A) 0.245 J
B) 2.45 J
C) 0.0245 J
D) 24.5 J
Correct Answer: D
Solution :
[d]: Here, \[\tau =pE\sin \theta \] \[10\sqrt{2}=p\times {{10}^{4}}\sin {{30}^{o}}={{10}^{4}}\times \frac{P}{2}\] \[p=\frac{20\sqrt{2}}{{{10}^{4}}}=2\sqrt{2}\times {{10}^{-3}}\] \[P.E.=pE\cos \theta \] \[=2\sqrt{2}\times {{10}^{-3}}\times {{10}^{4}}\cos {{30}^{o}}=20\sqrt{2}\times \frac{\sqrt{3}}{2}\] \[=10\sqrt{6}\times 10\times 2.45J=24.5J\]You need to login to perform this action.
You will be redirected in
3 sec