Two unequal masses are connected on two sides of a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped `1.0` second after the system is set into motion and then released immediately. The time elapsed before the string is tight again is: Take `g=10m//s^(2)`

Options

A. `1//4s`
B. `1//2s`
C. `2//3s`
D. `1//3s`

Answer

D

Explanation

Acceleration of the system: `a=(m_(2)-m_(1))/(m_(2)+m_(1))g=(10)/(3)m//s^(2)` Velocity of both the block at `t=1s` will be
`v_(0)=at=(10)/(3)xx1=(10)/(3)m//s`
Now at this moment, velocity of `2kg` block becomes zero, while that of `1kg` block is `(10)/(3)m//s` upwards. Hence, string becomes tight again when
Displacement of `1kg` block=displacement of `2kg` block
or `v_(0)t-(1)/(2)gt^(2)=(1)/(2)gt^(2)` .