Crackiitjee.in.Phy.ch23

MECHANICS: ILLUSTRATIONS

1. A particle moves along a path ABCD. Then the magnitude of net displacement of the particle from position A to D is:

(A) 10 m (B)

5 2 m

(C) 9 m (D)

7 2 m

Sol. As can be seen

The displacement is

   

AF 2  FD2  7 2 m

2. A truck travelling due north at 50 km/hr turns west and travels at the same speed. What is the in velocity.

(A) 50 km/hr north-west (B) 50 2 km/hr north-west (C) 50 km/hr south-west (D) 50 2 km / hr south-west Sol. V_i  50jˆ   f ˆ V 50 i

 

V

V

_f



V

_i =



50 i 50 j

ˆ



ˆ

crackiitjee.in



 



 V 50 2  50 2

=

50 2

 

V 50 2 SW

3. A boy start towards east with uniform speed 5m/s. After t = 2 second he turns right and travels 40 m with same speed. Again he turns right and travels for 8 second with same speed. Find out the displacement; average speed, average velocity and total distance travelled.

Sol. The particle starts from point A & reaches point D passing through B & C Now, AE = 40 m DE = 30 m  Displacement = AD =

AE

2



DE

2 =

40

2



30

2 = 5 cm

Time taken in the motion

= tAB + tBC + tCD

=

2



40



8

8

= 18 sec

Total distance travelled

= AB + BC + CD

= 10 + 40 + 40

= 90 m

Average velocity =

Displacement

time

=

50

18

=

25

m / s

9

Average speed =

Distance

time

=

90

18

= 5 m/s

A radius vector of point A relative to the origin varies with time t as



ˆ



2

ˆ

r

at i bt j

where a and b are constants. Find the equation of point's trajectory.



ˆ



2

ˆ

r

at i bt j

X = at y = – bt2

crackiitjee.in

t =  _{ }   x a    _{   }   2 x y b a







2 2

b

y

x .

a

4. A man moves on his motor bike with 54 km/h and then takes a u turn (180°) and continues to move with e speed. The time of u turn is 10s. Find the magnitude of average acceleration during u turn.

(A) 0 (B) 3 ms–2 (C) 1.5 2 ms–2 (D) none of these Sol. V_i 15

 

ˆi  f ˆ V 15 i

 





 V 15 iˆ 15 ˆi 54 km/h =

54



5

18

= 15 m/s <a> = 15 



15



10

crackiitjee.in

= 3 m/s2

A particle whose speed is 50 m/s moves along the line from A(2, 1) to

B(9, 25). Find its velocity vector in the from of

ai b j.

ˆ



ˆ



ˆ

ˆ

14 i 48 j

Position vector of point A =

2 i

ˆ ˆ



j

Position vector of point B =

9 i 25 j

ˆ



ˆ



 



 AB  9 i 25 jˆ ˆ  2 iˆ ˆ j

=

7 i 24 j

ˆ



ˆ

Unit vector in the direction of AB  AB AB AB =

7 i 24 j

ˆ



ˆ

25



vector

vector

unit vector



V 50 AB

=

14 i 48 j m / s

ˆ



ˆ

11. A particle is moving with initial velocity u   ˆ ˆi j k.ˆ What should be its acceleration so that it can remain moving in the same straight line without change of direction?

(A) a 2 i 2 j 2k ˆ ˆ ˆ

(B) a  2 i 2 j 2kˆ ˆ ˆ

(C) a 3 i 3 j 2k ˆ ˆ ˆ

(D)

a 1 i 1 j



ˆ



ˆ

Sol. To move in a straight line

a || u

In A,



a 2u

In B,

 

a

2u

 After sometime direction will get reversed of same straight.

12. Two balls are moving on the same smooth horizontal plane. Their velocity components along one edge of the square plane are 10 3 & 20 m/s. Their velocity components along a perpendicular edge are 30 & 20 m/s. Find the angle between their directions of motion.

Sol. V_A 10 3 i 30 j



ˆ ˆ







  B ˆ ˆ V 10 2 i 2 j   A B A B V .V cos V V =





 

  100 2 3 6 10 3 9 10 2 2





 

   2 3 3 cos 2 3 2 2 = 3 3 2 6 = 3 2 2  = 15°

13. An insect starts from rest from point (3, 4) and moves with an acceleration 2 2 m/s2 in x – y plane along a line, equally inclined to both the axis. After 3 sec insect turns towards right in perpendicular direction without wasting any time and keeping speed same at the movement of turning. For the further motion acceleration is 2 2 m/sec2 in the direction of motion. Find the position of insect after 5 seconds from the starting.

Sol. In 1st three seconds





 



 

 ˆ ˆ  1  ˆ ˆi j  2 S 3 i 4 i 2 2 3 2 2 =

12 i 13 j

ˆ



ˆ

V =

 

  ˆ ˆ_{i j} 2 2 3 2 =

6 i 6 j

ˆ



ˆ



ˆ



ˆ

v' 6 i 6 j

a' =

 

 ˆ ˆ 2 2 i j 2 =

2 i 2 j

ˆ



ˆ

In next 2 seconds





 

 

_S

_{6 i 6 j}

ˆ



ˆ

 

₂

1

_{2 i}

ˆ ˆ



_j



₂

2

2

=

12 i 12 j 4 i 4 j

ˆ



ˆ



ˆ



ˆ

=

16 i 16 j

ˆ



ˆ

  

S' S

S

=

12 i 13 j 16 i 16 j

ˆ



ˆ



ˆ



ˆ

crackiitjee.in



ˆ



ˆ

S' 28 i 3 j

 –

3 i 4 j

ˆ



ˆ

(A) It speed of a body is varying, its velocity must be varying and it must have zero acceleration

(B) It velocity of a body is varying, its speed must be varying (C) A body moving with varying velocity may have constant speed

(D) A body moving with varying speed may have constant velocity if its direction of motion remains constant.

Sol. If speed of a particle changes, the velocity of the particle definitely changes and hence the acceleration of the particle is nonzero.

Velocity of a particle change without change in speed. (In uniform circular motion)

When speed of a particle varies, its velocity cannot be constant.

