First slide

Elastic behaviour of solids

Question

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of 10N, then the increase in length is 0.5 mm. The ratio of the energy of the wire
and the work done in displacing it through 1.5 mm by the weight is

Moderate

A

$\frac{1}{3}$
B

$\frac{1}{4}$
C

$\frac{1}{2}$
D

$1$

Solution

Energy stored in stretching a wire
$U = \frac{1}{2} F l = \frac{1}{2} \times 10 \times 0.5 \times 10^{- 3} = 2.5 \times 10^{- 3} J$
Work done to displace it through 1.5 mm
$\begin{array}{l} W = F \times l = 10 \times 0.5 \times 10^{- 3} = 5 \times 10^{- 3} J \\ The ratio of energy and work = \frac{U}{W} = \frac{2.5 \times 10^{- 3}}{5 \times 10^{- 3}} = 1 : 2 \end{array}$

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