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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2026 · 22 January · Morning Shift · Physics · Waves and Sound

Two loudspeakers ( L_ 1 and L_ 2 ) are placed with a separation of...

Two loudspeakers (L1L_{1} and L2L_{2}) are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point AA, at equidistance to both loud speakers, is moved by 25 m along the line ABA B while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is ____\_\_\_\_ Hz (Speed of sound in air is 324 m/s324 \mathrm{~m} / \mathrm{s} and 5=2.23\sqrt{5}=2.23)
Two loudspeakers ( L_ 1 and L_ 2 ) are placed with a separation of 10 m, as shown in figure. Both speakers are fed with

Options

  1. ABoth Statement I and Statement II are true
  2. BBoth Statement I and Statement II are false
  3. CStatement I is true but Statement II is false
  4. DStatement I is false but Statement II is true

Correct answer

D. Statement I is false but Statement II is true

Step-by-step solution

Point B will 10th 10^{\text{th }} maxima
Δx=L2B−L1B\Delta x=L_2 B-L_1 B
L1B=202+402=205 m=44.6 mL_1 B=\sqrt{20^2+40^2}=20 \sqrt{5} \mathrm{~m}=44.6 \mathrm{~m}
L2B=402+302=50 mL_2 B=\sqrt{40^2+30^2}=50 \mathrm{~m}
Δx=50−44.6=5.4 m\Delta x=50-44.6=5.4 \mathrm{~m}
Δx=nλ\Delta x=n \lambda
5.4=10×λ5.4=10 \times \lambda
λ=0.54 m\lambda=0.54 \mathrm{~m}
V=fλV=f \lambda
f=3240.54=600 Hzf=\frac{324}{0.54}=600 \mathrm{~Hz}
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