8. The position vector of a particle is given as r = (t2 – 4t + 6) ˆi 

 

t j.2 ˆ The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to

(A) 1 sec (B) 2 sec (C) 1.5 sec (D) not possible Sol. r 



t2 4t 6 i

  

ˆ t j;2 ˆ r-



dr

V

dt

=



2t 4 i 2t j, a



ˆ ˆ =

dv



2 i 2 j

ˆ



ˆ

dt

If

a

and v are perpendicular



a.v 0









 



2 2t 4

2 2t

0

   4t 8 4t 0 t = 1 sec.  t = 1 sec.

9. The velocity time relation of an electron starting from rest is given by v = K t, where K = 2 m/s2. The distance traversed in 3 sec is :

(A) 9 m (B) 16m (C) 27 m (D) 36m Sol. a =

dv

dt

= k = 2 m/s2 a = constant S =

0



1

  

2 3

2

2

crackiitjee.in

= 9 m.

10. A particle has an initial velocity (i.e., at t = 0) of

3 i 4 j

ˆ



ˆ

(m/s) and an acceleration of 0.3

ˆi

+ 0.4

ˆj

(m/s2). The speed of particle at t = 10 sec is : (A) 7 m/s (B)

7

m / s

2

(C)

7 2 m / s

(D)

14 2 m / s

Sol.

v u at

 

when

a

is constant  velocity at t = 10 sec. is v 



3 i 4 jˆ ˆ

 

10 0.4 i 0.3 jˆ ˆ



=

7 i 7 j

ˆ



ˆ

 

v 7 2 m / s

14. A particle P is moving with a constant speed or 6 m/s in a direction   ˆ

ˆ ˆ

2i j 2k. When t = 0, P is at a point whose position vector is   ˆ

ˆ ˆ

3i 4j 7k. Find the position vector of the particle P after 4 seconds.

Sol. V  62iˆ ˆ j 2kˆ

3

= 2 2i



ˆ ˆ j 2kˆ



The equation of unit vector along 2iˆ ˆ j 2kˆ is

=     2 2 2 ˆ ˆ ˆ 2i j 2k 2 1 2 = 2iˆ ˆ j 2kˆ 3  ₀  r r vt = 3i 4j 7k 4 2 2iˆ ˆ ˆ 



ˆ ˆ j 2kˆ



 r 19i 4j 23kˆ ˆ ˆ

15. The velocity of a car moving on a straight road increases linearly according to equation, v = a + bx, where a & b are positive constants. The acceleration in the course of such motion : (x is the distance travelled) (A) increases (B) decreases (C) stay constant (D) becomes zero

crackiitjee.in

Sol. V = a + bx (V increases as x increases)





dV

dx

b

bV

dt

dt

hence acceleration increases as V increases with x.

16. The displacement of a body from a reference point is given by,





x

2t 3,

where 'x' is in meters and t in seconds. This shows that the body: (A) rest at t =

3

2

(B) is accelerated (C) is decelerated (D) is in uniform motion Sol. x = (2t – 3) x = (2t – 3)2 V =

dx



2 2t 3 2





 

dt

= 4(2t – 3) = 86 – 12 V = 4(2t – 3) = 0  2t – 3 = 0

crackiitjee.in

 

t

3

sec

2

rest at t = 8 a = 8 m/s a =

dV



8 m / s

2

dt

17. The position of a particle moving rectilinearly is given by X = t3 – 3t2 – 10. Find the distance travelled by the particle in the first 4 seconds starting from t = 0. Sol. X = t3 – 3t2 – 10 v =

dx

dt

= 3t2 – 6t v = 0; gives t = 0 & t = 2 sec.

Velocity will become zero at t = 2 sec., so particle will change direction after t = 2 sec.

At t = 0

x(0 sec) = – 10

At t = 2 sec.

X(2 sec) = 23 – 3(2)2 – 10 = 8 – 12 – 10 = – 14 At t = 4 sec. x(4 sec) = 43 – 3(4)2 – 10 = 64 – 48 – 10 = 6 Distance travelled = x1 + x2 = |– 14 – (– 10)| + |6 – (–14)| = 4 + 20 = 24

Distance Travelled = 24 units.

18. A particle moving with uniform acceleration along x - axis has speed v

m/s at a position x metre given by _   _

  180 0 x 16   V 180 16x

The acceleration of the particle is m/s2 is (A) – 16 (B) – 8 (C) 164 (D)





8

160 16x

Sol. v2 = 180 – 16x

crackiitjee.in

Taking

d

dx

of both sides

 

2









d

d

v

180 16x

dx

dx



2v

dV

 

16

dx

or a =

vdv

dx

= – 8 m/s2. Hence acceleration is – 8 m/s2.

19. Three vectors of equal magnitude A are inclined at an angle of 60° with each other. The magnitude of the resultant will be:

(A) zero (B) A (C) A 6 (D) cannot be calculated Sol.

R



A B C

 

R =



A B C . A B C 

 

 



= A2 B2 C2 2A.B 2B.C 2A.C 

crackiitjee.in

=  2 2 6A 3A 2 =

6A

2 = A 6

20. A man in a lift ascending with an upward acceleration 'a' throws a ball vertically upwards with a velocity 'v' with respect to himself and catches it after 't1' seconds. After wards when the lift is descending

with the same acceleration 'a' acting downwards the man again throws the ball vertically upwards with the same velocity with respect to him and catches it after 't2' seconds ?

(A) the acceleration of the ball w.r.t. ground is g when it is in air

(B) the velocity v of the ball relative to the lift is



1  2



1 2

g t t

t t

(C) the acceleration 'a' of the lift is







 2 1 1 2

g t t

t t

(D) the velocity 'v' of the ball relative to the man is



₁ 1 2₂



gt t

t t

Sol. For first case (when lift is ascending)

t1 =



2v

g a

…(i)

for second case (when lift is descending)

t2 =



2v

g a

…(i)

on solving equation (i) and (ii) we get

V = 1 2 1 2 gt t t t    1 2 t g a t g a gt1 + at1 = gt2 – at2 a(t1 + t2) = g(t2 – t1) a =











2 1 1 2 g t t t t

21. A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes 20 m/s and in tenth second it travels a distance of 10m. Then the acceleration of the particle will be–

(A) 10 m/s2 (B) 20 m/s2 (C)

1

5

m/s2 (D) 3.8 m/s2 Sol. V = u + at  20 = u + a × 10 …(1) Sn =









a

u

2n 1

2

crackiitjee.in







10 u

 

a

2 10 1





2

…(2)

on solving (1) and (2) we get

a = 20 m/s2

22. A particle moving along a straight line with a constant acceleration of – 4 m/s2 passes through a point A on the line with a velocity of + 8 m/s at some moment. Find the distance travelled by the particle in 5 seconds after that moment.

