IIT JEE PHYSICS REVISION

Paragraph for Question Nos. 57 to 59 57 59 A swimmer can swim with a speed v in still water. v [\\ashok\S-09-10\XI\Test paper\PHYSICS\VIKAAS (A)\02-08-09_CT-2_Paper-2_Q.62,63,64]

[Made PSA 2009]

57.

If the swimmer crosses a swimming pool 'd' from A to directly opposite point B on other side in time t1 as shown in figure (i) and in a flowing river (river velocity 'u') of same width d from A to directly opposite point B on other bank in time t2 , then (t1/t2) is equal to : (Assume v > u) A B 'd' (swimming pool) t1 (

(i))

u

A

B

t2

(t1/t2)

: (

v > u)

v2 (A) 1 – 2 u

Sol.

(B)

d d  t1 = v ; v sw

t2 =

1

u2 v2

1–

u2 v2

(D) 1

d v 2  u2

 v 2  u2 d/ v t1   t = 2 2 =  v d/ v  u 2 

58.

(C*)

   = 

1–

u2 v2

.

If the minimum time taken in swimming pool to reach opposite bank is t1 and minimum. time to reach t 1 opposite bank in river is t2 , then the ratio  will have a value : t2

t1 t2

t1 t 2

:

(A*) 1 d ; v

(B)

t2 =

d v

v 2  u2 v

(C)

v 2  u2

(D)

u

u v

t1 = 1. t 2

Sol.

t1 =

59.

If the time taken by swimmer to reach opposite point on other bank in river is T1 and the time taken to travel



an equal distance upstream (against the water current) in the river is T2 , then ratio T1

T2 will have a value : T1

T2 T1

T2

(A)

1– u / v 1 u / v

1 u / v 1– u / v

(B*)

d Sol.

T1 =

2

v u

T2 = T1

so,

53.

and

2

(C)

: v 2  u2 ( v  u)

(D)

v 2  u2 v

d ( v  u)

T2 =

v 2  u2 = v u

v u = v u

1 u / v 1– u / v

Two particles A & B are projected as shown in fig in x–y plane. Under the effect of force which provide a constant acceleration a =11 m/s2 in negative y-direction. Then match situation in column-I with the  corresponding results in column-II (All positions are given in metre) ( V AB = velocity of A w.r.t. B; rAB = Position of A w.r.t. B) . [Made RAG Sir 2010]

Column–I

Column–II

(A)

Separation between the two particles is minimum at ....sec.

(p) 0

(B) (C)

Minimum separation between the two particles is ........m. Time when velocities of both particles are perpendicular each other at .... sec.

(q) 0.5 (r) 0.9

(D)

At the time of minimum separation V AB . r AB =

(s) 2 (t) 2 5

A a =11 m/s2

B, x–y ) ( V AB = A

Sir 2010]

y -I B

 rAB = A

-II B

)

( [Made RAG

Column–I

Column–II

(A) (B) (C)

....sec. ........m

(D)

V AB . r AB =

(p) 0 (q) 0.5 (r) 0.9 (s) 2 (t) 2 5

Ans.

(A) – (r), (B) – (t); (C) – (p), (s); (D) – (p)

tan  =

5 1 = 20 4

dmin =

425 sin   425 sin(   )

Sol.

tan  =

10 1 = 20 2

= 5 17 sin  cos   cos  sin  = 5 17  1  4  1  1  17 5 17   5 =

t=

t=

5 5

[2]  2 5

425 cos  500

=

5 17  2 4 1 1       10 5  5 17 5 17 

1 8  1  9 sec 10 10

 v1  (8ˆi  6ˆj)  (11t )ˆj  v 2  (12ˆi  16ˆj )  (11t )ˆj  v1  8ˆi  (6  11t )ˆj  v 2  12ˆi  (16  11t )ˆj   v 1.v 2 = –96 + 96 – 66 t – 176 t + 121 t2 0 = –242 t + 121 t2 = 0

t = 0 and

t=

242 2 121

52.

