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NVAEasyJEE 2023Ohm's Law & Resistance

JEE Physics 2023 Question with Solution

The length of a metallic wire is increased by 20%20\% and its area of cross-section is reduced by 4%4\%. The percentage change in resistance of the metallic wire is:

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: The length of the metallic wire is increased by 20%20\% and the area of cross-section is reduced by 4%4\%.

Find: The percentage change in resistance.

The resistance of a conductor is given by

R=ρlAR = \frac{\rho l}{A}

After the changes,

R′=ρ⋅1.2l0.96A=1.25ρlAR' = \frac{\rho \cdot 1.2l}{0.96A} = \frac{1.25\rho l}{A}

So,

R′=1.25RR' = 1.25R

Therefore, the percentage change in resistance is 25%25\%.

So, the required numerical answer is 25.

Using percentage change formula

Given: R=ρLAR = \frac{\rho L}{A}, L′=1.2LL' = 1.2L, and A′=0.96AA' = 0.96A.

Find: The percentage change in resistance.

The new resistance is

R′=ρL′A′=ρ⋅1.2L0.96AR' = \frac{\rho L'}{A'} = \frac{\rho \cdot 1.2L}{0.96A}

Thus,

R′=1.25(ρLA)=1.25RR' = 1.25 \left(\frac{\rho L}{A}\right) = 1.25R

Now use the percentage change formula:

R′−RR×100%\frac{R' - R}{R} \times 100\%

Substituting R′=1.25RR' = 1.25R,

1.25R−RR×100%=25%\frac{1.25R - R}{R} \times 100\% = 25\%

Therefore, the percentage change in resistance is 25%25\%. Hence, the correct numerical answer is 25.

Common mistakes

  • Using the changed area as 1.04A1.04A instead of 0.96A0.96A. A reduction of 4%4\% means multiplying by 0.960.96, not adding 4%4\%.

  • Assuming resistance changes only with length. Resistance depends on both LL and AA through R=ρLAR = \frac{\rho L}{A}, so both changes must be included.

  • Writing the answer as 25%25\% in the answer field. For a numerical value answer, only the number 25 should be entered, without the percent sign.

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