rt5

PAGE # 1 XI (PTP,P1, P2 & Q) REVIEW TEST-5_[JEE MAIN]_PAPER-1_PHYSICS_ANSWER KEY, HINT & SOLUTION

XI (PTP,P1, P2 & Q) REVIEW TEST-5_JEE_[MAIN]_PAPER-1_TIME : 1 HOURS.

DATE :11-10-2015

SYLLABUS : FOR XI(PTP)BATCH_REVIEW TEST-5_JEE_[MAIN]_PAPER-1

Units & Dimensions, Kinematics, Newton�s Laws of motion, Friction, Circular Motion, Work Power Energy, Centre

of mass, Momentum, Impulse., Collision, Rotational dynamics, Fluid Mechanics.

SYLLABUS : FOR XI(P1 & P2)BATCH_REVIEW TEST-5_JEE_[MAIN]_PAPER-1

Units & Dimensions, Kinematics, Newton�s Laws of motion, Friction, Circular Motion, Work Power Energy, Centre

of mass, Momentum, Impulse., Collision, Rotational dynamics (taught till date).

SYLLABUS : FOR XI(Q)BATCH_REVIEW TEST-5_JEE_[MAIN]_PAPER-1

Units & Dimensions, Kinematics, Newton�s Laws of motion, Friction, Circular Motion, Work Power Energy, Centre

of mass, Momentum-Impulse-Collision

PART-A[SINGLE CORRECT CHOICE TYPE]

Q.61 to Q.90 has four choices (1), (2), (3), (4) out of which ONLY ONE is correct.

[30 questions × 4 marks (�1) = 120 Marks]

Q.61 (2)

[Sol. KE = 2

1

2

MR 2 2

R

V

=

4

Mv2

] [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.61 (3) [TOPIC : WPE]

Q.62 (3)

[Sol.

2

2

R10

M

2

MR = '

2

MR 2

10

1

2

1 =

2

' =

10

15 =

2

'

' = 5

6 ] [TOPIC : ROTATIONAL DYNAMICS]

ONLY FOR "P1 & P2, Q" BATCH

Q.62 (3)

[Sol. =

m ; µ ( � 1) g = 1 g ] [TOPIC : FRICTION]

Q.63 (2)

[Sol. F = st

sp =

5.1

225.4 = 66 N ] [TOPIC : COLLISION]

PAGE # 2 XI (PTP,P1, P2 & Q) REVIEW TEST-5_[JEE MAIN]_PAPER-1_PHYSICS_ANSWER KEY, HINT & SOLUTION

Q.64 (4)[Sol. mv = mv' cos + mv' cos

v' = cos2

v ; v' >

2

v] [TOPIC : COLLISION]

Q.65 (3) [TOPIC : COLLISION]

Q.66 (1) [TOPIC : CENTRE OF MASS]

Q.67 (2) [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.67 (4)[Sol. spring forces will be equal in both springs

cons.F2k

F

2k

xkkx

2

1U

2222

1

3

k

k

U

U

k

1Uá

1

2

2

1 ] [TOPIC : WPE]

Q.68 (1) [TOPIC : CONSERVATION OF LINEAR MOMENTUM]

Q.69 (4) [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.69 (3)

[Sol. (3); Limiting friction for block AfL = µ N = 753

1 ; fL = 25 N

In equilibrium conditions FBD of B block T = 25 NFBD of ASo system is in equilibrium. Net contact force

22 NfFnet 22 )75()25( 22 )25(9)25( =

Fnet = N1025

[TOPIC : NLM, FRICTION]

Q.70 (2) [TOPIC : CONSERVATION OF LINEAR MOMENTUM]

Q.71 (1) [TOPIC : C IRCULAR MOTION]

Q.72 (4) [TOPIC : C IRCULAR MOTION]

Q.73 (1) [TOPIC : WORK POWER ENERGY]

Q.74 (4) [TOPIC : MOTION IN VERTICAL PLANE]

