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

Answer 1

Answer: (D)

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

Answer 2

Free body diagram of the two blocks are
Given, $\frac{l_{1}}{l_{2} } = a, \frac{r_{1}}{r_{2}} = b , \frac{Y_{1}}{Y_{2}}$
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)
Diving E (i) by E (ii), we get
$\frac{Y_{1}}{Y_{2} } = \frac{\frac{F_{1}l_{1}}{A_{1}l_{1}}}{\frac{F_{2}.l_{2}}{A_{2} . \Delta l_{2}}}$
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
$\frac{Y_{1}}{Y_{2} } = \frac{2g}{4g} . \frac{\pi r^{2}_{2}}{\pi r^{2}_{1}} . \frac{l_{2}}{l_{2}} . \frac{\Delta l_{2}}{\Delta l_{1}}$
Or $c = \frac{1}{2} \frac{1}{b^{2} } . a \frac{\Delta l_{2}}{\Delta l_{1}}$
Or $\frac{\Delta l_{1}}{\Delta l_{2}} = \frac{a}{2b^{2} c }$