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. MgL
• C. \(\frac12Mgl\)
• D. \(\frac12MgL\)

Answer 1

The Correct Option is C

To determine the elastic potential energy stored in the extended wire when a block of mass \(M\) is suspended, we need to consider the properties of the wire and the forces involved.

The wire extends from its original length \(L\) to \(L + l\), showing an elongation of \(l\). The tensile force causing this elongation is due to the weight of the block, which is \(Mg\), where \(g\) is the acceleration due to gravity.

The formula for elastic potential energy (\(U\)) stored in a wire or spring due to elongation is:

U = \frac{1}{2} \times \text{Force} \times \text{Elongation}

Here, the force involved is the weight of the block, \(Mg\), and the elongation is \(l\). Therefore, the formula for the elastic potential energy becomes:

U = \frac{1}{2} \times Mg \times l

This calculation shows that the elastic potential energy stored in the wire is indeed \(\frac{1}{2} Mgl\).

Thus, the correct option is:

\(\frac{1}{2} Mgl\)

Justification for Other Options:

• Mgl: This value would imply that the entire energy is transferred directly to elongation without respecting the energy distribution stored in the wire, according to principles of elasticity.
• MgL: This would imply that the wire entirely stretches by its own length \(L\), which is not the case here.
• \(\frac{1}{2} MgL\): This implies elasticity involving the whole initial length \(L\), which isn’t appropriate in the context of merely considering elongation \(l\).

Hence, with a step-by-step analysis, the elastic potential energy is determined correctly as \(\frac{1}{2} Mgl\).