Sol. u = + 8 m/s a = – 4 m/s2

V = 0

 0 = 8 – 4t or t = 2 sec.

displacement in first 2 sec.

S1 = 8 × 2 +

1

2

. (– 4). 22 = 8m S1 = 8 × 2 +

1

2

(– 4) (2)2 = 16 – 8 = 8 m

displacement in next 3 sec.

S2 = 0 × 3 +

1

2

(– 4) 32 = – 18 m. S2 = 0 × 3 +

1

2

(– 4) (3)2 = – 18 m. distance travelled = |S1| + |S2| = 26 m. Ans. 26 m ALTER: total distance =

1

2

× 2 × 8 +

1

2

× 3 × 12 = 8 + 18 = 26 m

23. A person throws a ball vertically up in air. The ball rises to maximum height and then falls back down such that the person catches it. Neglect the friction due to air. While the ball was in air, three statements are given below. (g = 9.8 m/s2)

Statement 1:

Just after the bait leaves the persons hand, the direction of its acceleration is upwards.

Statements 2:

The acceleration of ball is zero when it reaches maximum height. Statement 3:

The acceleration of ball is g = 9.8 m/s2 downwards while the ball is falling down.

Then which of the above statement or statements are correct in the options below.

(A) Statements 1 only (B) Statement 2 only (C) Statement 3 only

(D) Both statement 2 and statement 3

Sol. (C) The acceleration of ball during its flight is g = 9.8 m/s2 downwards.

24. A stone is projected vertically upwards at t = 0 second. The net displacement of stone is zero in times interval between t = 0 second to t = T seconds. Pick up the INCORRECT statement.

(A) From time t =

T

4

second to t =

3T

4

second, the average

velocity is zero.

(B) The change in velocity from time t = 0 to t =

T

4

second is same

as change in velocity from t =

T

8

second to t =

3T

8

second

(C) The distance travelled from t = 0 to T =

T

4

second is larger than

distance travelled from t =

T

4

second t =

3T

4

second

(D) The distance travelled from t =

T

2

second to t =

3T

4

second is

half the distance travelled from t =

T

2

second to t = T second.

Sol. (D) At t =

T

4

and t =

3T

4

, the stone is at same height,

Hence average velocity in this time interval is zero.

Change in velocity in same time interval is same for a particle moving with constant acceleration.

Let H be maximum height attained by stone, then distance travelled from

t = 0 to t =

T

4

is

3

4

H and from t =

T

4

to t = T distance travelled is

H

.

2

From t =

T

2

to t = T sec distance travelled is H and from t =

T

2

to t =

3T

4

distance travelled is

H

.

4

crackiitjee.in

25. The velocity displacement graph of a particle moving along a straight line is shown in figure.

Then the acceleration displacement graph is. Sol. (C) From the graph, we can write

V = – x + 2 a =

dv

dt

 





vdv

v 1

dx

= – v = – (– x + 2) = x – 2

26. A point moves in a straight line under the retardation a v2, where 'a' is a positive constant and v is speed. If the initial velocity is u, the distance covered in 't' seconds is:

(A) a u t (B)

1

a

log (a u t) (C)

1

a

log (1 + a u t) (D) a log (a u t)

Sol. The retardation is given by

 

2

dv

_av

dt

integrating between proper limits

 



v ₂ 



b u 0 dv _{a dt} V or

1



at



1

v

u



dt



at



1

dx

u







u dt

dx

1 aut

integrating between proper limits

  



s



t 0 0 u dt dx 1 aut







S



1

ln 1 aut



a

27. A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as a =

k v,

where k is a positive constant. At the initial movement the velocity of the point is equal to v0. What distance will it traverse before it stops ? What time will it take

to cover that distance ?

Sol. t0 = 0

2

_{v ,}

k

S =

2

v

3 /2₀

3k

28. A particle starts moving rectilinearly at time t = 0 such that its velocity 'v' changes with 't' according to the equation v = t2 – 1 where t is in seconds and v is in m/s. Find the time interval for which the particle retards.

Sol. Acceleration of the particle a = 2t – 1

The particle retards when acceleration is opposite to velocity.

 a.v < 0

 (2t – 1) (t2 – 1) < 0

 t (2t – 1) (t – 1) < 0 now t is always positive

 (2t – 1) (t – 1) < 0

 either 2t – 1 < 0 & t – 1 > 0

 t <

1

2

& t > 1

This is not possible.

or 2t – 1 < 0 & t – 1 < 0



1

 

t 1

2

Ans. or 2t – 1 < 0 & t – 1 < 0  2t > 1 & t < 1 t >

1

2



1

 

t 1

2

Ans.

29. A particle starts moving along a straight line at t = 0. Its acceleration - time graph is shown. Correct relation between magnitudes of displacements between time durations t = 2s to t = 3s and t = 7s to t = 8s represented by S23 and S78 respectively is :

(A) S23>S78 (B) S78 > S23

(C) S23 = S78 (D) S78 > S23

Sol. (A) At t = 6s, particle again is at rest having same acceleration but in opposite direction.

Thus S23 > S78.

30. A particle is projected vertically upwards in vacuum with a speed u. (A) When it rises to half its maximum height, its speed becomes

u

_.

2

(B) When it rises to half its maximum height, its speed becomes

u

_.

2

(C) The time taken to rise to half its maximum height is half the time taken to reach its maximum height.

(D) The time taken to rise to three - fourth of its maximum height is half the time taken to reach its maximum height.

Sol. At maximum height, velocity = 0

H = 2 u & 2g At height h =

H

2

V2 = u2 – 2gh V2 =   2 2 2 u u u 2g 4g 2  2 2 u V 2



V



u

2

Time taken to rise to maximum height u

T =

u

g

for height h =

H

,

2

t =















u u

2

g

=





 2 1 u 2g

Time taken to rise to

3

4

H = T – time taken to fall down by

H

4

=

T



T



T

2

2

31. A particle has initial velocity,

v 3i 4j



ˆ



ˆ

and a constant force



ˆ



ˆ

F

4i 3j

acts on the particle. The path of the particle is : (A) Straight line

(B) parabolic (C) circular (D) elliptical Sol. (B)

For constant acceleration if angle between initial velocity makes an oblique angle with acceleration then path will be parabolic.

