Q134 / 160 · TS EAMCET 2019 (06 May Shift 1) · Physics
A copper wire and an aluminium wire have lengths in the ratio $5: 2$, diameters in the ratio $4: 3$ and forces applied in the ratio $4: 5$. Find the ratio of increase in length of the copper wire to that aluminium wire. (Let $Y_{\mathrm{Cu}}=1.1 \times 10^{11} \mathrm{Nm}^{-2}, Y_{\mathrm{Al}}=0.7 \times 10^{11} \mathrm{Nm}^{-2}$ )
- A$\frac{176}{63}$
- Correct answerB$\frac{63}{88}$
- C$\frac{189}{11}$
- D$\frac{33}{89}$
Solution
Given, ratio of length of copper wire and
aluminium wire, $l_1: l_2=5: 2$
and diameter ratio of the wire $=d_1: d_2=4: 3$
$\therefore \quad r_1: r_2=2: \frac{3}{2}$
If $\Delta l_1$ and $\Delta l_2$ are increase in the length of copper and aluminium wire, respectively then for copper wire,
$\Delta l_1=\frac{F_1 l_1}{A_1 y_1}$
For aluminium wire,
$\Delta l_2=\frac{F_2 l_2}{A_2 y_2}$
From Eqs. (i) and (ii), we get
$\frac{\Delta_1}{\Delta l_2}=\frac{l_1}{l_2} \cdot \frac{A_2}{A_1} \cdot \frac{y_2}{y_1} \cdot \frac{F_1}{F_2} \quad\left[\because \frac{y_2}{y_1}=\frac{y_{\mathrm{Al}}}{y_{\mathrm{Cu}}}=\frac{0.7}{1.1}\right]$
$=\frac{5}{2} \cdot \frac{\pi r_2^2}{\pi r_1^2} \cdot \frac{0.7 \times 10^{11}}{1.1 \times 10^{11}} \cdot \frac{4}{5} \quad\left[\because\right.$ Given, $\left.\frac{F_1}{F_2}=\frac{4}{5}\right]$
$=\frac{5}{2} \cdot \frac{\left(\frac{3}{2}\right)^2}{(2)^2} \times \frac{7}{11} \times \frac{4}{5}=\frac{63}{88}$
Hence, the ratio of increase in length of the copper and aluminium wire will be $\frac{63}{88}$.
Hence, the correct answer is option: 'B'.