Question

Below is the graph for $\tau =2\sin \theta$. calculate work done by the torque till $\theta =\pi .$

• A. $2\text{J}$
• B. $4\text{J}$
• C. $0\text{J}$
• D. $2\pi \text{J}$

Answer 1

The correct option is B $4\text{J}$

We know that, $W=\int \tau d\theta$
Also, area under $\tau -\theta$ graph gives the amount of workdone.
From graph, we can see that torque $\left(\tau \right)$ varies with $\theta$ as $\tau =2\sin \theta$
$\therefore$ Area under $\tau -\theta$ graph
$W={\int }_0^{\pi }\text{2 sin}\theta d\theta$
$=2{\int }_0^{\pi }\sin \theta d\theta$
$=-2\cos \theta {|}_0^{\pi }$
$=-2[\cos \pi -\cos 0]$
$=4\text{J}$