32. A particle moves rectilinearly with a constant acceleration 1 m/s2. Its speed after 10 seconds is 5 m/s. The distance covered by the particle in this duration is (Initial & final velocities are in opposite direction)

Sol. For the given problem, velocity time graph will be as below:

Form the graph distance = _   _

 

1

2 5 5

2

= 25 m

33. A particle moves through the origin of an xy - co-ordinate system at t = 0 with initial velocity u = 4i – 5 j. The particle moves in the xy plane with an acceleration a = 2i m/s2. Speed of the particle at t = 4 second is : (A) 12 m/s (B) 8 2 m/s (C) 5 m/s (D) 13 m/s Sol. Using vx = ux + axt = 4i + (2i) 4 = 12 i

As ay = 0, velocity component in y - direction remains unchanged. Final

velocity = 12 i - 5j speed at t = 4 sec. = 122  

 

52 13 m / s.

34. Two particles at a distance 5m apart, are thrown towards each other on an inclined smooth plane with equal speeds 'V'. It is known that both particles moves along the same straight line. Find the value of v if they collide at the point from where the lower particle is thrown. Inclined plane is inclined at an angle of 30° with the horizontal. [take g = 10 m /s2]

(A) 2.5 m/sec (B) 5 m/sec (C) 7.5 m/sec (D) 10 m/sec Sol. Down the plane

5 = V.t + t2 …(1) 5 =

Vt



1

g sin t



2

2

up the plane 0 = V – g sin q t1 =



V

gsin

0 = V – t' =



V

gsin

crackiitjee.in

 _{t 2t}1 t = 2t1 =



2V

gsin

5 =     2 2 2 2 2.V 1 gsin .4v gsin 2 g sin 10 g sin  = 8v2 v =

 



_{  }

 

1

10 10

2

8

= 100 16 =

10



2.5 sec.

4

Alternate 2vt = 5 …(1)





2V

t

gsin









2v

2v

5

gsin

  v  5gsin  5 4 4

crackiitjee.in

35. A stone is dropped from the top of a tower. When it is gallen by 5 m from the top, another stone is dropped from a point 25 m below the top. If both stones reach the ground at the same moment, then height of the tower from ground is : (take g = 10 m/s2)

(A) 45m (B) 50m

(C) 60m (D) 65m

Sol. Vel of 1st stone when passing at A  V2 = 0 + 2.10.5 V = 10 m/s S1 – S2 = 20 m.  _  _  _      2  2 1 1 10.t 10.t .10.t 20 2 2 t = 2s S2 =

1

_.10.4

2

= 20 m Ht = 25 + 20 = 45 m.

36. Two bikes A and B start from a point. A moves with uniform speed 40 m/s and B starts from rest with uniform acceleration 2 m/s2. If B start at t = 0 and A starts from the same point at t = 10s, then the time during the journey in which A was ahead of B is:

(A) 20s (B) 8s (C) 10s

(D) A is never ahead of B Sol. A will be ahead of B when XA > XB

40 (t – 10) > (0) t +

1

2

(2)t2

as A is 10 sec. late than B.

 t2 – 40 t + 400 < 0

 (t – 20)2 < 0

Which is not possible? So A will never be ahead of B.

38. A tiger running 100 m race, accelerates for one third time of the total and then moves with uniform speed Then the total time taken by the tiger to run 100m if the acceleration of the tiger is 8m/s2 is:

(A) 3 5 s (B) 3 5 s (C) 12 s (D) 9 s

Sol. Let the total time of race be t seconds and the distance be S = 100 m. The velocity vs time graph is

Area of OAD

 

s

5

    _{ }   2 s 1_a T 5 2 3 =  _{ }   2 1₈ T 2 3 or T = 3 5 m/s.

40. Velocity (v) versus displacement (S) graph of a particle moving in a straight line, corresponding acceleration (a) versus velocity (v) graph will be

Sol. From the given relation between velocity (V) and displacement (S) is given by v = S





d

1

dS

Hence, acceleration a =



d



dS

a =  × 1 a = 

Therefore, graph between acceleration and velocity will be as shown.

42. A particle is projected-at angle 60° with speed

10 3,

from the point 'A'. At the same time the wedge is made with speed 10 3 towards

right as shown in the figure. Then the time after which particle will strike with wedge is (g = 10 m/sec2):

(A) 2 sec (B) 2 3 sec

(C)

4

3

sec (D) none of these

Sol. Suppose particle strikes wedge at height 'S' after time t. S = 15t –

1

2

10t2

= 15 t – 5 t2. During this time distance travelled by particle in horizontal

direction = 5 3 t. Also wedge has travelled extra distance X =



S

tan30

=  2 15t 5t 1 3

Total distance travelled by wedge in time

t = 10 3t

= 5 3 t + 3 (15 – 5t2)

 t = 2 sec. Alternate

(by Relative Motion)

T = 1 1 2u g =

10 3 sin 30

 

5 3

= 2 10 3 1 10 3 = 2 sec.

43. A particle is projected from a point (0, 1) on y - axis (assume + Y direction vertically upwards) aiming towards a point (4, 9). It fell on ground along x axis in 1 sec.

Taking g = 10 m/s2 and all coordinate in meters. Find the X - coordinate where it fell (A) (3, 0) (B) (4, 0) (C) (2, 0) (D) ( 2 5, 0) Sol. tan  =







9 1

₂

4 0

now, – 1 = u sin (1) –

1

2

g(1)2 u sin  = 4 and sin  =

2

5

 u 2 5

crackiitjee.in

horizontal direction & Y axis is vertically upwards. Take g = 10 m/s2. Find

(i) Y component of initial velocity and (ii) X component of initial velocity Sol. From graph (1) Vy = 0

at t =

1

2

sec.

i.e., time taken to reach maximum height H is

t = uy  1

g 2



u

_y



5 m / s

…Ans. (i) from graph (2): Vy = 0 at X = 2m

i.e. when the particle is at maximum height, its displacement along horizontal X = 2 m  _x  X u t



2 u



_x



1

2

 U_x  4 m / s …Ans. (ii)

