Question

A wire extends by ‘l’ on the application of load ‘mg’. Then, the energy stored in it:
A) $mgl$.
B) $\dfrac{{mgl}}{2}$.
C) $\dfrac{{mg}}{l}$.
D) $mg{l^2}$.

Answer 1

Hint: When a wire is stretched then till the elastic limit the wire can get back to its original shape but if the wire is stretched move than the elastic limit then the wire gets permanently deformed and energy is stored in the wire in the form of potential energy.

Formula Used: The formula of the energy stored per unit volume is:
$U = \dfrac{1}{2} \times \sigma \times \varepsilon \times V$
Where energy stored per volume is $\dfrac{U}{V}$ the stress is $\sigma $ and the strain is equal to $\varepsilon $.

Complete step by step answer:
It is given in the problem that a wire extends by l on the application of load mg and we need to find the energy stored in the wire.
The formula of the energy stored per unit volume is:
$ \Rightarrow \dfrac{U}{V} = \dfrac{1}{2} \times \sigma \times \varepsilon $
Where energy stored per volume is $\dfrac{U}{V}$ the stress is $\sigma $ and the strain is equal to$\varepsilon $.
The formula of the energy stored is equal to,
$ \Rightarrow U = \dfrac{1}{2} \times \sigma \times \varepsilon \times V$………eq. (1)
We have rearranged the terms here to get the formula of the energy stored.
The stress is defined as the ratio of applied force and area so the stress is equal to,
$ \Rightarrow \sigma = \dfrac{F}{A}$………eq. (2)
The strain is defined as the ratio of change of length and original length so the strain will be equal to,
$\varepsilon = \dfrac{{\Delta l}}{l}$………eq. (3)
Replacing the value of stress and strain from equation (2) and equation (3) into equation (1) we get.
$ \Rightarrow U = \dfrac{1}{2} \times \left( {\dfrac{F}{A}} \right) \times \left( {\dfrac{{\Delta l}}{l}} \right) \times \left( {Al} \right)$
$ \Rightarrow U = \dfrac{1}{2} \cdot \left( {F \times \Delta l} \right)$
$ \Rightarrow U = \dfrac{1}{2} \cdot \left( {mgl} \right)$
The energy stored in the wire is equal to$U = \dfrac{1}{2} \cdot \left( {mgl} \right)$.

The correct answer for this problem is option B.

Note: It is advisable for students to understand and remember the formula of the energy stored in any material as it can be helpful in solving these types of problems. The work done on the wire gets stored in wire in the form of potential energy.