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# Work Done In Stretching A Wire - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Work done in stretching a wire is considered one the most difficult concept.

• 32 Questions around this concept.

## Solve by difficulty

A wire fixed at the upper end stretches by length $l$ by applying a force $F$. The work done in stretching is:

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy (in Joule) stored in the wire is

## Concepts Covered - 1

Work done in stretching a wire

When a body is in its natural shape, its potential energy corresponding to the molecular forces is minimum. When deformed, internal forces appear and work has to be done against these forces. Thus, the potential energy of the body is increased. This is called the elastic potential energy.

Suppose a wire having natural length L and cross-sectional area A is fixed at one end and is stretched by an external force applied at the other end. When the extension is x,the wire is under a longitudinal stress F/A. The strain is x/L.

If Young's modulus is Y, then

$\frac{F/A}{x/L}=Y\Rightarrow F=\frac{AY}{L}x$

So, the work done for an additional small increase dx in length will be:-

$dW=Fdx\Rightarrow dW=\frac{AY}{L}xdx$

The total work done by the extermal force in increasing the length from 0 to $\Delta L$ will be:-

$W=\int_{0}^{\Delta L}\frac{AY}{L}xdx=\frac{1}{2}\frac{YA}{L}(\Delta L)^2$

This work done is stored into the wire as its elastic potential energy. So, the elastic potential energy of the stretched wire is:

$U=\frac{1}{2}\frac{YA}{L}(\Delta L)^2$

We can also write,

$W=\frac{1}{2}\left [ \frac{YA \Delta L}{L} \right ]\Delta L=\frac{1}{2}(maximum\ stretching\ force)\times extension$

$W=\frac{1}{2}Y(AL)\left [ \frac{ \Delta L}{L} \right ]^2=\frac{1}{2}\times Y \times Volume\times (strain)^2$

$W=\frac{1}{2}\left [ \frac{ \Delta L}{L} \right ]\left [ Y\frac{ \Delta L}{L} \right ](AL)=\frac{1}{2}\times strain\times (Ystrain)\times Volume$

$W=\frac{1}{2}\left [ \frac{ \Delta L}{L} \right ]\left [ Y\frac{ \Delta L}{L} \right ](AL)=\frac{1}{2}\times strain\times stress \times Volume$

$Also, Potential\ energy\ per\ unit\ volume= \frac{1}{2}\times strain\times stress$

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Work done in stretching a wire

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## Books

### Reference Books

#### Work done in stretching a wire

Physics Part II Textbook for Class XI

Page No. : 244

Line : 34