47. A particle P is projected from a point on the surface of smooth inclined plane (see fig.) Simultaneously another particle Q is released on the

smooth inclined plane from the same position. P and Q collide after t = 4 second. The speed of projection of P is

(A) 5 m/s (B) 10m/s (C) 15 m/s (D) 20 m/s Sol. (B)

It can be observed from figure that P and Q shall collide if the initial component of P along incline. uII = 0 that is particle is projected perpendicular to incline.  Time of flight T = 



2u

gcos

=



2u

gcos



 

u

gT cos

2

= 10 m/s

49. A stone is projected from a horizontal plane. It attains maximum height 'H' & strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assuming the collision to be elastic the height of the point on the wall where ball will strike is

(A)

H

2

(B)

H

4

(C)

3H

4

(D) None of these Sol. H =

1

g 2t

 

2

2

= 2gt2 …(1) h =

1

gt

2

2

…(2) By (1) & (2) h =

H



H

4

=

3H

4

50. Some students are playing cricket on the roof of a building of height 20 m. While playing, ball falls on the ground. A person on the ground returns their ball with the minimum possible speed at angle 45° with the horizontal find out the speed of projection.

Sol. Let us assume that person throws ball from distance x. Taking point of projection origin.

By equation of trajectory y = X tan  –  2 2 2 1 gx 2 U cos

crackiitjee.in

20 = x tan 45° – 2 2 2 1 10x 2 U cos 45 U2 =  2 10x x 20 …(1)

For 'U' to be minimum

du



0

dx

On differentiation w.r.t. 'X'









       2 2 10 2x x 20 10x 1 du 2U 0 dx _{x 20} = 10 (x – 40) x = 40

For x greater than 40, slope is positive & x less than 40, slope in negative.

So at x = 40 There is minima

Required minimum velocity from Eq. 1

   2 2 min 10 40 U 40 20 Umin = 800 Umin =

20 2 m / s

51. A train is standing on a platform, a man inside a compartment of a train drops a stone. At the same instant train starts to moves with constant

acceleration. The path of the particle as seen by the person who drops the stone is:

(A) parabola

(B) straight line for sometime & parabola for the remaining time (C) straight line

(D) variable path that cannot be defined

(E) Relative to the person in the train, acceleration of the stone is 'g' downward, (acceleration of train backwards).

Sol. According to him :

x =

1

2

at2, y =

1

2

gt2



X



a

Y

g



Y



g

x

a

 straight line.

52. Two men P and Q are standing at corners A & B of square ABCD of side 8 m. They start moving along the track with constant speed 2 m/s and 10 m/s respectively. Find the time when they will meet for the first time.

(A) 2 sec (B) 3 sec (C) 1 sec (D) 6 sec Sol. a = 8 m

They meet when Q displace 8 × 3 m

more that P [relative displacement

= relative velocity × time.]

8 × 3 = (10 – 2) t

t = 3 sec. Ans.

53. A coin is released inside a lift at a height of 2 m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of 9 m/s2 downwards. The time after which the coin will strike with the lift: (g = 10 m/s2) (A) 4 s (B) 2 s (C)

4

s

21

(D)

2

_s

11

Sol. Relative to lift Initial velocity and acceleration of coin are 0 m/s and 1 m/s2 downward

 

 

₂

1

_{1 t}

2

2

or t = 2 second

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54. A particle is thrown up inside a stationary lift of sufficient height. The time of flight is T. Now it is thrown again with same initial speed v0

with respect to lift. At the time of second throw, lift is moving up with speed v0 and uniform acceleration g upwards (the acceleration due to

gravity). The new time of flight is –

(A)

T

4

(B)

T

2

(C) T (D) 2T

Sol. With respect to lift initial speed = v0 acc = – 2g displacement = 0  S = ut +

1

2

at2 0 = v0 T' +

1

2

× 2g × T' 2  T' =



0

g

=

1 2





0



1

T

2

5

2

55. Two particles P and Q are moving with velocities of

 

ˆ ˆi  j and



 ˆi 2jˆ



respectively. At time t = 0, P is at origin and Q is at a point

with position vector



2iˆ ˆ j .



Find the equation of the path of Q with respect to P. Sol. r_P 

 

ˆ ˆi  j t



 



     Q ˆ ˆ ˆ ˆ r 2i j i 2j t   QP Q P r r r = 2iˆ ˆ  j



2i t jˆ ˆ





 



    QP ˆ ˆ r 2 2t i 1 t j X = 2 – 2t y = 1 + t  x = 2 – 2 (y – 1) x + 2y = 4 Ans.

55. When two bodies move uniformly towards each other, the distance between them diminishes by 16 m every 10s, If bodies move with velocities of the same magnitude and in the same magnitude and in the same direction as before the distance between then will decrease 3m every 5s. Calculate toe velocity of each body.

Sol. Let velocity of bodies be V1 + V2.

V1 + V2 =

16

10

…(1)

V1 – V2 =

3

5

…(2) 2V1 =





16 3

22

10 5

10

= 2.2 m/s V1 = 1.1 m/s V2 = 1.6 – 1.1 = 0.5 m/s

After solving we have

V1 = 1.1 m/s

and V2 = 0.5 m/s.

56. A weight W is supported by two strings inclined at 60° and 30° to the vertical. The tensions in the strings are T1 & T2 as shown. If these

tensions are to be determined in terms of W using a triangle of forces, which of these triangles should you draw? (block is in equilibrium) Sol. AB  W,



₁

BC

T ,



₂

CA

T







AB BC CA 0

Ans. (E)

crackiitjee.in

58. A 2 kg toy car can move along an x axis. Graph shows force Fx, acting on

the car which beings at rest a time t = 0. The velocity of the particle at T = 10 s is :

(A) – i m/s (B) – 1.5 i m/s (C) 6.5 i m/s (D) 13 i m/s Sol.



dp p _f P_i

=



Fdt

= Area under the curve.

Pi = 0 Net Area = 16 – 2 – 1 = 13 N-s = Vf =

13

2

= 6.5 i m/s

[As momentum is positive, particle is moving along positive x axis.]

59. A particle is moving in a straight line whose acceleration versus time graph is given. Assume that initial velocity is in the direction of acceleration. Then which of the statement is correct between time t = 0 to t = t0.

(A) Velocity first increases then decreases, displacement always increases (B) Velocity and displacement both, first increases and then

decreases

(C) Displacement increases and velocity decreases (D) Displacement and velocity both always increases.