Q.75 (1) [TOPIC : PRESSURE]

ONLY FOR "P1 & P2, Q" BATCH

Q.75 (3)[Sol. (3); Initial diagram of A and B with respect to person

So path of A and B with respect to person

BA

] [TOPIC : K INEMATICS]

PAGE # 3 XI (PTP,P1, P2 & Q) REVIEW TEST-5_[JEE MAIN]_PAPER-1_PHYSICS_ANSWER KEY, HINT & SOLUTION

Q.76 (1) [TOPIC : FLUID MECHANICS]

ONLY FOR "P1 & P2, Q" BATCH

Q.76 (4)[Sol. Force required is same in all condition because normal contact force is same in all the cases. ]

[TOPIC : NLM, FRICTION]

Q.77 (1)

[Sol. mgx + mgx � 0)x2(0k2

1 22 ; k

mgx ] [TOPIC : WORK POWER ENERGY]

Q.78 (4)

[Sol. 10 × d � 2

1kd2 = 0 ; d =

100

20 = 0.2 m; U =

2

1kd2 =

2

1 × 100 × (0.2)2 = 2J ]

[TOPIC : WORK POWER ENERGY]

Q.79 (2)

[Sol. W = mgR = same = 2

1I2

I1 = 2

3MR2 I2 =

4

5mR2 2 > 1 ]

[TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.79 (2)[Sol. (2); With respect to ground acceleration of ball is g downwards so path of ball is parabolic as seen from

ground.

Relative acceleration LbbL aaa ; )(

^^

jgjgabL ; 0bLa ]

[TOPIC : K INEMATICS]

Q.80 (3) [TOPIC : WORK POWER ENERGY]

Q.81 (2) [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.81 (3)

[Sol.

mg

Body is at rest as. ] [TOPIC : FRICTION]

Q.82 (3) [TOPIC : WORK POWER ENERGY]

Q.83 (4) [TOPIC : FLUID MECHANICS]

ONLY FOR "P1 & P2, Q" BATCH

Q.83 (1)[Sol. (1); Student B is wrong since in this case changes with time.

Here, y = x tan (y, x, all three variables) ] [TOPIC : NLM, FRICTION]

PAGE # 4 XI (PTP,P1, P2 & Q) REVIEW TEST-5_[JEE MAIN]_PAPER-1_PHYSICS_ANSWER KEY, HINT & SOLUTION

Q.84 (1) [TOPIC : WPE]

Q.85 (1) [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.85 (3)

[Sol. mg cos = R

mv2

mg

mg (R � R cos ) = 2

1mv2

cos = 3

2

at = g sin = g × 3

5] [TOPIC : C IRCULAR MOTION]

Q.86 (2) [TOPIC : FLUID MECHANICS]

ONLY FOR "P1 & P2, Q" BATCH

Q.86 (3)

[Sol. Area under curve = work done = 160 × 15 + 2

10160

= 3200

2

1 × m × 102 = 3200 m = 64 kg ] [TOPIC : WORK POWER ENERGY]

Q.87 (4) [TOPIC : K INEMATICS]

Q.88 (2) [TOPIC : NEWTON�S LAWS]

Q.89 (2) [TOPIC : K INEMATICS]

Q.90 (1) [TOPIC : K INEMATICS]

PAGE # 1 XI (PTP,P1-P2 & Q) REVIEW TEST-5_[JEE ADVANCED]_PAPER-2_PHYSICS_ANSWER KEY, HINT & SOLUTION

XI (PTP,P1-P2 & Q) REVIEW TEST-5_JEE[ADVANCED]_PAPER-2

TIME : 1 HOURS. DATE :11-10-2015

SYLLABUS : FOR XI(PTP)BATCH_REVIEW TEST-5_JEE_[ADV.]_PAPER-2

Units & Dimensions, Kinematics, Newton�s Laws of motion, Friction, Circular Motion, Work Power Energy, Centre

of mass, Momentum, Impulse., Collision, Rotational dynamics, Fluid Mechanics.

