1.

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by mm. Then the elastic energy stored in the wire :-

Answer»

0.1 J
0.2 J
10 J
20 J

Solution :`U = 1/2 xx F xx L = 1/2 xx 200 xx 10^(-3) = 0.1J`.


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