Q146 / 160 · TS EAMCET 2023 (12 May Shift 2) · Physics
Two wires made of the same material have lengths in the ratio $2: 3$ and radii in the ratio $8: 9$. If the same potential difference is applied across the ends of the wires, the ratio of the electric currents flowing through them is
- A$5: 6$
- B$6: 5$
- C$4: 3$
- Correct answerD$32: 27$
Solution
Ratio of length, $\frac{\ell_1}{\ell_2}=\frac{2}{3}$
Ratio of radii, $\frac{r_1}{r_2}=\frac{8}{9}$
Resistance of wire is given as, $R=\rho \frac{\ell}{A}$
Resistance of $\mathrm{I}^{\text {st }}$ wire, $\mathrm{R}_1=\rho \frac{\ell_1}{\mathrm{~A}_1}$
$$
\mathrm{R}_1=\frac{\rho \ell_1}{\pi \mathrm{r}_1^2}
$$
Resistance of II ${ }^{\text {nd }}$ wire, $R_2=\rho \frac{\ell_2}{A_2}$
$$
\mathrm{R}_2=\frac{\rho \ell_2}{\pi \mathrm{r}_2^2}
$$
Dividing (i) and (ii):
$$
\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\rho \ell_1}{\pi \mathrm{r}_1^2} \times \frac{\pi \mathrm{r}_2^2}{\rho \mathrm{l}_2}=\frac{2}{3} \times \frac{81}{64}=\frac{27}{32}
$$
Using ohm's law, $\mathrm{V}=\mathrm{IR}$
$$
\begin{aligned}
& I=\frac{V}{R} \\
& \Rightarrow \quad \frac{I_1}{I_2}=\frac{V}{R_1} \times \frac{R_2}{V}=\frac{R_2}{R_1}=\frac{32}{27} \\
& \therefore I_1: I_2=32: 27
\end{aligned}
$$
Hence, the correct answer is option: 'D'.