JEE Physics 2024 Question with Solution

Given: Percentage error in length $L$ is $0.1\%$ and percentage error in diameter $d$ is $0.1\%$.

Find: Percentage error in resistance $R$ of the wire.

The resistance of a wire is

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

For a circular cross-section,

$$A = \frac{\pi d^2}{4}$$

So,

$$R = \frac{4\rho L}{\pi d^2}$$

Hence $R \propto \frac{L}{d^2}$.

For error propagation in products and powers,

$$\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$$

Substituting the given percentage errors,

$$\text{Percentage error in } R = 0.1\% + 2 \times 0.1\%$$ $$\text{Percentage error in } R = 0.3\%$$

Therefore, the percentage error in measuring the resistance is $0.3\%$. The correct option is B.

Given: $R \propto \frac{L}{d^2}$, with percentage errors in $L$ and $d$ equal to $0.1\%$ each.

Find: Percentage error in $R$.

Because $L$ is in the numerator with power $1$, its percentage error is added once. Because $d$ is in the denominator with power $2$, its percentage error is added twice.

So directly,

$$\text{Error in } R = 0.1\% + 2(0.1\%) = 0.3\%$$

Therefore, the correct option is B.