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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 isa) 0.2Jb) 10 Jc) 20 Jd) 0.1 J

satendra , 10 Years ago
Grade 12th pass
anser 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,

askIITians faculty

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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