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Two wires made of same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If $x$ is the increase in the length of the first wire, then the increase in the length of the second wire is
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The correct answer is:
$3 x$
As wires are of same material so both are having same young's modulus
$\Rightarrow \quad Y_1=Y_2$
$\Rightarrow \quad \frac{\frac{F_1}{A_1}}{\frac{\Delta L_1}{L_1}}=\frac{\frac{F_2}{A_2}}{\frac{\Delta L_2}{L_2}}$
$\Rightarrow \quad \frac{F_1 L_1}{\Delta L_1 A_1}=\frac{F_2 L_2}{\Delta L_2 A_2}$
Now given,
$\begin{aligned} L_1 & =3 L_2 \text { and } r_1=3 r_2 \\ \Delta L_1 & =x\end{aligned}$
Also, $\quad F_1=F_2$
So from eq. (i), we have
$\frac{3 L_2}{x\left(3 r_2\right)^2}=\frac{L_2}{\Delta L_2 r_2^2} \Rightarrow \Delta L_2=3 x$
$\Rightarrow \quad Y_1=Y_2$
$\Rightarrow \quad \frac{\frac{F_1}{A_1}}{\frac{\Delta L_1}{L_1}}=\frac{\frac{F_2}{A_2}}{\frac{\Delta L_2}{L_2}}$
$\Rightarrow \quad \frac{F_1 L_1}{\Delta L_1 A_1}=\frac{F_2 L_2}{\Delta L_2 A_2}$
Now given,
$\begin{aligned} L_1 & =3 L_2 \text { and } r_1=3 r_2 \\ \Delta L_1 & =x\end{aligned}$
Also, $\quad F_1=F_2$
So from eq. (i), we have
$\frac{3 L_2}{x\left(3 r_2\right)^2}=\frac{L_2}{\Delta L_2 r_2^2} \Rightarrow \Delta L_2=3 x$
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