Question

The length of two wires are in the ratio $3:4$. Ratio of the diameters is $1:2$ young's modulus of the wires are in the ration $3:2$; if they are subjected to same tensile force, the ration of the elongation produced is

• A. 1:1
• B. 1:2
• C. 2:3
• D. 2:1

Answer 1

The correct option is D 2:1

$\begin{array}{c}\text{ratio of lengths}\frac{l_1}{l_2}=\frac{3}{4}\\ \text{ratio of diameters}\frac{d_1}{d_2}=\frac{1}{2}\\ \text{ratio of young's modulus}\frac{Y_1}{Y_2}=\frac{3}{2}\end{array}$

$\begin{array}{c}\text{Given, Forces are same so}{F}_1=f_2\\ \text{Formula for elongation}\Delta l=\frac{Fl}{Ay}\\ \text{ratio of elongation}\frac{\Delta l_1}{\Delta l_2}=\frac{F_1{l}_1}{A_1{y}_1}\times \frac{A_2{y}_2}{F_2{l}_2}\end{array}$

$\begin{array}{c}=\left(\frac{l_1}{l_2}\right)\times \left(\frac{A_2}{A_1}\right)\times \left(\frac{Y_2}{Y_1}\right)\\ =\frac{3}{4}\times \left(\frac{\pi d_2^2}{\pi d_1^2}\right)\times \frac{2}{3}\\ =\frac{3}{4}\times 4\times \frac{2}{3}\\ =2:1\end{array}$