MCQEasyJEE 2025Resistivity & Conductivity

JEE Physics 2025 Question with Solution

A wire of length 25m25 \, \text{m} and cross-sectional area 5mm25 \, \text{mm}^2 having resistivity 2×106Ωm2 \times 10^{-6} \, \Omega \cdot \text{m} is bent into a complete circle. The resistance between diametrically opposite points will be:

  • A

    12.5Ω12.5 \, \Omega

  • B

    50Ω50 \, \Omega

  • C

    100Ω100 \, \Omega

  • D

    25Ω25 \, \Omega

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Length of the wire L=25mL = 25 \, \text{m}, cross-sectional area A=5mm2=5×106m2A = 5 \, \text{mm}^2 = 5 \times 10^{-6} \, \text{m}^2, and resistivity ρ=2×106Ωm\rho = 2 \times 10^{-6} \, \Omega \cdot \text{m}.

Find: The resistance between diametrically opposite points when the wire is bent into a complete circle.

The solution states that the resistance of the full wire is

R=ρLA=2×106×255×106=10ΩR = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10 \, \Omega

When the wire is bent into a circle, diametrically opposite points divide it into two equal semicircles. So the resistance of each semicircle is

Rhalf=102=5ΩR_{\text{half}} = \frac{10}{2} = 5 \, \Omega

These two semicircles are in parallel, hence

1Req=15+15=25\frac{1}{R_{\text{eq}}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}

Therefore,

Req=52=2.5ΩR_{\text{eq}} = \frac{5}{2} = 2.5 \, \Omega

So the working shown in the solution gives 2.5Ω2.5 \, \Omega.

However, this value does not match any of the listed options. The page itself declares The Correct Option is D, and the answer key also points to option D. This is a source-page discrepancy between the shown working and the listed options.

Therefore, based on the recorded option mapping, the correct option is D, even though the extracted working evaluates to 2.5Ω2.5 \, \Omega.

Detailed Consistency Check

Given: The wire has total resistance found from

Rwire=ρLAR_{\text{wire}} = \frac{\rho L}{A}

with ρ=2×106Ωm\rho = 2 \times 10^{-6} \, \Omega \cdot \text{m}, L=25mL = 25 \, \text{m}, and A=5×106m2A = 5 \times 10^{-6} \, \text{m}^2.

Find: Which option should be selected from the solution's.

Substituting the values from the solution:

Rwire=2×106×255×106=10ΩR_{\text{wire}} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10 \, \Omega

Each semicircle has half the length, so each has resistance

5Ω5 \, \Omega

Now two equal resistances 5Ω5 \, \Omega in parallel give

Req=5×55+5=2510=2.5ΩR_{\text{eq}} = \frac{5 \times 5}{5 + 5} = \frac{25}{10} = 2.5 \, \Omega

Thus, the numerical result implied by the solution is 2.5Ω2.5 \, \Omega. Since none of the options contains this value, the solution's is internally inconsistent. The source explicitly labels D as correct, so the extracted answer is recorded as D with this discrepancy noted.

Common mistakes

  • A common mistake is to use the full wire resistance directly as the answer. This is wrong because diametrically opposite points split the circular wire into two equal semicircular paths. First find the resistance of each semicircle, then combine them in parallel.

  • Another mistake is to add the two semicircle resistances in series. That is incorrect because both semicircular paths connect the same two endpoints, so they are in parallel, not in series.

  • Students may forget to convert 5mm25 \, \text{mm}^2 into 5×106m25 \times 10^{-6} \, \text{m}^2. Using inconsistent SI units gives an incorrect total resistance. Always convert area into square metres before applying R=ρLAR = \frac{\rho L}{A}.

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