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:

Question

Which has the highest first ionization energy?

Answer 1

To determine the elastic potential energy stored in the extended wire, let's analyze the problem using the concepts of elasticity and potential energy.

The wire is stretched by a length \( l \) when a block of mass \( M \) is suspended. The force exerted by the mass on the wire is given by the weight of the block:

\[ F = Mg \]

where \( g \) is the acceleration due to gravity.

The elastic potential energy (U) stored in a wire, or any elastic object, when stretched is given by the formula:

\[ U = \frac{1}{2} \times \text{Force} \times \text{Extension} \]

Substituting the values:

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

This formula shows that the elastic potential energy stored in the wire is proportional to both the weight of the block and the amount of stretch.

Therefore, the correct answer is \(\frac{1}{2} Mgl\), which matches option (C).

Let's briefly analyze the incorrect options:

Mgl: This represents the total work done or energy corresponding to the full stretch force and distance, not halved as elastic energy requires.
MgL and \(\frac{1}{2} MgL\): These involve the original length of the wire, \( L \), rather than the extension \( l \). The extension determines the elastic potential energy.

In elastic energy problems where a force causes an extension or compression, remember that energy stored is typically halved from initial force considerations, aligning with the formula derived.

Answer 2