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 Δ l

$= \frac{1}{2} F \times ∆ l$

[Where F is applied force and

Δ l

$∆ l$ is the extension in the wire]

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

W \propto Δ l

$W \propto ∆ l$

Young's modulus (Y) =

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

$\frac{F}{A} \times \frac{l}{∆ l} \Rightarrow ∆ l = \frac{F}{A} \times \frac{l}{Y}$ ...(i)

As both springs are identical,

Δ l \propto \frac{1}{Y}

$∆ l \propto \frac{1}{Y}$ ...(ii)

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

W \propto \frac{1}{Y}

$W \propto \frac{1}{Y}$

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.