A) \[0.245\text{ }J\]
B) \[2.45\times {{10}^{-4}}J\]
C) \[0.0245\text{ }J\]
D) \[245.0\text{ }J\]
E) \[24.5\times {{10}^{-4}}J\]
Correct Answer: E
Solution :
Torque, \[\tau =pE\sin \theta \] or \[P=\frac{\tau }{E\sin \theta }\] Potential energy, \[U=pE\cos \theta \] \[=\frac{\tau }{E\sin \theta }.E\cos \theta \] \[=\frac{\tau }{E\tan \theta }=\frac{10\sqrt{2}}{{{10}^{4}}\tan {{30}^{o}}}\] \[=24.5\times {{10}^{-4}}J\]You need to login to perform this action.
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