Q1JEE Main 2026MCQ

In an open organ pipe $ν_3$ and $ν_6$ are $3^{\;\text{rd }\;}$ and $6^{\;\text{th }\;}$ harmonic frequencies, respectively. If $ν_6-ν_3=2200 Hz$ then length of the pipe is $\_\_\_\_$ mm. (Take velocity of sound in air is $330 m / s$ .)

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📖 Explanation

For an open organ pipe, the $n^{\text{th}}$ harmonic frequency is given by $\nu_n = \frac{nv}{2L}$ .
The given harmonic frequencies are $\nu_3 = \frac{3v}{2L}$ and $\nu_6 = \frac{6v}{2L}$ .
The difference is $\nu_6 - \nu_3 = \frac{6v}{2L} - \frac{3v}{2L} = \frac{3v}{2L} = 2200 \text{ Hz}$ .
Substituting $v = 330 \text{ m/s}$ , we get $\frac{3 \times 330}{2L} = 2200 \implies \frac{990}{2L} = 2200$ .
Solving for $L$ , we find $L = \frac{990}{4400} = 0.225 \text{ m}$ .
Converting to millimeters, $L = 0.225 \times 1000 \text{ mm} = 225 \text{ mm}$ .

Q2JEE Main 2026NAT

Two loudspeakers $(L_1.$ and $L_2)$ 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 $A$ , at equidistance to both loud speakers, is moved by 25 m along the line $A B$ 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 $324 m / s$ and ${\sqrt{5}}=2.23$ )

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📖 Explanation

The path difference at any point $P$ on line $AB$ at a distance $y$ from $A$ is given by $\Delta x = \sqrt{D^2 + (y + d/2)^2} - \sqrt{D^2 + (y - d/2)^2}$ , where $D = 40\text{ m}$ and $d = 10\text{ m}$ .
At point $B$ ( $y = 25\text{ m}$ ), the path difference is:
$\Delta x = \sqrt{40^2 + (25 + 5)^2} - \sqrt{40^2 + (25 - 5)^2} = \sqrt{1600 + 900} - \sqrt{1600 + 400} = 50 - \sqrt{2000}$
Using $\sqrt{5} = 2.23$ , we have $\sqrt{2000} = 20\sqrt{5} = 20(2.23) = 44.6\text{ m}$ , so $\Delta x = 50 - 44.6 = 5.4\text{ m}$ .
For $10$ cycles of interference, the total path difference corresponds to $10$ wavelengths, so $10\lambda = 5.4\text{ m}$ , which gives $\lambda = 0.54\text{ m}$ .
The frequency of the input signal is $f = \frac{v}{\lambda} = \frac{324}{0.54} = 600\text{ Hz}$ .

[CORRECT_OPTION: 600]

Q3JEE Main 2026MCQ

A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

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📖 Explanation

For a point source, Intensity varies as $I\propto\frac{1}{r^2}$ If volume of spherical detector increases 8 \times , $V'=8V$ For sphere, $V=\frac{4}{3}\pi r^3$ so $r'^3=8r^3$ $r'=2r$ Now intensity becomes $I'=\frac{1}{(2r)^2}$ $=\frac{1}{4}\cdot\frac{1}{r^2}$ So $I'=\frac{I}{4}$ Hence intensity becomes one-fourth of original.

Q4JEE Main 2026NAT

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 original frequency of tuning fork $A$ is $\_\_\_\_$ Hz.

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📖 Explanation

The initial beat frequency is $f_b = \frac{8}{2} = 4 \text{ Hz}$ . Since $f_B = 380 \text{ Hz}$ , the original frequency of $A$ is $f_A = 380 \pm 4 \text{ Hz}$ , giving $f_A = 384 \text{ Hz}$ or $376 \text{ Hz}$ .
When $A$ is loaded, its frequency $f_A$ decreases to $f_A'$ , and the new beat frequency becomes $f_b' = \frac{4}{2} = 2 \text{ Hz}$ .
This implies $|f_A' - 380| = 2 \text{ Hz}$ , so $f_A' = 382 \text{ Hz}$ or $378 \text{ Hz}$ .
If $f_A = 376 \text{ Hz}$ , loading $A$ would decrease its frequency below $376 \text{ Hz}$ , moving it further from $380 \text{ Hz}$ and increasing the beat frequency, which contradicts the result.
If $f_A = 384 \text{ Hz}$ , loading $A$ decreases its frequency towards $380 \text{ Hz}$ (specifically to $382 \text{ Hz}$ ), which is consistent with the beat frequency reducing to $2 \text{ Hz}$ .
Therefore, the original frequency $f_A = 384 \text{ Hz}$ .