For the velocity–time graph shown in figure, in a time interval from t = 0 to t = 15 s, match the following : t = 0 t = 15 s [Arihant_New Pattern IITJEE_DC Pandey_Pg.22_Q.4]

Column–I (A) Change in velocity (in m/s) (B) Average acceleration (in m/s2) (C) Total displacement (in m) (D) Acceleration at t = 7 s (in m/s2) –I (A) (B) (C) (D) t = 7 s

Sol.

(m/s ) (m/s2 ) (m ) (m/s2 )

–4/3

Ans. (A)q, (B)p, (C)t, (D)s  v = 0 – 20 = – 20 m/s  v 20 4   a  = t = 15 = 3 m/s2 Displacemen

t=

Acceleration 6.

Column–II (p) –4/3 (q) –20 (r) –10 (s) –4 (t) –50 –II (p) (q) –20 (r) –10 (s) –4 (t) –50

=

1 1 × 5 × 20 + × 10 × (–20) = 50 m 2 2 20 = – 4 m/s2 . 5

Figure shows four situations in which a small block of mass 'm' is released from rest (with respect to smooth fixed wedge) as shown in figure. Column-II shows work done by normal reaction on the block (with respect to an observer who is stationary on ground) till block reaches at the bottom of inclined wedge, match the appropriate column. [Made CSS 2011-12] 'm' ( )

-II ( ):

Column–I

(A)

[Made CSS 2011-12] Column–II

v = 2gh

h

(p) Positive 45°

v = 2gh

(B) h

(q) Negative 45°

h 45°

(C)

(r) equal to mgh in magnitude

mgh

45° v = 2gh

h

(D) v = 2gh

(s) equal to zero 45°

Ans. 7.

(A) – p ; (B) – p ; (C) – s ; (D) – q

In all cases in column–I, the blocks are placed on the smooth horizontal surface. [Made CPG sir 2010-11] Column–I Column–II (A)The initial velocities given to the blocks (p) Centre of mass of the complete system shown when spring is relaxed are as shown will not move horizontally (friction is absent)

System (two blocks + spring) (B) A constant force is applied on 2 kg block. shown Springs are initially relaxed & friction is absent

(q) Centre of mass of the complete system will move horizontally

System (three blocks + two springs) (C) There is no friction between plank and ground be and initially system is at rest. Man starts moving on a large plank with constant velocity.

(r) Mechanical energy of the system will conserved

System (man and plank) (D) Two trolleys are resting on a smooth horizontal crease surface and a man standing on one of the trolleys jumps to the other with relative velocity of 4 m/s

(s) Mechanical energy of the system will in-

System (two trolleys + man) –I, [Made CPG sir 2010-11] –II

–I (A)

(p)

(

+

)

(B) 2 kg

(q)



( (C)

+

) (r)

(

)

(D)

(s) 4 m/s

(

Sol.

+

)

Ans. (A) – p,r ; (B) – q, s ; (C) – p,s ; (D) – p,s A – p,r,t F = 0 So, linear momentum conservation and centre of mass will not move. B – q,s So, linear momentum will not be conserved and centre of mass will accelerate W ext = E. , Wext = E. C – p,s,t D – p,s,t

Paragraph for Question Nos. 57 to 59

57

59

A block of mass m slides down a wedge of mass M as shown. The whole system is at rest, when the height of the block is h above the ground. The wedge surface is smooth and gradually flattens. There is no friction between wedge and ground. m M m h [Aakash(IIT–JEE 2009)_Success Point(Physics)_Pg.66_C-25]

57.

As the block slides down, which of the following quantities associated with the system remains conserved? (A) Total linear momentum of the system of wedge and block (B*) Total mechanical energy of the complete system (C) Total kinetic energy of the system (D) Both linear momentum as well as mechanical energy of the system

(A) (B*) (C) (D)

(

)

58.

If there would have been friction between wedge and block, which of the following quantities would still remain conserved ? ? (A*) Linear momentum of the system along horizontal direction (B) Linear momentum of the system along vertical direction (C) Linear momentum of the system along a tangent to the curved surface of the wedge (D) Mechanical energy of the system (A*) (B) (C) (D)

59.

