In the graph shown below, find the amount of work done.
A
2J
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B
4J
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C
0J
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D
2πJ
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Solution
The correct option is B4J We know that, W=∫τdθ Also, area under τ−θ graph gives the amount of workdone. From graph, we can see that torque (τ) varies with θ as τ=2sinθ ∴ Area under τ−θ graph W=∫π02 sinθdθ =2∫π0sinθdθ =−2cosθ|π0 =−2[cosπ−cos0] =4J