First slide

Elastic Modulie

Question

Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be

Easy

A

1 : 2
B

1 : 4
C

2 : 1
D

1 : 1

Solution

Work done, $W = \frac{1}{2} F l$
$∴$ Work done $W \propto l$ (If F is constant) area of cross section and material same so elongation is proportional to original length.
$\Rightarrow \frac{W_{1}}{W_{2}} = \frac{l}{l_{2}} = \frac{l}{2 l} = \frac{1}{2}$

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