Question

A wire of length $L$ and radius $r$ is fixed at one end. When a stretching force $F$ is applied at free end, the elongation in the wire is $l$. When another wire of same material but of length $2L$ and radius $2r$, also fixed at one end is stretched by a force $2F$ applied at free end, then elongation in the second wire will be
(A) $l/2$
(B) $l$
(C) $2l$
(D) $l/4$

Answer 1

To solve this question, we need to determine the Young’s modulus of the material of the given wire from the given information. As the material of the second wire is the same as that of the first wire, so we can find out the required elongation from the value of the Young’s modulus.

Formula used: The formula which has been used to solve this question is given by
$Y = \dfrac{{Fl}}{{A\Delta l}}$, here $Y$ is the young’s modulus of a string of length $l$ and area of cross section $A$, $F$ is the force applied on it due to which its length gets changed by $\Delta l$.

Complete step by step answer
We know that the Young’s modulus of a wire is given as
$Y = \dfrac{{Fl}}{{A\Delta l}}$ (1)
We know that the area of cross section of a wire is given by
$A = \pi {R^2}$ (2)
Putting (2) in (1) we get
$Y = \dfrac{{Fl}}{{\pi {R^2}\Delta l}}$ (3)
According to the question, the length of the wire is $l = L$, the elongation is $\Delta l = l$, the radius is $R = r$, and the applied force on the wire is $F$. Putting these in (3) we get
$Y = \dfrac{{FL}}{{\pi {r^2}l}}$ (4)
Now, the second wire is of length $2L$ and radius $2r$. Also, it is stretched by a force of $2F$. Let the elongation in the second wire be $l'$, and its Young’s modulus be $Y'$. Putting these values in (3) we get
$Y' = \dfrac{{\left( {2F} \right)\left( {2L} \right)}}{{\pi {{\left( {2r} \right)}^2}l'}}$
$\Rightarrow Y' = \dfrac{{FL}}{{\pi {r^2}l'}}$ (5)
According to the question, the second wire is of the same material. This means that it’s Young’s modulus is the same as that of the first wire, that is,
$Y = Y'$
From (4) and (5)
$\dfrac{{FL}}{{\pi {r^2}l}} = \dfrac{{FL}}{{\pi {r^2}l'}}$
$\Rightarrow l' = l$
Thus, the elongation in the second wire is equal to that in the first wire, that is, $l$.
Hence, the correct answer is option B.

Note
We should not use the Hooke’s law for solving this question. In this question, the geometrical parameters of the two wires are not the same and hence their force constants will be different.