First slide

Elastic behaviour of solids

Question

A 5-metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

Moderate

A

Zero
B

0.05 J
C

100 J
D

500 J

Solution

$\begin{matrix} W = \frac{1}{2} \times F \times l where l = increase in length of the wire \\ = \frac{1}{2} mgl = \frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0 .05 J \end{matrix}$

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