Question

When a block of mass $M$ is suspended by a long wire of length $L$ , the length of the wire becomes $(L + l)$. The elastic potential energy stored in the extended wire is:
A) $MgL$
B) $\dfrac{1}{2}Mgl$
C) $\dfrac{1}{2}MgL$
D) $Mgl$

Answer 1

Hint: First calculate the stress and strain on the long wire when the block is suspended by it. Then, use them in the formula for Energy stored per unit volume and after solving the equation formed, we will get the answer for elastic potential energy stored in the entire wire.

Formula Used: Stress $ = \dfrac{F}{A}$ where $F$ is the force acting on the wire and $A$ is the cross sectional area of the wire
Strain $ = \dfrac{l}{L}$ where $l$ is the elongation in the wire and $L$ is the initial length of the wire
Energy stored in a wire per unit volume, $\dfrac{E}{V} = \dfrac{1}{2} \times stress \times strain$

Complete Step by Step Solution:
The initial length of the wire is $L$ . when the block of mass $M$ is suspended by it, the wire will experience a force equal to $Mg$ where $g$ is the acceleration due to gravity and on experiencing this force, it gets elongated by length $l$
Therefore, stress in the wire will be forced per unit area, i.e. $\dfrac{{Mg}}{A}$ where $A$ is the cross-sectional area of the wire.
Similarly, strain in the wire will be elongation in the wire per unit length, i.e. $\dfrac{l}{L}$
Now, we know that the energy stored in a wire per unit volume, $\dfrac{E}{V} = \dfrac{1}{2} \times stress \times strain$ where, $V$ is the volume of the wire.
When the block of mass is suspended by the wire, this energy will be equivalent to the elastic potential energy of the wire, therefore, we get
$\dfrac{E}{{A \times L}} = \dfrac{1}{2} \times \dfrac{{Mg}}{A} \times \dfrac{l}{L}$ (volume of the wire is equal to cross-sectional area time length of the wire)
On simplifying, we get $E = \dfrac{1}{2} \times Mg \times l$ (like terms get cancelled out)
Or, $E = \dfrac{1}{2}Mgl$
This will be the final answer.

Hence, option (B) is the correct answer.

Note: Energy per unit volume, on multiplying by volume, will give energy of the wire. Depending on the situation, this energy can be called by different terms. For example, when the wire is stretched it is called elastic potential energy. Don’t get confused by these names, the total energy in the wire will remain the same, whatever form it may take, as per the laws of physics.