Physics · Current Electricity

JEE Main 2024 — 8 April, Shift 2 — Question 39

Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be $\qquad$ to $\qquad$ times of its initial length if the water is to be boiled in 15 minutes.

Option A:
increased, $\frac{3}{4}$
Option B:
increased, $\frac{4}{3}$
Option C:
decreased, $\frac{3}{4}$
Correct
Option D:
decreased, $\frac{4}{3}$

Answer: C

Step-by-step solution

$\mathrm{P}=\frac{\mathrm{v}^{2}}{\mathrm{R}}, \mathrm{R}=\frac{\rho \ell}{\mathrm{A}}$ $\mathrm{P} \propto \frac{1}{\ell}$

$\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{15}{20}=\frac{\ell_{2}}{\ell_{1}}$

$\ell_{2}=\frac{3}{4} \ell_{1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Heating Effects of Current and Thermal Powe

A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Omega and a…

Answer: C

Step-by-step solution

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