Sol. It is clear from the figure that acceleration does not change sign, i.e., does not change in direction. Only the magnitude of acceleration first increases and then decreases.

 Velocity keeps on increasing.

Hence displacement also keeps on increasing.

59. Surface inside which on  constant force F starts acting on the trolley as a result of which the string stood at an angle of 37° from the vertical (bob at rest relative to trolley) then.

(A) acceleration of the trolley is

40

3

m/sec2

(B) force applied is 60 N (C) force applied is 75 N

(D) tension in the string is 25 N. Sol. F = (8 + 2) a

a =

F

10

T cos 37° = mg T sin 37° = ma tan 37° =

a

g

a = g tan 37∟ =

10



3

4

= 7.5 m/s2  F = 10 × 7.5 = 75 N T =



mg

cos37

=

2 10 5





4

=

100



25 N

4

60. A cylinder of mass M and radius R is resting R is resting on two corner edges A and B as shown. The normal reaction at the edges A and B are: (neglect friction).

(A) N_A  2 N_B (B) N_B  3 N_A

(C)

N

_B



2 3 mg

5

(D) A



Mg

N

2

Sol. (B) (D)

For equilibrium NA cos 60° + NBcos 30°

= Mg and NA sin 60° = NB sin 30°





A B

N

N 3

_Mg

2

2



A B

N 3

N

2

2

 N 3_A  N_B On Solving NB = 3 N ;A NA =

Mg

.

2

61. An insect moving along a straight line, travels in every second distance equal to the magnitude of the elapsed. Assuming acceleration to be constant, and the insects starts at t = 0. Find the magnitude of initial velocity of insect.

Sol. Distance travelled from time 't – 1' sec to 't' sec is

S =

u



a



2t 1





2

…(1)

From given condition S = t …(2) 2t = 2u + 2at – a  2u = 2t – 2at 2u = 2t (1 – a) (1) & (2)





  

t u

a

2t 1



2





 

u

a



t 1 a .



2

Since u and a are arbitrary constants, and they must be constant for every time.

 coefficient of t must be equal to zero.

 1 – a = 0  a = 1 for a = 1, u =

1

2

unit Initial speed is

1

2

units.

62. In the figure if blocks A and B will move with same acceleration due to external agent, there is no friction between A and B then the magnitude of interaction force between the two blocks will be :

(A) 2 mg/ cos  (B) 2 mg cos  (C) mg cos  (D) none of these

63. Three rigid rods are joined to from an equilateral triangle ABC of side 1m. Three particles carrying charges 20 µC each are attached to the vertices of the triangle. The whole system is at rest always in an inertial frame. The resultant force on the charged particle at A has the magnitude.

(A) zero (B) 3.6 N (C) 3.6 3 N (D) 7.2 N Sol. Fnet  ma

a = acceleration of charge of particle at A = 0 Fnet  0.

64. Two stones are projected simultaneously from a tower at different angles of projection with same speeds 'u' the distance between two stones is increasing at constant rate 'u'. Then the angle between the initial velocity vectors of the two stone is:

(A) 30° (B) 60°

(C) 45° (D) 90°

Sol. (B)

To an observer who starts falling freely under gravity from rest at the instant stones are projected, the motion of stone A and B is seen as



dx

_u

dt

….(1)



dl

_u

dt

…(2)  x = l and BOA = 60°

65. A rope of negligible mass passes over a pulley of negligible mass attached to the ceilling, as shown in figure. Open end of the rope is held by student A of mass 70 Kg. Who is at rest on the floor. The opposite end of the rope is held by student B of mass 60 Kg. who is suspended at rest above the floor. The minimum acceleration a0 with

which the student B should climb up the rod to lift the student A upwards off the floor.

(A)

1

3

m/s2 (B)

2

3

m/s2 (C)

4

3

m/s2 (D)

5

3

m/s2 Sol. (D)

For student A to just lift off the floor, tension 1 in string must

The F.B.D. of student B is

Applying Newton's second law

T – mg = mg

 700 – 600 = 60 a or a =

5

m / s

2

3

66. Force exerted by the supporting table on the pulley (in Newton)

Sol. Let

T T and F

_{1 2 S} be the force exerted by the horizontal string, vertical string and the support on the mass-less respectively. Then







1 2 S

T

T

F

0

or F_S  T₁  T₂ =

2 2 mg

( Tension in each string is T₁  T₂  2 mg)

68. The system starts from rest and A attains a velocity of 5 m/s after it has moved 5 m towards right. Assuming the arrangement to be frictionless every where pulley & strings to be light, the value of the constant force F applied on A is:

(A) 50 N (B) 75 N (C) 100 N (D) 96 N

Sol. (B)

a = 2 V 2S =

25

10

= 2.5 m/s2 For 6 kg: – F – 2T = 6A For 2 kg: – T – 2g = 2 (2a) Feon (1) & (2) F = 75 N

69. A rod of length 2l is moving such that its ends A and b move in contact with the horizontal floor and wall respectively as shown in figure O is the intersection point of the vertical wall and horizontal floor velocity vector of the centre of rod C is always directed along tangent drawn at C to the

(A) circle of radius

2

whose centre lies at O

(B) circle of radius l whose centre lies at O (C) circle of radius 2l whose centre lies at O (D) None of these

Sol. (B)

At any instant of time the rod makes an angle with horizontal, the x & y coordinates of centre of rod are.

x = l cos  y = l sin 

 x2 + y2 = l2

Hence the centre C moves along a circle of radius l with centre at O.

 velocity vector of C is always directed along the tangent drawn at C to the circle of radius l who centre lies at O.

70. A system is as shown in the figure. All speeds shown are with respect to ground. Then the speed of B with respect to ground is:

(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 7.5 m/s Sol. (B)  1 2 2 = constant



d

1



2d

2



₀

dt

dt

(5 + 5) + 2 (5 + V0) = 0 or VB = 10 m/s

crackiitjee.in

(A) The acceleration of the cart is

2g

7

towards right.

(B) The cart comes to momentary rest after 2.5 s.

