A bob of mass `m` is suspended from a fixed support with a light string and the system with bob and support is moving with a uniform horizontal acceleration. The breaking strength of the string is `mg sqrt(2)`. Find the workdone by the tension in the string in the first one second :

Options

A. `2 mg^(2)`
B. `(mg^(2))/(sqrt(2))`
C. `(mg^(2))/(2)`
D. `mg^(2) sqrt(2)`

Answer

C

Explanation

`T sin theta = ma` and `T cos theta = mg`
So, `a = g tan theta`
Now, `T = (mg)/(cos theta) = mg sqrt(2)` (given)
`cos theta =(1)/(sqrt(2))`, or `theta = 45^@`
and `a = g tan theta = g tan 45^@ = g`
workdone, `W =T.S =T((1)/(2)at^(2)) sqrt(2)`
=`(mg sqrt(2)) ((1)/(2))(g) (1)^(2) sqrt(2) = mg^(2)//2`.