A) \[\sqrt{2}L\]
B) \[\sqrt{3}L\]
C) \[2L\]
D) L
Correct Answer: D
Solution :
On bending the magnet, the length of the magnet \[AC=AB+BC\] \[=L\sin \left( \frac{\theta }{2} \right)+L\sin \left( \frac{\theta }{2} \right)\] \[=L\sin \left( \frac{\theta }{2} \right)\] \[=2L\sin 30{}^\circ \] \[=2L\times \frac{1}{2}=L\]You need to login to perform this action.
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