[CORRECT_OPTION: 384]

Q5JEE Main 2026MCQ

The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is $5 / x$ . The value of $x$ is $\_\_\_\_$

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📖 Explanation

The fifth harmonic of a closed organ pipe of length $L_c$ is $f_c = \frac{5v}{4L_c}$ .
The first harmonic of an open organ pipe of length $L_o$ is $f_o = \frac{v}{2L_o}$ .
Equating the two frequencies for unison: $\frac{5v}{4L_c} = \frac{v}{2L_o}$ .
Canceling $v$ and rearranging the terms gives the ratio of lengths: $\frac{L_c}{L_o} = \frac{5 \times 2}{4} = \frac{10}{4} = \frac{5}{2}$ .
Given the ratio is $\frac{5}{x}$ , we find $x = 2$ .

Q6JEE Main 2026MCQ

Two strings $(A, B)$ having linear densities $\mu _A=2 \times 10^{-4} kg / m$ and, $\mu _B=4 \times 10^{-4} kg / m$ and lengths $L_A=2.5 m$ and $L_B=1.5 m$ respectively are joined. Free ends of $A$ and $B$ are tied to two rigid supports $C$ and $D$ , respectively creating a tension of 500 N in the wire. Two identical pulses, sent from $C$ and $D$ ends, take time $t_1$ and $t_2$ , respectively, to reach the joint. The ratio $t_1 / t_2$ is:

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📖 Explanation

The speed of a transverse wave in a string is $v = \sqrt{\frac{T}{\mu}}$ , and the time taken to travel distance $L$ is $t = \frac{L}{v} = L \sqrt{\frac{\mu}{T}}$ .
For the pulses to reach the joint from supports $C$ and $D$ , the times are $t_1 = L_A \sqrt{\frac{\mu_A}{T}}$ and $t_2 = L_B \sqrt{\frac{\mu_B}{T}}$ .
Taking the ratio, we get:
$\frac{t_1}{t_2} = \frac{L_A}{L_B} \sqrt{\frac{\mu_A}{\mu_B}} = \frac{2.5}{1.5} \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}}$
$\frac{t_1}{t_2} = \frac{5}{3} \times \sqrt{\frac{1}{2}} = \frac{5}{3 \times 1.414} \approx 1.1785$
Rounding to two decimal places, we obtain $1.18$ .

Q7JEE Main 2026NAT

The velocity of sound in air is doubled when the temperature is raised from $0^{\circ} C$ to $\alpha ^{\circ} C$ . The value of $\alpha$ is $\_\_\_\_$ 。

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📖 Explanation

The velocity of sound $v$ is proportional to the square root of the absolute temperature $T$ , expressed as $v \propto \sqrt{T}$ .
Let $v_1$ be the velocity at $0^\circ C$ ( $T_1 = 273 K$ ) and $v_2$ be the velocity at $\alpha^\circ C$ ( $T_2 = \alpha + 273 K$ ).
Given $v_2 = 2v_1$ , we have the ratio $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$ , which implies $2 = \sqrt{\frac{\alpha + 273}{273}}$ .
Squaring both sides gives $4 = \frac{\alpha + 273}{273}$ , leading to $\alpha + 273 = 4 \times 273 = 1092$ .
Solving for $\alpha$ , we get $\alpha = 1092 - 273 = 819$ .

[CORRECT_OPTION: 819]

Q8JEE Main 2025MCQ

Displacement of a wave is expressed as $x(t)=5 \cos (628 t+\;\frac{\pi }{2}) {\text{m}}$ . The wavelength of the wave when its velocity is $300 {\text{m}} / {\ s}$ is : $(\pi =3.14)$

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📖 Explanation

Given the wave equation $x(t) = 5 \cos(628t + \frac{\pi}{2})$ , the angular frequency $\omega$ is $628 \text{ rad/s}$ .
The frequency $f$ of the wave is calculated as:
$f = \frac{\omega}{2\pi} = \frac{628}{2 \times 3.14} = \frac{628}{6.28} = 100 \text{ Hz}$
Using the relationship between wave velocity $v$ , frequency $f$ , and wavelength $\lambda$ , which is $v = f \lambda$ , we find:
$\lambda = \frac{v}{f} = \frac{300 \text{ m/s}}{100 \text{ Hz}} = 3 \text{ m}$

Q9JEE Main 2025MCQ

A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the $x$ -direction in 0.3 sec . The crest $P$ is at $x=0$ at $t=0 {\ sec}$ and maximum displacement of the wave is 2 cm . Which equation correctly represents this wave?