If there is no friction any where, the speed of the wedge, as the block leaves the wedge is :

(A*) m

2gh (M  m) M

(B) M

2gh (M  m) m

(C) ( 2gh )

m Mm

(D) ( 2gh )

M Mm

Sol.(57,58,59) (57) Linear momentum is conserved only in horizontal direction. (58)

Net Fext on system is zero in horizontal direction therefore linear momentum is conserved only in horizontal direction. F

(59)

mv 1 = Mv 2 1 1 mv 12 + Mv 22 = mgh 2 2 From (i) & (ii),

v2 = m 51.

.......(i) .......(ii)

2gh . (M  m) M

A spring balance reads w1 when a ball is suspended from it. A weighing machine reads w2 when a tank containing liquid is kept on it. When the ball is immersed in the liquid, the spring balance reads w3 and weighing machine reads w4 : [Made CPG sir 2010-11] w1 w2 w3 w4 (A*) w1 > w3 (B*) w2 < w4 (C*) w1 + w2 = w3 + w4 (D) w1 + w4 = w2 + w3

Sol.

Only ball w3 + FB = w1 w1 > w3 Only tank w2 + F B = w4 w4 > w2 Both w2 + w4 = w1 + w2 . 57.

Match the column : Column–I

[Made CPG sir 2010-11] Column–II

(A)

(p) Speed of component travelling wave is portion

Two strings each of length  and linear mass

AP will be

T 

density  and 9 are joined together and system is oscillated such that joint P is node T is tension in the strings. A and B are fixed ends.

(B)

(q) Speed of component travelling wave in the

Two strings each of length  and linear mass density  and 9 are joined together and system is oscillated such that joint P is antinode. T is tension in each string.A and B are fixed ends.

portion AP will be more than that in portion BP.

(C)

(r) Frequency of oscillation of the system AB can

P is the mid–point of the string fixed at both ends.

be

1 T 2 

T is tension in the string and  is its linear mass density.

(D)

(s) Frequency of oscillation of the system AB can

T is the tension in the string fixed at A and B is free

be

1 4

T 

end. P is mid–point. is its the linear mass density. (t) Wavelength of the wave in the portion PB can

be :

2 . 3

[Made CPG sir 2010-11]

–I

–II

(A)

(p) AP

T 

  9 P (

T,

A

B

)

(B)

(q) AP



BP

 9 P A

B

T, (

)

(C)

(r)

AB

1 T 2 

(s)

AB

1 4

P T



(D) A

B

T

T 

P

 2 3

(t) PB Ans.

(A) – p,q,r,t ; (B) – p,q,s ; (C) – p,r,s,t ; (D) – p,s

Paragraph for Question Nos. 19 to 20 19 20 A metal ball (Neutral) with radius r is concentric with hollow metal sphere of radius ‘R’, having charge ‘Q’ as shown in figure, Now ball is connected with a very long wire to earth. Then : r

‘R’ ‘Q’

, :

made by nsb sir 2012-13

Q,R r

19.

Potential difference between sphere and metal ball, after grounding is :

(A)

kQ R

(B) 

kQr R

2

kQ  r (C*) R 1  R   

Ans. (c) 20.

After grounding : (A) net electric field between sphere and ball is zero.

(B) electric field between ball and sphere is zero due to ball only.

(C*) electric field between sphere and ball due to ball is non–zero.

(D) electric field between sphere and ball is non-zero, due to sphere

Ans. (c) Q,R

Sol.

r x

Vball  0 KX KQ  0 r R x

Qr R

(D)

kQ  R  1  r  r

KQ Kx  R R because potential difference depend only on charge on inner surface after electrostatic condition is reached after grounding. Vs  Vb 

Potential difference

Vs  Vb 

KQ  r 1   R  R

Paragraph for Question Nos. 23 to 24 23 24

COMPREHENSION

[M_Bank _C.E._9.7]

A car battery with a 12 V emf and an internal resistance of 0.04  is being charged with a current of 50 A. 12 V

0.04 

50 A

(charged)

[Made A.K.S. sir]

23.

