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2. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1 mm . Then the elastic energy stored in the wire is a) 0.2J b) 10 J c) 20 J d) 0.1 J
The elastic energy stored U = ½ x F x Dl = ½ × 200 ´ 10–3 J =0.1 J Thanks & Regards, Vasantha Sivaraj, askIITians faculty
The elastic energy stored U = ½ x F x Dl
= ½ × 200 ´ 10–3 J
=0.1 J
Thanks & Regards,
Vasantha Sivaraj,
askIITians faculty
Dear student,Please find the solution to your question below. The elastic energy stored U = ½ x F x ΔL= ½ × 200 x 10–3 J= 0.1 J Thanks and regards,Kushagra
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