
(A) (B*)
(C)
+ + + + + + + + + + +
� � � � � � � � � � �
(D)
+ + + + + + + + + + +
� � � � � � � � � � �
[Sol. An important point to remember about electric fields is that they emanatefrom positive charges and point towards negative charges. With this inmind, we know choices A and C are incorrect. The next question is tojudge the straightness or curvature of the lines between the plates.
+ + + + + + +
� � � � � � �
Imagine a point placed directly between two parallel plates (see the diagram to the right). It will have anet electric field determined by all the individual electric fields generated by every charge on the twoplates. For clarity, only five electric field lines are shown in the diagram. If these field lines are addedvectorially, all the horizontal components will cancel each other out; a net field oriented verticallydownward will remain (the field lines shown in the choices are the possible net electric fields). Thecancellation is due to the symmetry of the point in space with respect to the charges on the plates.However, this horizontal component cancellation continues only until one approaches the left and rightedges of the plates. Here, the field lines bow out, much like those around a dipole (the parallel plates arelike several dipoles positioned next to each other). Remember, between parallel plates of oppositecharge, the field lines are straight. They are also constant in strength throughout the interplate region.Thus, a parallel plate arrangement is a good design for capacitors, electrophoresis equipment, and chargeaccelerators. The correct choice is B. ]
XII RT-5 [JEE Mains] Code-A Page # 7
PHYSICSQ.16 How much work is done by electric field on a positively charged particle moving along the 15 cm path
shown by dotted line, if the electric field i�C/N10E
and the charge on the particle is a 8C ?
0 cm 5 cm 10 cm
(A) 0.8 J (B*) 8 J (C) 12 J (D) 1200 J[Sol. W = qEd
= 8 × 10 × 10 × 10�2
= 8 J]Q.17 In the given circuit, the capacitor of capacitance C is charged by closing key K at t = 0. Find the
time required to charge the capacitor upto maximum charge for the given circuit, if it were to becharged with the constant initial charging rate at t = 0 in the given circuit.
2R
R
R
C
E K
(A) 3
RC(B)
5
RC2(C)
3
RC2(D*)
3
RC5
[Sol. Initial charging rate means initial current.
Current through battery R5
E3
R32
R
Ei0
Initial current through capacitor i = R5
E2i
3
2i 0
Steady state potential difference across capacitor
E3
2V0 ; EC
3
2CVq 00 ; Now,, RC
3
5
R5E2
EC32
i
qt 0 ]
XII RT-5 [JEE Mains] Code-A Page # 8
PHYSICSQ.18 A wire loop ABCDE carrying a current I is placed in x-y plane as shown in figure. A particle of
mass m and charge q is projected from origin with smj�i�2
vv 0
. The instantaneous acceleration
acts along (r = radius of circular arc ABC)
(A) AO (B*) OA (C) x-axis (D) OP
[Sol. Net field at k�45sin45sin2r
I
4k�
r2
I
4
1O 00
i.e., net field at O is along negative z-axis. ; Now,
m
Bvq
m
Fa
]
Q.19 A square of side 1 m is placed in a uniform magnetic field T2 in a direction perpendicular to the
plane of the square inwards. Equal current i = 1A is flowing in the direction shown in figure. Themangnetic force on the loop is :
S R
P Q
B
(A*) 6 N (B) 4 N (C) 2 N (D) zero
[Sol
PRPQRPSR FFF ; N62213BPRi�3Fnet ]
Q.20 Two soap bubbles of different radii are formed at the two ends of the tube and are in communicationwith each other through the tube. What will be happen to them?(A) Smaller bubble will grow and larger will collapse(B*) Larger bubble will grow until smaller bubble collapses(C) Smaller bubble will grow till both have the same radius(D) Both the bubbles will remain as it is
[Sol.R
1P i.e., pressure inside smaller bubble is more than the pressure inside larger drop, i.e., air will
rush from smaller drop to larger drop.]
For F1 & F2Q.20 A man of 80 kg attempts to jump from the small boat of mass 40 kg on to the shore. He can generate a
relative velocity of 6 m/s between him and boat. His velocity towards shore is(A) 4 m/s (B) 8 m/s (C*) 2 m/s (D) 3 m/s.
XII RT-5 [JEE Mains] Code-A Page # 9
PHYSICSQ.21 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude
of the magnetic field due to the current 1.5 m below the line?(A) 2 × 10�5 T (B*) 1.2 × 10�5 T (C) 2.5 × 10�5 T (D) none of these
[Sol. Given, I = 90 A and r = 1.5 m, Here point P is below the power line, where we have to find the magneticfield and its direction. The magnitude of magnetic field
B r
I2
4
µ0
5.1
90210 7
= 1.2 × 10�5 TThe direction of magnetic field is given by Maxwell�s right hand rule. So, the direction of magnetic field
at point P due to the flowing current is perpendicularly outwards to the plane of paper.]
PASSAGE FOR Q.22 & Q.23A parallel plate capacitor of capacitance C is connected between two horizonatal metallic railswhere uniform vertical magnetic field B is present. A metallic rod of length l which slide freely on therails has mass m. The distance between the rails is also l . A constant horizontal force F acts on therod. If magnetic field acts into the plane and if resistance of the system is neglected, answer thefollowing questions.
m, l F
Q.22 During the motion of the rod :(A) the current flow keep on increasing (B) the current flow keep on decreasing(C*) the current flow remains constant (D) no current flows through the rod
Q.23 During the motion of the rod the charge on the capacitor:(A*) keep on increasing (B) keep on decreasing(C) remains constant (D) is zero
[Sol. Let v be the velocity of rod at some instant.In that case,
PD, BvlV vCBlCVq dt
dvCBl
dt
dqi
Due to this current a magnetic force will act on the rod in opposite direction of the applied force F.
mnet fFF or dt
dvlCBFilBF
dt
dv.m 22
22lCBm
Fa
dt
dv constant
Due to this constant acceleration v and q will go on increasing, but i will remain constant.]