(C) The distance travelled by the cart in the first 5s is 17.5m. (D) The velocity of the cart after 5s will be same as initial velocity. Sol.  0.2 g = 0.7a

 

_a

2g

_{m / s}

2

7

For the case, it comes to rest when V = 0

    _{ }   2g 0 7 t 7

 

t

49

2g

= 2.5 s

Distance travelled till it comes to rest

    _ _   2 2g 0 7 2 s 7 S = 8.75 m

So in next 2.5s, it covers 8.75 m towards right

Total distance = 2 × 8.75 = 17.5 m

After 5s, it speed will be same as that of initial (7 m/s) but direction will be reversed.

72. A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a mass-less rope and pulley arrangement. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s2, then: (A) tension in the rope is 1125 N

(B) tension in the rope is 2250 N

(C) force of contact between the painter and the floor is 375 N (D) none of these

Sol. For painter: R + T – mg = ma

R + T = m(g + a) …(1)

For the system:

2T – (m + M) g = (m + M) a 2T = (m + M) (g + a) …(2) where: m = 100 kg M = 50 kg a = 5 m/sec2

crackiitjee.in





T



150 15

2

= 1125 N

and; R = 375 N.

73. A person wants to raise a block laying on the ground to height h. In both the case time required is same then in which case he has to exert more force. Assume pulleys and strings light

(A) (i) (B) (ii)

(C) Same is both

(D) Cannot be determined

Sol. Since, h =

1

2

at2 a should be same in both cases, because h and t are

same in both cases as given

In (i) F1 – mg = ma  F1 = mg + ma. In (ii) 2 F – mg = ma  F2 =



mg ma

2

F₁  F .₂

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74. A particle is projected from the horizontal x – z plane, in vertical x – y plane where x - axis is horizontal and positive y - axis vertically upwards. The graph of 'y' coordinate of the particle v/s time. The range of the particle is 3m. Then the speed of the projected particle is: (A)

3 m / s

(B) 403 4 m/s (C)

2 5 m / s

(D)

28 m / s

Sol. (D) From graph uy = tan 60° = 3 m/s range R = 2u ux y g or 3 2 ux 3 g    or ux = 5 m/s 2 2 x y u u u   

crackiitjee.in

=

28 m / s

(A)

3 m / s

(B) 2 m/s (C)

2 3 m / s

(D) 3 m/s Sol. Let AB = l, B = (x, y) B x

ˆ

y

ˆ

v



v i v j



B

ˆ

y

ˆ

v



3 i v j



…(i) x2 + y2 = l2 2x vx + 2y vy = 0 x vx + 2y vy = 0 y

y

3

v

0

x











y

3

tan 60 v

0









vy = – 1

Hence from (i)

B ˆ ˆ

v  3 i  j

Hence

B

v



2 m / s

76. Mass m is in equilibrium. If it is displace further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.

(A)

4kx

5m

(B)

2kx

5m

(C)

4kx

m

(D) none of these Sol. (C)

Initially the block is at rest under action of force 2T upward and mg down wards. When the blocks is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx.

 acceleration of blocks is =

4kx

m

77. Assume that cylinder remains in contact with the wedges. The velocity of cylinder is – (A)

19 4 3

u

2



m/s (B)

13 u

2

m/s

crackiitjee.in

(C)

3 u

m/s (D)

7 u

m/s Sol. (D)

Method-I

As cylinder will remains in contact with wedge A

Vx = 2u

As it also remain in contact with wedge B

u sin 30° = Vy = cos 30° – Vx sin 30° vy = x

sin30

usin30

v

cos30

cos30



_







vy = vx tan 30° + u tan 30° Vy = 3u tan 30° =

3 u

V = v2_x  v2_y =

7 u

Method-II In the frame of A

crackiitjee.in

3u sin 30° = Vy cos 30°  Vy = 3 u tan 30° =

3 u

and Vx = 2u 2 2 x y v  v  7 u

78. Two stones are thrown vertically upwards simultaneously from the same point on the ground with initial speed u1 = 30 m/sec and u2 = 50

m/sec. which of the curve represent correct variation (for the time interval in which both reach the ground) of

(x2 – x1) = the relative position of second stone with respect to first with

time (t).

(v2 – v1) = the relative velocity of second stone with respect to first with

time (t).

Assume that stones do not rebound after hitting.

Sol. While both the stones are in flight, a1 = g and a2 = g

So arel = 0

 Vrel = constant

 xrel = (const) t

 Curve of xrel v/s t will be straight line.

78. After the first particle drops on ground, the separation (xrel) will

decrease parabolically (due to gravitational acceleration), and finally becomes zero.

and Vrel = slope of xrel v/s t

So

79. Two springs are in a series combination and are attached to a block of mass 'm' which is in equilibrium. The spring contents and the extensions in the springs are as shown in the figure. Then the force exerted by the spring on the block is:

(A) 1 2



_{1 2}



1 2 k k _{x x} k k  (B) k x_{1 1} k x_{2 2} (C) k1 (D) None of these Sol. (A)

Tension in both springs are same

both the springs are in series,

keq = 1 2 1 2

k k

k k

80. A & B are free to move. All the surfaces are smooth. (A) the acceleration of A will be more than g sin (B) the acceleration of A will be less than g sin

(C) normal force on A due to B will be more than mg cos (D) normal force on A due to B will be less than mg cos Sol. ma0 sin  + N = mg cos 

 mg cos   N = mg cos  – ma0 sin   N < mg cos  Hence, (D) is true. ma0 cos  + mg sin  = ma  a = g sin  + a0 cos  Hence acceleration of A

=



a a cos ₀  

 

2 a sin₀  



2 gsin

81. There are three particle A, B & C are lying in a horizontal plane the particle b is situated 5m due North from A and particle C is situated 30° East of North at a distance of 2 3m from A. These three particles start moving simultaneously along straight lines and collide after 2 second of

their start. Particle A moves with constant velocity 5 m/s due 30° South of East. Find out the velocity vectors (constant) of the particles B & C. Assume unit vector

ˆi

in the east and

ˆj

is the north.

Sol. Taking A at the origin and east direction as the direction x - axis and positive y - axis along north.

