Question

A wire fixed at upper end is stretched by a length l by applying a force F. The work done in stretching is:
(A) $\dfrac{F}{{2l}}$
(B) $Fl$
(C) $2Fl$
(D) $\dfrac{{Fl}}{2}$

Answer 1

Here, wire is considered a massless rod. Since, we know a massless rod behaves as a spring and we can find the spring constant using Young's modulus which describes the stiffness of the material. Here the stress by strain formula is used in which force is replaced by the spring force so that we can get a relation between the spring constant and the Young’s constant.

Formula used:
$\gamma = \dfrac{{\dfrac{F}{A}}}{{\dfrac{l}{L}}}$ ,
$F = k \times l$ and
$dW = Fdl$
Here,$\gamma$ is the young’s modulus,
$dW$ represents small work done,
$dl$ represents small change in length,
$F$ is the force applied on the rod or spring,
$A$ is the area of the cross-section of rod,
$k$ is the spring constant,
$l$ is the change in the length of the rod or spring and
$L$ is the Original length of the rod.

Complete step by step solution:
When the force is applied on the rod it gets elongated to the length $L + l$ .Now when we apply the formula to get the spring constant for rod we get,
$F = k \times l$ and $\gamma = \dfrac{{\dfrac{F}{A}}}{{\dfrac{l}{L}}}$ so, $F = \dfrac{{\gamma {\rm A} \times l}}{L} = k \times l$
Now the spring constant is:
$k = \dfrac{{\gamma {\rm A}}}{L}$
If we now apply work done formula we get,
$dW = Fdl = kl \times dl = \dfrac{{\gamma {\rm A}}}{L} \times l \times dl$
Now on integrating both sides,

$$\int {dW = \dfrac{{\gamma {\rm A}}}{L}\int {ldl} } \\\ \Rightarrow W = \dfrac{{\gamma {\rm A}}}{L} \times \dfrac{{{l^2}}}{2} \\\ \\\$$

If we replace $\gamma$ from the above equation we get,
$W = \dfrac{{Fl}}{2}$
So, by the above solution we reach on a conclusion that the work done by force on stretching the rod is
$\dfrac{{Fl}}{2}$ .
Therefore the correct option is ‘D’.

Note:
The above solution is only valid for massless rod since, there is no mass in the rod here extra force applied is considered as the force applied on the spring as rod have a similar action on stretching or compressing. So, we can get the relation of spring constant for the rod to find the solution of this problem.