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

Answer 1

Answer: (D)

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 Eq. (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
$F_2=T^{\prime }=T+2g=\left(4g\right)$ newton
Now, putting the value of $F_1,F_2,$ in Eq. (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 $\frac{\Delta l_1}{\Delta l_2}=\left(\frac{a}{2b^{2}c}\right)$