The potential difference V across the terminals of the battery are V (A) 10 V (B) 12 V (C*) 14 V

(D) 16 V

Sol.

V = E + ir (during charging) ( = 14 V.

)

24.

The rate at which energy is being dissipated as heat inside the battery is :

Sol.

(A*) 100 W (B) 500 W 2 P = I r (Due to internal resistance) ( = 502 × 4 × 10–2 = 100 W

(C) 600 W

(D) 700 W )

Paragraph for Question Nos. 13 to 14 13 14 In the arrangement shown in the figure when the switch S2 is open, the galvanometer shows no deflection 5 L . The internal 12 resistance (r) of 6 V cell, and the emf E of the other battery are respectively. Wire AB is potentiometer wire and resistance of other conducting wires is negligible. (Internal resistance of cell E is negligible) S2  = L/2

for  = L/2. When the switch S2 is closed, the galvanometer shows no deflection for  =

=

S2 E )

5 L 12 AB

6V

(r) (

E

[Only (JP,JF,JR)]

13.

Calculate emf of cell E : (A) 6 V (B) 5 V

E (C*) 12 V

14.

Calculate the internal resistance 'r' : (A) 1  (B*) 2  Sol.(13 to 14) case -I S2 is open

Potential gradient so

6=

E = 12 V case -II

i=

6 10  r

=

E L L 2

S2 is closed

E L

(D) 10 V 'r'

(C) 3 

(D) zero

6 – ir = 6–

E 5L L 12

6 r=5 10  r

6r =1 r  10 6r = r + 10 r = 2

15.

If the current in 8  resistance is 2A then the current through resistance 'R' (in ampere) would be : 8 2A 'R' [Made OP sir 2010-11]

Ans.

6

Sol.

4

Current in 4 resistance = 4A Total current = 4 + 2 = 6A

Paragraph for Question Nos. 11 to 12 11 12 A small particle of mass m = 1kg and charge of 1C enters perpendicularly in a triangular region of uniform magnetic field of strength 2T as shown in figure : 2T m = 1kg 1C :

[USED P,F,R 2012]

[Made Kamal sir 2010-11]

[Only (JP,JF,JR)]

11.

Calculate maximum velocity of the particle with which it should enter so that it complete a half–circle in magnetic region : : (A) 2 m/s (B) 2.5 m/s (C*) 3 m/s (D) 4 m/s

Sol.

In triangle PMC cos53º =

PMC MP MC

3 R = 5 4R 12 = 8R

R=

3 m (R is the maximum radius of half–circle) ( 2

Rmax = 12.

mumax qB



R

)

Umax = 3 m/s.

In previous question, if particle enters perpendicularly with velocity 48 m/s in magnetic region. Then, how much time will it spend in magnetic region : 48 m/s :

Sol.

(A)

11 sec. 360

R=

mu = 24 m qB

(B*)

7 sec. 360

(C)

13 sec. 360

(D)

17 sec. 360

Let, MPQ =  By geometry, CPO = (37 – ) In CPO ,

OC OP = sin(CPO) sin(PCO) 20 24 = sin(37º ) sin(180 º 37º ) 5 56 = sin(37º ) 3 sin(37º – ) = =

1 2

7 rad. 180

qB m  = 2 rad/sec.

=

t= 11.

7 sec. 360

Three indentical bulbs each of resistance 2 are connected as shown. The maximum power that can be consumed by individual bulb is 32W, then the maximum power consumed by the combination is : 2 32W , : [Made RKS sir 2012-13]

Sol.

12.

(A*) 48 W (B) 96 W P = i2 R 32 = i2 (2) i = 4A Pmax = (2)2 (2) + (22)2 + (4)2 (2) = 8 + 8 + 32 = 48 W

(C) 128 W

(D) 160 W

Two coaxial rings are separated by a distance x. R is radius of larger ring having charge Q uniformly distributed on it.The other ring is having very small radius r (