Question

A metal wire of length L, area of cross section A and Young's modulus Y is stretched by a variable force F such that F is always slightly greater than the elastic forces of resistance in the wire. Then the elongation of the wire is $l$.

• A. The work done by F is $\frac{YA{l}^2}{2L}$
• B. The work done by F is $\frac{YA{l}^2}{L}$
• C. The elastic potential energy stored in the wire is $\frac{YA{l}^2}{2L}$
• D. No heat is produced during the elongation

Answer 1

Answer: (A), (C), (D)

The elastic potential energy stored in the wire is $\frac{YA{l}^2}{2L}$
The work done by F is $\frac{YA{l}^2}{2L}$
No heat is produced during the elongation

We have
$l=\frac{FL}{AY}$
or
$F=\frac{AY}{L}l$
Work done is given using the relation
$dW=Fdx$
or
$dW=\frac{AY}{L}xdx$
or
$W=\frac{AY}{L}{\int }_0^{l}xdx$
or
$W=\frac{AY{l}^2}{2L}$
The elastic potential energy is given as $\Delta U=W$
or
$\Delta U=\frac{AY{l}^2}{2L}$ (This is the energy stored in the wire).
All the work done on the wire is stored as its potential energy, thus there is no heat produced during elongation.