SYLLABUS : FOR XI(P1 & P2)BATCH_REVIEW TEST-5_JEE_[ADV.]_PAPER-2

Units & Dimensions, Kinematics, Newton�s Laws of motion, Friction, Circular Motion, Work Power Energy, Centre

of mass, Momentum, Impulse., Collision, Rotational dynamics (taught till date).

SYLLABUS : FOR XI(Q)BATCH_REVIEW TEST-5_JEE_[ADV.]_PAPER-2

Units & Dimensions, Kinematics, Newton�s Laws of motion, Friction, Circular Motion, Work Power Energy, Centre

of mass, Momentum-Impulse-Collision

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which only one is correct.

[4 questions × 3 marks (�1) (Moderate)]

Q.1 (B)

[Sol. tan = x

y

U

U;

3

1=

x

y

U

U=

9

Uy

Uy = 3 32yV = yy

2x Sa2U

9 = 27 + 2(�10)SySy = 0.9 Ans. ] [TOPIC : PROJETILE]

Q.2 (D) [TOPIC : NEWTON�S LAWS OF MOTION]

Q.3 (B)[Sol. Com of both sides will move with same acceleration

10

a2)aa(8 0 =

10

a10 a =

5

2a6

Acceleration of monkey with respect to ground = a0 � a = 0.6 a0 = 1.2 ms�2

t = 2.1

)4.2(2= 2 sec. ] [TOPIC : NEWTON�S LAWS OF MOTION]

Q.4 (D)

[Sol. amax = m4

mg=

4

g=

Mm4m

Mg

M =

4

5m] [TOPIC : FRICTION]

PAGE # 2 XI (PTP,P1-P2 & Q) REVIEW TEST-5_[JEE ADVANCED]_PAPER-2_PHYSICS_ANSWER KEY, HINT & SOLUTION

Q.5 to Q.7 has four choices (A), (B), (C), (D) out of which only one is correct.

[3 questions × 4 marks (�1) (Tough)]

Q.5 (A)

[Sol.16

1 = tan 37° = 4

3

37° 53°16

tt

0 = 12

t

12 =

t0 = tan 53° =

3

4

t = 9 sec. = area under curve.

= 2

1 × [50 + 25] × 12 = 450

<> = t

=

50

450 = 9 rad/s. ] [TOPIC : ROTATIONAL DYNAMICS]

Q.6 (B)[Sol. (50 + 50 + 95 + 5) × 2 = 195 v + 5(v + 4)

50 50

25 2m/s

v = 1.9

t = s/m4

m4 =

rel

rel

u

s = 1 sec.

1.9 × 100 × 1 = 190 ][TOPIC : CONSERVATION OF LINEAR MOMENTUM]

Q.7 (B)

[Sol. MgH = KE = 2

1MV

c2

2c

MR

I1 ] [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.7 (B)[Sol. When the spring undergoes maximum compression the relative velocity between the block and car is

zero

(4m + 2m) v = mv0 ; v = 6v0

20

2

v)m2(

2

1

=

21

kx2 + 21

(6m)2

0

6

v

]

[TOPIC : CONSERVATION OF LINEAR MOMENTUM]

Paragraph for question nos. 8 to 9Q.8 (B) [TOPIC : COLLISION]

Q.9 (D) [TOPIC : COLLISION]

PAGE # 3 XI (PTP,P1-P2 & Q) REVIEW TEST-5_[JEE ADVANCED]_PAPER-2_PHYSICS_ANSWER KEY, HINT & SOLUTION

Q.10 to Q.12 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct.

