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. $\frac{1}{2}Mgl$
• B. $\frac{1}{2}MgL$
• C. $mgl$
• D. $MgL$

Answer 1

The Correct Option is A

To find the elastic potential energy stored in the extended wire, we will use the formula for elastic potential energy. The elastic potential energy \( (U) \) stored in a wire or a spring is given by the formula:

U = \frac{1}{2} k x^2

where \( k \) is the spring constant, and \( x \) is the extension in the length of the wire.

In this scenario, the force causing the extension in the wire is the weight of the block, which is \( Mg \). The extension in the wire is given as \( l \), so we can express the spring constant \( k \) in terms of the force and the extension as:

F = k \cdot x \Rightarrow Mg = k \cdot l \Rightarrow k = \frac{Mg}{l}

Now substitute the value of \( k \) into the elastic potential energy formula:

U = \frac{1}{2} \cdot \frac{Mg}{l} \cdot l^2 = \frac{1}{2} \cdot Mg \cdot l

Therefore, the elastic potential energy stored in the extended wire is:

\frac{1}{2}Mgl

Thus, the correct answer is:

\frac{1}{2}Mgl

Other given options do not satisfy the conditions of elastic potential energy stored due to the applied force (weight) and the resulting elongation in this setup.