Question

A force F doubles the length of wire of cross-section ‘a’. The Young modulus of wire is
(A) $\dfrac{F}{a}$
(B) $\dfrac{F}{{3a}}$
(C) $\dfrac{F}{{2a}}$
(D) $\dfrac{F}{{4a}}$

Answer 1

Hint: In this question, we need to determine the Young modulus of the wire such that it elongates to double its length when a force is applied on it. For this, we will use the relation between the Young modulus, force and length of the wire.

Complete step by step answer:
The product of the ratio of the force and the area of the cross-section of the wire and the ratio of the length of the wire to the change in the length of the wire while applying a force along the axis of the cross-section of the wire results in the Young modulus of the wire. Mathematically, $Y = \dfrac{F}{a} \times \dfrac{L}{{\vartriangle L}}$ where, F is the force applied on the wire, ‘a’ is the area of the cross-section of the wire, ‘l’ is the initial length of the wire (before applying the force), $\vartriangle l$ is the change in the length of the wire on applying a force and ‘Y’ is the young modulus of elasticity of the wire.
According to the question, when a force is applied on the wire, then the length of the wire doubles. So, change in the length of the wire is given as $\vartriangle l = 2l - l = l$.
Now, substituting $\vartriangle l = l$ in the equation $Y = \dfrac{F}{a} \times \dfrac{L}{{\vartriangle L}}$ to determine the expression for the young modulus of elasticity of the wire.
$
\Rightarrow Y= \dfrac{F}{a} \times \dfrac{L}{{\vartriangle L}} \\
\Rightarrow Y= \dfrac{F}{a} \times \dfrac{L}{L} \\
\Rightarrow Y= \dfrac{F}{a} \\
$
Hence, the young modulus of the wire such that it elongates when a force (F) is applied on it is $\dfrac{F}{a}$.
Hence,option A is the correct answer.

Note: The Young modulus or the modulus of elasticity in tension, is a mechanical property that measures the tensile stiffness of a solid material. In other words, it is the elasticity measurement of material.