First slide

Elastic Modulie

Question

A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is I metre above the floor. The wire was elongated by I mm. The energy stored in the wire due to stretching is

Moderate

A

zero
B

0.05 joule
C

100 joule
D

500 joule

Solution

$W = \frac{1}{2} \times F \times l = \frac{1}{2} mgl$
= $\frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 1} = 0.05 J$

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