Question

In the system shown in figure, masses of the blocks are such that when the sytem is released, the acceleration of pulley $P_1$ is upwards and the acceleration of block 1 is $a_1$ upwards. It is found that the acceleration of block 3 is same as that of 1 both in magnitude and direction.
Given $a_1>a>\frac{a_1}{2}$, match the following:


$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{i.}& \text{Acceleration of 2}& \text{a.}& 2a+a_1\\ \text{ii.}& \text{Acceleration of 4}& \text{b.}& 2a-a_1\\ \text{iii.}& \text{Acceleration of 2 with respect to 3}& \text{c.}& \text{Upwards}\\ \text{iv.}& \text{Acceleration of 2 with respect to 4}& \text{d.}& \text{Downwards}\end{array}$

• A. i - b,c; ii - a; iii - d; iv - c
• B. i - b; ii - a,d; iii - d; iv - c
• C. i - b,c; ii - a,d; iii - d; iv - c
• D. i - b,c; ii - a; iii - c; iv - c

Answer 1

The correct option is C i - b,c; ii - a,d; iii - d; iv - c

Acceleration of block 1 wrt P1 is $a_1-a$ upwards.
So acceleration of block 2 wrt P1 is $a_1-a$ downwards.
So acceleration of block 2 wrt ground is $a_1-2a$ downwards. Given that, $a>\frac{a_1}{2}$, $a_1-2a$ is negative. So, block 2 has an acceleration of $2a-a_1$ upwards.

Acceleration of block 3 is $a_1$ upwards. Acceleration of the pulley $P_2$ is $a$ downwards.
So, the acceleration of block 3 wrt P2 is $a_1+a$ upwards.
Therefore, the acceleration of block 4 wrt P2 is $a_1+a$ downwards, and it is $a_1+2a$ downwards wrt ground.
Acceleration of block 2 wrt 3 is $2a-a_1-a_1=2(a-a_1)$ upwards. Given that, $a_1>a$, so Acceleration of block 2 wrt 3 is negative in the upward direction or simply downwards.

Acceleration of block 2 wrt 4 is $2a-a_1+(a_1+2a)=4a$ upwards.