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

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)

$\displaystyle T \le \frac{2}{3}m g$

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

a = g/3

## 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….

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/s2

Thus, acceleration of A is 2/3 m/s2 in rightward direction.

Sol:(ii)
From FBD of A,

F -T = m1 a = 1 × 2/3 ….(i)

From FBD of B, ….(ii)

T – m2 g sinθ = m2 a

= 2 × 2/3 = 4/3

or,  T-10 = 4/3

T= 10 + 4/3 = 34/3

=11.33 N

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

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 = t1 block is accelerated with maximum acceleration , then decelerated with deceleration g .
$\displaystyle h_1 = \frac{1}{2}(\frac{4mg-mg}{m})t_1^2$

vo = 3gt1
Also
0 = vo – g(t-t1)
t1 = t/4 ,
h2 = vo2/2g
h = h1 + h2
On solving
t= 4sec

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

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:
$\displaystyle Mg sin\theta -T = Ma$ …(i)

$\displaystyle T-mg sin\theta = ma$ …(ii)

$\displaystyle a = \frac{(M-m)g sin\theta}{M+m}$

$\displaystyle T= Mg sin\theta - Ma$

$\displaystyle T = Mg sin\theta - M\frac{(M-m)gsin\theta}{M+m}$

$\displaystyle T = Mg sin\theta (1-\frac{M-m}{M+m})$

$\displaystyle T = Mg sin\theta(\frac{2m}{M+m})$

$\displaystyle T = \frac{2 mg sin\theta}{1+m/M}$

As , M >> m

$\displaystyle T = 2m g sin\theta$

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

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 T1 & in other part is T2

For 1 kg block ,

T1 – 1×g = 1×a   ..(i)

For 2kg block ;

T2 – 2g sin37° -T1 = 2a   …(ii)

since T2 = F = 38 N

from (ii)

38 – 2×10 ×(3/5) – (g+a) = 2a

(As , T1 = g+a & sin37° = 3/5)

38 – 12 -10 = 3a

16 = 3a

a = 16/3 m/s2

Since , T1 = g+a

T1 = 10 + 16/3 = 46/3 N

$\displaystyle \frac{T_1}{T_2} = \frac{46/3}{38}= \frac{23}{57}$