XII RT-5 [JEE Mains] Code-A Page # 10
PHYSICSQ.24 A container has a vertical tube, connected to it at its side. An unknown liquid reaches level A in the
container and level B in the tube level A being 5.0cm higher than level B. The liquid supports a 20.0 cmhigh column of oil, between levels B and C, whose density is 500 kg/m3. In figure, density of unknownliquid is
For F1 & F2Q.24 Sachin (55kg) and Kapil (65kg) are sitting at the two ends of a boat at rest in still water. The boat weighs
100 kg and is 3.0 m long. Sachin walks down to Kapil and shakes hand. The boat gets displaced by:(A) zero m (B*) 0.75 m (C) 3.0 m (D) 2.3 m
Q.25 Two blocks of mass M and m are in equilibrium as shown in figure. If mass m is replaced by 8m, find theacceleration of M.
(A*) 57
g7(B)
35
g49(C)
57
g71(D) None of these
[Sol. Draw the FBD of the various blocks as shown in figure.
A B C
8T 4T 2T
4T 2T T4T 2T T
7T = Mg [For equilibrium]T = mg
XII RT-5 [JEE Mains] Code-A Page # 11
PHYSICSM = 7mWhen m replaced by 8m, the blocks would have some acceleration.From constraint theory, it can be found out that when acceleration of8m is a in downward direction, then acceleration of M is a/7 in upwarddirection.
8 mg � T = 8 ma
7T � 7 mg = 7m × a/7
a = 49mg / 57m = 49g/57 Acceleration of M = 7g/57
Q.26 Time constant of the below circuit is
R
CR R
(A) RC2
3(B*)
3
RC(C) RC
3
2(D) 3RC
[Sol.
Req = 3
R, t = Req C =
3
RC]
Q.27 An astronaut weighs 650 N on the surface of Earth. If orbiting at a distance of twice the radius of Earthfrom the centre of Earth, how does his weight change.(A) His weight doubles (B) His weight remains the same(C) His weight is a factor of two smaller (D*) His weight is a factor of four smaller
[Sol. gs = 2R
GM ; gh = 4
gs
)R2(
GM2 ]
For F1 & F2Q.27 The velocities of a particle executing S.H.M. are 30 cm/s and 16 cm/s when its displacements are 8 cm
and 15 cm from the equilibrium position. Then its amplitude of oscillation in cm is :(A) 25 (B) 21 (C*) 17 (D) 13
Q.28 Which of the following changes to a parallel plate capacitor would not increase its capacitance?(A) decreasing the distance between the plates (B) increasing the area of the plates(C) increasing the dielectric constant (D*) increasing the voltage across the plates
[Sol. C = d
AK 0
It is not dependent on voltage.]
XII RT-5 [JEE Mains] Code-A Page # 12
PHYSICSQ.29 A galvanometer G deflects full scale when a potential difference of 0.50 V is applied. The internal
resistance of the galvanometer rg is 25 ohms. An ammeter is constructed by incorporating the galvanometerand an additional resistance RS. The ammeter deflects full scale when a measurement of 2.0 A is made.The resistance RS is closest to(A*) 0.25 (B) 2.5 (C) 0.45 (D) 0.1
[Sol. 1.98 × RS = 0.02 × 25 = 0.5 ]Q.30 In a potentiometer (see figure) a balance is obtained at a length of 400 mm when using a known battery
of emf 1.6 volts. After removing this battery. Another battery of unknown emf is used and balance isobtained at a length of 650 mm. The emf of unknown battery is
G
l
(A*) 2.6 volt (B) 1.6 volt (C) 3.4 volt (D) 4.7 volt
[Sol. a400
650
1
2
40
652 × 1.6 = 2.6V]
Q.31 Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in thepositive x-direction, is produced by charges on the plates. The center plane at x = 0 m is an equipotentialsurface on which V = 0. An electron is projected from x = 0 m, with an initial kinetic energy K = 300 eV,in the positive x-direction, as shown. KE of electron when it reaches plate A is
E = 1500 V/m
�K = 300 eV
V =
0
x = �1.2 m x = +1.2 m
x = 0 m
A B
(A) 300 eV (B*) 2100 eV (C) 1800 eV (D) 600 eV
[Sol. Kf � Ki = WE ; Kf = 300 eV + 1500 ×2
4.2 eV = 2100 eV]
Q.32 A nonuniform electric field is directed along the x-axis at all points in space. This magnitude of the fieldvaries with x, but not with respect to y or z. An imaginary cylinder of length L and diameter d is made asshown. The electric field at left face is E1, electric field at right face is E2. The value of charge enclosedby cylinder does not depend on
L
E1 E2
d
y
x
z
(A) E1 (B) E2 (C*) L (D) d
[Sol.0
inqdA·E
; E2A � E1A = 0
inq
; qin = (E2 � E1) 4
d20
]
XII RT-5 [JEE Mains] Code-A Page # 13
PHYSICSQ.33 Water rises to a height h1 in a capillary tube in a stationary lift. If the lift moves up with uniform acceleration,
it rises to a height h2 then acceleration of the lift is
