$\begin{array}{1 1}(A)\;2mm\\(B)\;1.25mm\\(C)\;1mm\\(D)\;3.25mm\end{array} $

Given : $\large\frac{l_1}{l_2}=\frac{3}{5}$

$\large\frac{a_1}{a_2}=(\frac{r_1}{r_2})^2=(\frac{1}{2})^2=\frac{1}{4}$

$\large\frac{F_1}{F_2}=\frac{2}{3}$

$\Delta l_1$=2mm

As 2 wires are of same material their young modulus is same

$Y=(\large\frac{F}{a})\frac{l}{\Delta l}$

$\Rightarrow (\large\frac{F_1}{a_1})\frac{l_1}{\Delta l_1}=(\large\frac{F_2}{a_2})\frac{l_2}{\Delta l_2}$

$\Delta l_2=(\large\frac{F_2}{F_1})(\frac{a_1}{a_2})(\frac{l_2}{l_1})$$\Delta l_1$

$\Rightarrow \large\frac{3}{2}\times \frac{1}{4}\times \frac{5}{3}$$\times 2$mm

$\Rightarrow \large\frac{5}{4}$mm

$\Rightarrow$ 1.25mm

Hence (B) is the correct answer.

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