On stretching a wire what is the elastic energy stored per unit volume? (A) \([\mathrm{F} \ell / 2 \mathrm{AL}]\) (B) [FA/2L] (C) \([\mathrm{FL} / 2 \mathrm{~A}]\) (D) \([\mathrm{FL} / 2]\)

The elastic potential energy (U) stored in a stretched wire is given by the formula: U = (1/2) * F * ∆l where F is the applied force and ∆l is the change in length of the wire.

Hooke's Law relates the force (F) applied on the wire to the extension (∆l) and the spring constant (k): F = k * ∆l

The spring constant (k) is related to the Young's modulus (Y), the length (l), and the area (A) of the wire as follows: k = (Y * A) / l

Substituting the expressions for spring constant (k) and force (F) in the equation for elastic potential energy (U), we get: U = (1/2) * (Y * A * ∆l^2) / l

Divide the elastic potential energy (U) by the volume (V) of the wire to find the energy stored per unit volume (u), where V = A * l: u = U / V = ((1/2) * (Y * A * ∆l^2) / l) / (A * l) Simplifying the expression, we get: u = (Y * ∆l^2) / (2 * l^2) Now let's compare this expression with the given options. (A) [\(\mathrm{F} \ell / 2 \mathrm{AL}\)] - This option is incorrect as it does not include the Young's modulus (Y) and has the incorrect relationship. (B) [FA/2L] - This option is incorrect as it does not involve the Young's modulus (Y) and the extension (∆l). (C) [\(\mathrm{FL} / 2 \mathrm{~A}\)] - This option is incorrect as it does not involve the Young's modulus (Y) and the extension (∆l). (D) [\(\mathrm{FL} / 2\)] - This option is incorrect as it does not have the correct relationship for the energy stored per unit volume. None of the given options match our derived expression for the elastic energy stored per unit volume. Therefore, the exercise does not provide the correct option. The correct expression should be: \(u = (Y * ∆l^2) / (2 * l^2)\).