JEE Physics 2023 Question with Solution

Given: Current $I = 2 \, \text{A}$, potential difference $V = 3.4 \, \text{V}$, mass $m = 8.92 \times 10^{-3} \, \text{kg}$, density $\rho_d = 8.92 \times 10^3 \, \text{kg/m}^3$, resistivity $\rho = 1.7 \times 10^{-8} \, \Omega \, \text{m}$.

Find: The length $L$ of the wire.

Using Ohm's law,

$$R = \frac{V}{I} = \frac{3.4}{2} = 1.7 \, \Omega$$

For the wire,

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

so

$$1.7 = \frac{\rho L}{A}$$

and hence

$$L = \frac{1.7A}{\rho}$$

Now use mass = density $\times$ volume:

$$m = \rho_d (AL)$$

Therefore,

$$AL = \frac{m}{\rho_d} = \frac{8.92 \times 10^{-3}}{8.92 \times 10^3} = 10^{-6} \, \text{m}^3$$

From

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

we get

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

Substituting into $AL = 10^{-6}$,

$$\frac{\rho L}{R} \cdot L = 10^{-6}$$ $$\frac{\rho L^2}{R} = 10^{-6}$$ $$L^2 = \frac{R}{\rho} \times 10^{-6}$$

Now substitute the values:

$$L^2 = \frac{1.7}{1.7 \times 10^{-8}} \times 10^{-6} = 10^2$$

Thus,

$$L = 10 \, \text{m}$$

Therefore, the computed length is $10 \, \text{m}$.

However, the solution explicitly concludes "The Correct Option is A" while the numerical working gives $10 \, \text{m}$, which matches option B. the answer is marked as A.

Given: $R = 1.7 \, \Omega$ from Ohm's law and $AL = 10^{-6} \, \text{m}^3$ from mass and density.

Find: Express $L$ only in known quantities.

Start with

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

Rearrange for $A$:

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

Multiply by $L$:

$$AL = \frac{\rho L^2}{R}$$

But

$$AL = 10^{-6}$$

so

$$\frac{\rho L^2}{R} = 10^{-6}$$ $$L^2 = \frac{R}{\rho} \times 10^{-6}$$

Substitute $R = 1.7$ and $\rho = 1.7 \times 10^{-8}$:

$$L^2 = \frac{1.7}{1.7 \times 10^{-8}} \times 10^{-6} = 10^8 \times 10^{-6} = 10^2$$ $$L = 10 \, \text{m}$$

So the working supports option B, even though the solution labels the correct option as A.