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 isa) 0.2Jb) 10 Jc) 20 Jd) 0.1 J
satendra , 10 Years ago
Grade 12th pass
2 Answers
Vasantha Kumari
Last Activity: 10 Years ago
The elastic energy stored U = ½ x F x Dl
= ½ × 200 ´ 10–3 J
=0.1 J
Thanks
& Regards,
Vasantha
Sivaraj,
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Kushagra Madhukar
Last Activity: 4 Years ago
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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