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# 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) $\dfrac{F}{{4a}}$

Last updated date: 20th Jun 2024
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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.

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 $\dfrac{F}{a}$.