A thin bar magnet of length L is bent in the middle to form an angle of 600. The new length of the magnet is :
2L
L
L√2
L2
A
L2
B
L√2
C
L
D
2L
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Solution
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AD=DC=L2sin300 ∴AC=L2sin300+L2sin300=Lsin300=L2
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