Ncert Exercises & Solutions

Q.16 Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

Hint: The work done on the springs will depend on Young's modulus of the material of the springs.

Step 1: Find the work done in terms of Young's modulus.

Work done in stretching a wire is given by, W

= \frac{1}{2} F \times ∆ l

[Where F is applied force and

∆ l

is the extension in the wire]

As springs of steel and copper are equally stretched. Therefore, for the same force (F),

W \propto ∆ l

Young's modulus (Y) =

\frac{F}{A} \times \frac{l}{∆ l} \Rightarrow ∆ l = \frac{F}{A} \times \frac{l}{Y}

...(i)

As both springs are identical,

∆ l \propto \frac{1}{Y}

...(ii)

From Eq. (i) and (ii), we get,

W \propto \frac{1}{Y}

Step 2: Find the work done on both the springs.

∴ \frac{W_{steel}}{W_{copper}} = \frac{Y_{copper}}{Y_{steel}} < 1 (As, Y_{steel} > Y_{copper})

\Rightarrow W_{steel} < W_{copper}

Therefore, more work will be done for stretching the copper spring.