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Question

In the system shown in figure, masses of the blocks are such that when the sytem is released, the acceleration of pulley P1 is upwards and the acceleration of block 1 is a1 upwards. It is found that the acceleration of block 3 is same as that of 1 both in magnitude and direction.
Given a1 >a > a1/2, match the following:


Column IColumn IIi.Acceleration of 2a.2a+a1ii.Acceleration of 4b.2aa1iii.Acceleration of 2 with respect to 3c.Upwardsiv.Acceleration of 2 with respect to 4d.Downwards
  1. i - b,c; ii - a; iii - d; iv - c
  2. i - b; ii - a,d; iii - d; iv - c
  3. i - b,c; ii - a,d; iii - d; iv - c
  4. i - b,c; ii - a; iii - c; iv - c


Solution

The correct option is C i - b,c; ii - a,d; iii - d; iv - c
Acceleration of block 1 wrt P1 is a1a upwards.
So acceleration of block 2 wrt P1 is a1a downwards.
So acceleration of block 2 wrt ground is a12a downwards. Given that, a>a12a12a is negative. So, block 2 has an acceleration of 2aa1 upwards.

Acceleration of block 3 is a1 upwards. Acceleration of the pulley P2 is a downwards.
So, the acceleration of block 3 wrt P2 is a1+a upwards.
Therefore, the acceleration of block 4 wrt P2 is a1+a downwards, and it is a1+2a downwards wrt ground.
Acceleration of block 2 wrt 3 is 2aa1a1=2(aa1) upwards. Given that, a1>a, so Acceleration of block 2 wrt 3 is negative in the upward direction or simply downwards.

Acceleration of block 2 wrt 4 is 2aa1+(a1+2a)=4a upwards.
 

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