Q: At what minimum acceleration should a monkey should slide a rope whose breaking strength is 2/3 rd of its weight?

(a) 2g/3

(b) g

(c) g/3

(d) zero

Ans: (c)

Sol:

mg -T = ma …(i)

On putting the value of T in (i)we get

a = g/3

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# Tag: Laws of Motion

## At what minimum acceleration should a monkey should slide a rope whose breaking strength …

## Two smooth blocks A of mass 1 kg and B of mass 2 kg are connected by alight string passing over a smooth pulley as shown….

## A block of mass m is to be raised by a light rope from rest to rest through a height 60m, the maximum tension….

## Two blocks A and B having masses m and M (m << M) are on smooth inclined planes. They are connected....

## For the situation shown in figure, all pulleys are smooth and fixed. The inclined plane is also smooth….

Q: At what minimum acceleration should a monkey should slide a rope whose breaking strength is 2/3 rd of its weight?

(a) 2g/3

(b) g

(c) g/3

(d) zero

Ans: (c)

Sol:

mg -T = ma …(i)

On putting the value of T in (i)we get

a = g/3

Q: Two smooth blocks A of mass 1 kg and B of mass 2 kg are connected by alight string passing over a smooth pulley as shown. The block B is sliding down with a velocity 2 m/s. A force F is applied on the block A so that the block B will reverse its direction of motion after 3 s.

(i) Find the acceleration of block A.

(a) 1 m/s²

(b) 2/3 m/s²

(c) 3 m/s²

(d) 4/9 m/s²

Ans: (b)

(ii) Find the tension in the string.

(a) 10 N

(b) 11.33 N

(c) 13.67 N

(d) 40 N

Ans: (b)

Sol:(i) When direction of motion will reverse at the instant, the velocity becomes zero.

∵ v = u + at

Or 0 = 2 + a×3

a = -2/3 m/s^{2}

Thus, acceleration of A is 2/3 m/s^{2} in rightward direction.

Sol:(ii)

From FBD of A,

F -T = m_{1} a = 1 × 2/3 ….(i)

From FBD of B, ….(ii)

T – m_{2} g sinθ = m_{2} a

= 2 × 2/3 = 4/3

or, T-10 = 4/3

T= 10 + 4/3 = 34/3

=11.33 N

Q: A block of mass m is to be raised by a light rope from rest to rest through a height 60m, the maximum tension of which the rope safely bear is 4 mg. The least time in which the ascent can be made is

(a) 1 s

(b) 2 s

(c) 3 s

(d) 4 s

Ans: (d)

Sol: Let up to t = t_{1} block is accelerated with maximum acceleration , then decelerated with deceleration g .

v_{o} = 3gt_{1}

Also

0 = v_{o} – g(t-t_{1})

t_{1} = t/4 ,

h_{2} = v_{o}^{2}/2g

h = h_{1} + h_{2}

On solving

t= 4sec

Q: Two blocks A and B having masses m and M (m << M) are on smooth inclined planes. They are connected through light string passing over a smooth pulley. The tension in string is

(a) (M – m) g sinθ

(b) 2 Mg sinθ

(c) 2 mg sinθ

(d) (M + m) g sinθ

Ans: (c)

Sol:

…(i)

…(ii)

Adding (i) and (ii)

As , M >> m

Q: For the situation shown in figure, all pulleys are smooth and fixed. The inclined plane is also smooth. A constant force F of 38 N is applied at the end of the string. Find the ratio of magnitudes of tensions in the string A and B.

(a) 23 : 57

(b) 32 : 57

(c) 23 : 75

(d) 32 : 75

Ans: (a)

Sol:

Let tension in the string between 1kg and 2kg block is T_{1} & in other part is T_{2}

For 1 kg block ,

T_{1} – 1×g = 1×a ..(i)

For 2kg block ;

T_{2} – 2g sin37° -T_{1} = 2a …(ii)

since T_{2} = F = 38 N

from (ii)

38 – 2×10 ×(3/5) – (g+a) = 2a

(As , T_{1} = g+a & sin37° = 3/5)

38 – 12 -10 = 3a

16 = 3a

a = 16/3 m/s^{2}

Since , T_{1} = g+a

T_{1} = 10 + 16/3 = 46/3 N