A

5 3 ˆ

ˆ

v

i 2.5 j

2





Assume velocity of B as x

ˆ

y

ˆ

u, u u i u i





and C as

v, v



v i v i

_x

ˆ



_y

ˆ

 

B

ˆ

y

ˆ

x

ˆ

r t

5 j u t j u t j









 





B xˆ y ˆ r 2  2u i  2u 5 j …(1)

 

 

A ˆ ˆ r 2  5 3 i 5 j  2

 

C

ˆ

ˆ

x

ˆ

y

ˆ

r 2



3 i 3 j 2v i 2v j









3 2v i x



ˆ



3 2v j y



ˆ …(3) (1) & (2) x

5 3

u

,

2





crackiitjee.in

uy = – 5 (2) & (3) x V  2 3, y

V



…(4)

 the velocity vector of B and C are

5 3 ˆ ˆ

u

i 5 j

2





and

v 2 3 i 4 j



ˆ



ˆ

respectively.

82. Two inclined planes inter in a horizontal plane. Their inclinations to the horizontal being  and . If a particle is projected with velocity u at right to the former from a point on it, find the time after which the velocity vector will become perpendicular to the other inclined plane. Sol. Taking x - axis parallel to second inclined, we can say that when x -

component of velocity will become zero, than velocity vector will become perpendicular to the other inclined.

ux = u sin ( + ) ax = – g sin  vx – ux + axt

0 = u sin ( + ) – g sin t

t = usin





gsin     Ans. t = usin





gsin    

83. A man standing on the block is pulling the rope velocity of the point of string in contact with the hand of the man is 2 m/s downwards. The velocity of the block will be: [assume that the block does not rotate] (A) 3 m/s (B) 2 m/s (C)

1

2

m/s (D) 1 m/s Sol. (B) 1  2  3  4  C 1 2 3 4

d

d

d

d

₀

dt



dt



dt



dt



– v – v + 0 + v + 2 = 0 v = 2 m/s

84. The velocity of lift is 2 m/s while string is winding on the motor shaft with velocity 2 m/ s and block A is moving downwards with a velocity of 2 m/s, then find out the velocity of block B.

(A) 2 m/s



(B) 2 m/s 

(C) 4 m/s



(D) none of these Sol. (D) B, v  4 m / s  B, B

v



v , g v , g



4 m/s = v , g 2 m / s_B  B v , g 2 m / s 

85. Three stones A, B and C are simultaneously projected from same point with same speed. A is thrown upwards, B is thrown horizontally and C is thrown downwards from a building. When the distance between stone A and C becomes 10m, Then distance between A and B will be –

(A) 10 m (B) 5m

(C)

5 2 m

(D) 10 2 m

Sol. Let the stones be projected at t = 0 sec with a speed u form point O. Then an observer at rest at t = 0 and having constant acceleration equal to acceleration due to gravity, shall observe the three stones moves with constant velocity.

Sol. In the given time each ball shall travel a distance 5 metre as seen by this observer. Hence the required distance between A and B will be =

2 2

5



5



5 2

meter.

86. A stone is projected horizontally with speed v from a height h above ground. A horizontal wind is blowing in direction opposite to velocity of projection and gives the stone a constant horizontal acceleration f (in direction opposite to initial velocity). As a result the stone falls on ground at a point vertically below the point of projection. Then the value of height h in terms of f, g, v is (g is acceleration due to gravity)

(A) 2 2 gv 2f (B) 2 2 gv f (C) 2 2

2v

f

(D) 2 2 2gv f

Sol. Time taken to reach the ground is given by

h =

1

2

gt …(1)

Since horizontal displacement in time t is zero

2v

t

f

 

…(2) h = 2 2 2gv f

crackiitjee.in

87. Two blocks A and B connected to an ideal pulley string system. In this system when bodies are released then: (neglect friction and take g = 10 m/s2)

(A) Acceleration of block A is 1 m/s2 (B) Acceleration of block A is 2 m/s2

(C) Tension in string connected to block B is 40 N (D) Tension in string connected to block B is 80 N

Sol. Applying NLM on 40 kg block 400 – 4T = 40 a

For 10 kg block T = 10.4 a

Solving a = 2 m/s2

T = 80 N

Comprehension

A Weighing machine kept in a lift. Lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown be weighing machine and spring balance is 15 kg and 45 kg respectively. Answer the following question. Assume that the weighing machine can

measure weight by having negligible deformation due to block, while the spring balance requires larger expansion: (take g = 10 m/s2)

88. Mass of the object in kg is and the normal force acting on the block due to weighing machine are:

(A) 60 kg, 450 N (B) 40 kg, 150 N (C) 80 kg, 400 N (D) 10, kg, zero

Sol. FBD of Block in ground frame: applying N.L. 150 + 450 – 10 M = 5M  15 M = 600

600

M

15





 M = 40 Kg.

Normal on block is the reading of weighing machine i.e. 150 N.

89. In lift is stopped and equilibrium is reached. Reading of weighing machine and spring balance will be:

(A) 40 kg, zero (B) 10 kg, 20 kg (C) 20 kg, 10 kg (D) zero, 40 kg

Sol. If lift is stopped & equilibrium is reached then

450 + N = 400

 N = – 50

So block will loss the contact with weighing machine thus reading of weighing machine will be zero.

T = 40 g

So reading of spring balance will be 40 Kg.

90. Find the acceleration of the lift such that the weighing machine shows its true weight.

(A)

45

4

m/s2 (B)

85

4

m/s2 (C)

22

4

m/s2 (D)

60

4

m/s2 Sol. a =

950 400

40



91. Two messes m1 and m2 which are connected with a light string, are

placed over a frictionless pulley. This set up is placed over a weighing machine, as shown. Three combination of masses m1 and m2 are used,

in first case m1 = 6 kg and m2 = 2 kg, in second case m1 = 5 kg and m2 = 3

kg and in third case m1 = 4 kg and m2 = 4 kg. masses are held stationary

initially and then released. If the readings of the weighing machine after the release in three cases are W1, W2 and W3 respectively then:

(A) W1 > W2 > W3

(B) W1 < W2 < W3

(C) W1 = W2 = W3

(D) W1 = W2 < W3

Sol. (B)

Reading of the weighing machine = 2T + weight of the machine.

As weight of the machine is constant.

T = 1 2 1 2

2m m

m m

So reading is maximum for the case m1m2 is maximum as m1 + m2 is all cases is same.

92. In the following arrangement the system is initially at rest. The 5 kg block is now released. Assuming the pulleys and string to be massless and smooth, the acceleration of block 'C' will be

(A) zero (B) 2.5 m/s2

(C)

10

7

m/s2 (D)

5

7

m/s2

Sol. Block b will not move. 5 g – T = 5a …(1)

2T – 8g = 8

a

2

…(2)