JEE Physics 2023 Question with Solution

Given:

• Resistance, $R = 2 \times 10^{-3} \, \Omega$
• Resistivity, $\rho = 2.4 \times 10^{-8} \, \Omega \, \text{m}$
• Length, $\ell = 3.14 \, \text{m}$
• Inner diameter $= 4 \, \text{mm}$, so inner radius $a = 2 \times 10^{-3} \, \text{m}$
• Outer diameter $= 8 \, \text{mm}$, so outer radius $b = 4 \times 10^{-3} \, \text{m}$

Find: The value of $n$.

For a conductor,

$$R = \rho \frac{\ell}{A}$$

where $A$ is the cross-sectional area of the hollow cylinder.

The area is

$$A = \pi (b^2 - a^2)$$

Substituting the radii,

$$A = \pi \left[(4 \times 10^{-3})^2 - (2 \times 10^{-3})^2\right]$$ $$A = \pi \left[16 \times 10^{-6} - 4 \times 10^{-6}\right]$$ $$A = \pi \cdot 12 \times 10^{-6}$$

Now,

$$R = \rho \frac{\ell}{A} = \frac{2.4 \times 10^{-8} \cdot 3.14}{\pi \cdot 12 \times 10^{-6}}$$

Using the given values, this gives

$$R = 2 \times 10^{-3} \, \Omega$$

So the coefficient $n = 2$.

Therefore, the numerical answer is $2$.