Physics · Electrostatics

JEE Main 2026 — 28 January, Evening Shift — Question 46

Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s . When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s . If the original frequency of tuning fork B is 380 Hz , then the original frequency of tuning fork A is $\_\_\_\_$ Hz.

Answer: 384

Numerical answer — enter this value.

Step-by-step solution

$\left|f_{A}-f_{B}\right|=4$ $\left|\mathrm{f}_{\mathrm{A}}-380\right|=4$ So, $\mathrm{f}_{\mathrm{A}}=384 \mathrm{~Hz}$ or 376 Hz on loading with wax $\mathrm{f}_{\mathrm{A}}$ decreases So, $\mathrm{f}_{\mathrm{A}}=384 \mathrm{~Hz}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law