[4 questions × 4 marks (No Negative)]

Q.10 (BC) [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.10 (BCD) [TOPIC : C IRCULAR MOTION]

Q.11 (BD) [TOPIC : CONSERVATION OF LINEAR MOMENTUM]

Q.12 (AC) [TOPIC : COLLISION]

Q.1 have three statements (A,B,C) given in Column-I and four statements (P,Q,R,S) given in Column-II. Anygiven statement in Column-I can have correct matching with one or more statement(s) given in Column-II.

[(3 + 3 + 3) (No Negative)]

Q.1 [(A)-Q (B)-P (C)-S ][Sol. In all process there is no external force in Y-direction so linear momentum in Y-direction will conserve

similarly no resistive force hence K.E. will also conserve.

K.E. of tube will be max when ball will at mid of tube.At this instant <C.O. momentum in Y-direction>

(2m+m)Vc = mu Vc = u/3

(Kmax)tube = 9

))(2(2

1 22 mu

Vm c

<By C.O. energy>

balltube KKmu )()(2

1minmax

2

18

7

92

1)(

222

min

mumumuK ball [TOPIC : CONSERVATION OF LINEAR MOMENTUM]

Q.1 to Q.2 are "Integer Type" questions. (The answer to each of the questions are Single digits)

[2 question × 4 marks (Moderate), (No Negative)]

Q.1 [5]

[Sol. Isys = 2

mr2

+

2

2

)r2(m2

mr× 6

= 2

55mr2 = 55 ] [TOPIC : ROTATIONAL MOTION]

PAGE # 4 XI (PTP,P1-P2 & Q) REVIEW TEST-5_[JEE ADVANCED]_PAPER-2_PHYSICS_ANSWER KEY, HINT & SOLUTION

ONLY FOR "P1 & P2, Q" BATCH

Q.1 [2 ]

[Sol. t = g

h2 =

10

202 = 2 s ] [TOPIC : K INEMATICS]

Q.2 [1][Sol. 20 × 0.5 = 35 × 0.2 + 20 × v

10 � 7 = 20 v

v = + 20

3 m/s = 0.15 m/s

e = 05.0

)15.0(2.0

=

10

1] [TOPIC : COLLISION]

Q.3 to Q.5 are "Integer Type" questions. (The answer to each of the questions are Single digits)

[3 question × 5 marks (Tough), (No Negative)]Q.3 [5]

[Sol.2

Lmg =

3

mL2

, a = L2

g3,

2

LN =

12

mL2

= 12

mL2

L2

g3,

N = 4

mg = 10 ] [TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.3 [5][Sol. The speed just after jerking,

v = gL2 cos 30° = 2

gL3

heat = 10 sec in kinetic energy

= m2

1

2

gL3gL2

= 4

mgL = 5 J ] [TOPIC : IMPULSE]

PAGE # 5 XI (PTP,P1-P2 & Q) REVIEW TEST-5_[JEE ADVANCED]_PAPER-2_PHYSICS_ANSWER KEY, HINT & SOLUTION

Q.4 [1]

v0

0

[Sol. Fr = macm

FrN

mgR Fr = II = 1/2 mR2

N = mgFr = µN

acm

= µg

a = 2µg / R

v0 � µgt = Rt

R

µg20

t = 1 sec. ][TOPIC : ROTATIONAL MOTION]

ONLY FOR "P1 & P2, Q" BATCH

Q.4 [1]

[Sol. Here arod = 2 aball< For Ball>

2T � mg = ma ...(i)< For Rod>

mg � T = m2a

or 2mg � 2T = 4ma

Equation (i) + (ii), mg = 5ma a = g/5 ar = a + 2a = 3a = 3g/5

Sr =2

2

1tar g

l

g

l

a

lt

r

10

5/3

22

= 10[TOPIC : NEWTON�S LAWS OF MOTION]

Q.5 [2 cm]

[Sol. Vbottom = gr5

2

1kx2 =

2

1mv2 =

2

5mgr

x = k

mgr5 =

4500

18.0108 =

45000

18 =

50

1 = 2 cm ]

[TOPIC : MOTION IN VERTICAL PLANE]