  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.2a−a1iii.Acceleration of 2 with respect to 3c.Upwardsiv.Acceleration of 2 with respect to 4d.Downwardsi - b,c; ii - a; iii - d; iv - ci - b; ii - a,d; iii - d; iv - ci - b,c; ii - a,d; iii - d; iv - ci - b,c; ii - a; iii - c; iv - c

Solution

## The correct option is C i - b,c; ii - a,d; iii - d; iv - cAcceleration of block 1 wrt P1 is a1−a upwards. So acceleration of block 2 wrt P1 is a1−a downwards. So acceleration of block 2 wrt ground is a1−2a downwards. Given that, a>a12, a1−2a is negative. So, block 2 has an acceleration of 2a−a1 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 2a−a1−a1=2(a−a1) 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 2a−a1+(a1+2a)=4a upwards.  Suggest corrections   