Question

If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are $a, b$ and $c$ respectively, the ratio between the increase in lengths of brass and steel wires would be

• A. $\frac{b^{2}a}{2c}$
• B. $\frac{bc}{2a^2}$
• C. $\frac{ba^{2}}{2c}$
• D. $\frac{c}{2b^2c}$

Answer 1

Answer: (D)

$\frac{c}{2b^2c}$

Answer 2

Given, $\frac{l_{1}}{l_{2}} =a, \frac{r_{1}}{r_{2}} =b, \frac{Y_{1}}{Y_{2}} =c$
Free body diagram of the two blocks brass and steel are
Let Young's modulus of steel is $Y_1$ and of brass is $Y_2$.
$\therefore Y_{1} = \frac{F_{1} .l_{1}}{A_{1} .\Delta l_{1}}$ .....(i)
and $Y_{2} = \frac{F_{2} .l_{2}}{A_{2} .\Delta l_{2}}$ ....(ii)
Dividing E (i) by (ii),
$\frac{Y_{1}}{Y_{2}} = \frac{\frac{F_{1} .l_{1}}{A_{1} .\Delta l_{1}}}{\frac{F_{2} . l_{2}}{A_{2} .\Delta l_{2}}}$
or $\frac{Y_{1}}{Y_{2}} = \frac{F_{1} .A_{2} .l_{1} .\Delta l_{2}}{F_{2} .A_{1} .l_{2} .\Delta l_{1}} \,\,\,...(iii)$
Force on steel wire from free body diagram
$T =F_{1} = \left(2g\right)$ newton
Force on brass wire from free body diagram
$F_{2} =T' =T + 2g =\left(4g\right)$ newton
Now, putting the value of $F_1, F_2,$ in E (iii), we get
$\frac{Y_{1}}{Y_{2}} = \left(\frac{2g}{4g}\right). \left(\frac{\pi r_{2}^{2}}{\pi r_{1}^{2}}\right) . \left[\frac{l_{1}}{l_{2}}\right] .\left(\frac{\Delta l_{2}}{\Delta l_{1}}\right)$
or $c = \frac{1}{2} \left(\frac{1}{b^{2}}\right). a\left(\frac{\Delta l_{2}}{\Delta l_{1}}\right)$
or $\frac{\Delta l_{1}}{\Delta l_{2}} = \left(\frac{a}{2b^{2}c}\right)$