Q116 / 180 · JIPMER 2016 · Physics
A rod made up of metal is 1.2 m long and 0.8 cm in diameter. Its resistance is $3.5 \times 10^{-3} \Omega$. Another disc made of the same metal is 2.0 cm in diameter and 1.25 mm thick. What is the resistance between the round faces of the disc?
- A$1.35 \times 10^{-8} \Omega$
- B$2.70 \times 10^{-7} \Omega$
- Correct answerC$5.82 \times 10^{-7} \Omega$
- D$8.10 \times 10^{-5} \Omega$
Solution
Given, $I=12 \mathrm{~m}, d=0.8 \mathrm{~cm}$, $R=3.5 \times 10^{-3} \Omega$, thickness $=1.25 \mathrm{~mm}$, Resistivity of the material of the rod
$\begin{aligned} \rho & =\frac{R A}{I}=\frac{3.5 \times 10^{-3} \times \pi\left(0.4 \times 10^{-2}\right)^2}{1.2} \\ & =46.6 \times 10^{-9} \pi \Omega \mathrm{m}\end{aligned}$
$\begin{aligned} & \text { Resistance of disc, } R=\frac{\rho \text { (Thickness) }}{\text { Area of cross-section }} \\ &= \frac{46.6 \times 10^{-9} \pi \times\left(1.25 \times 10^{-3}\right)}{\pi\left(1 \times 10^{-2}\right)^2} \\ &= 5.82 \times 10^{-7} \Omega\end{aligned}$
Hence, the correct answer is option: 'C'.