Physics · Current Electricity

JEE Main 2024 — 31 January, Shift 2 — Question 41

By what percentage will the illumination of the lamp decrease if the current drops by $20 \%$ ?

Option A:
$46 \%$
Option B:
$26 \%$
Option C:
$36 \%$
Correct
Option D:
$56 \%$

Answer: C

Step-by-step solution

$\mathrm{P}=\mathrm{i}^{2} \mathrm{R}$ $\mathrm{P}_{\text {int }}=\mathrm{I}_{\text {int }}^{2} \mathrm{R}$ $\mathrm{P}_{\text {final }}=\left(0.8 \mathrm{I}_{\mathrm{int}}\right)^{2} \mathrm{R}$ $\%$

change in power $=$

$\frac{\mathrm{P}_{\text {final }}-\mathrm{P}_{\text {int }}}{\mathrm{P}_{\text {int }}} \times 100=(0.64-1) \times 100=-36 \%$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Heating Effects of Current and Thermal Powe

Answer: C

Step-by-step solution

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