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📖 Explanation

$\;\lambda =7.5 {\text{cm}}$ $\;\lambda =\;\frac{2 \pi }{K}$ $\;K=\;\frac{2 \pi }{7.5}=0.83 {\text{cm}}^{-1}$ $\;v=\;\frac{1.2}{0.3}=\;\frac{\omega }{K}$ $\;\omega =4 \times 0.83=3.35$ $\;\;\text{ At }\; t=0, x=0, \;\text{ there is a crest }\;$ $\; \therefore \;\; y=2 \cos (0.83 x-3.35 t) {\text{cm}}$

Q10JEE Main 2025MCQ

Two harmonic waves moving in the same direction superimpose to form a wave $x=a \cos (1.5 t) \cos$ (50.5t) where $t$ is in seconds. Find the period with which they beat. (close to nearest integer)

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📖 Explanation

$x=a \cos (1.5 t) \cos (50.5 t)$ Clearly, $x=\;\;\frac{a}{2} \cos (50.5 t+1.5 t)+\;\frac{a}{2} \cos (50.5 t-1.5 t)$ $\Rightarrow x=\;\frac{a}{2} \cos (52 t)+\;\frac{a}{2} \cos (49 t)$ $\;f_1=\;\frac{52}{2 \pi } \;\text{ and }\; f_2=\;\frac{49}{2 \pi }$ $\Rightarrow \Delta f=\;\frac{3}{2 \pi }$ $\;\Delta T=\;\frac{1}{\Delta f}=\;\frac{2 \pi }{3}=2.09$ $\Rightarrow \Delta T \approx 2 {\ s}$

Q11JEE Main 2025MCQ

The equation of a wave travelling on a string is $y= sin \;[20 \pi x+10 \pi f]$ , where $x$ and $t$ are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :

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📖 Explanation

$\;k=20 \pi =\;\frac{2 \pi }{\lambda }$ $\;\lambda =10 {\text{cm}}$ Minimum distance $=\;\frac{\lambda }{2}=5 {\text{cm}}$

Q12JEE Main 2025MCQ

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_1 (x, t)=4 sin \;(k x-\omega t)$ and $y_2(x, t)=2 sin \; (k x-\omega t+\;\frac{2 \pi }{3})$ , are : (Take the angular frequency of initial waves same as $\omega$ )

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📖 Explanation

$\;A={\sqrt{A_1^{2+}A_2^{2+}2 A_1 A_2 \cos \varphi }}$ $\;\varphi =\;\frac{2 \pi }{3}$ $\;A=\sqrt{4^{2+}2^{2+}2 \times 4 \times 2 \cos \;\frac{2 \pi }{3}}=2 {\sqrt{3}}$ $\;\tan \alpha =\;\frac{2 sin \;\varphi }{4+2 \cos \varphi }=\;{1}/{{\sqrt{3}}}$ $\;\alpha =\pi / 6$

Q13JEE Main 2025MCQ

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason ${R}$ Assertion A: A sound wave has higher speed in solids than gases.Reason R: Gases have higher value of Bulk modulus than solids.In the light of the above statements, choose the correct answer from the options given below

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📖 Explanation

Speed of sound in a medium depends on inertial and elastic properties as $v=\sqrt{\;\frac{B}{P}}$ for gases and $v=\sqrt{\;\frac{Y}{P}}$ for solids. Since the elastic property of solid happens to be many folds greater than that of gases, the speed of sound in solids i5 higher than in gases.Also, bulk modulus of gases varies between 0 and $\infty (B=-v \;\frac{d P}{d v})$ hence reason is false.

Q14JEE Main 2025MCQ

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities $\rho _1$ and $\rho _2$ , respectively. The frequency of $9^{\;\text{th }\;}$ harmonic of closed tube is identical with $4^{\;\text{th }\;}$ harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is $\rho _1: \rho _2=1: 16$ , then the length of the open tube is:

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📖 Explanation

$\;9 (\;\frac{v_c}{4 I_c}) =4 (\;\frac{v_o}{2 I_o})$ $\;v=\sqrt{\;\frac{B}{\rho }}$ $\therefore \;\; \;\frac{v_c}{v_o}=\sqrt{\;\frac{\rho _2}{\rho _1}}=4$ $I_o=\;\frac{8}{9} I_c (\;\frac{v_o}{v_c})$ $\;\; =\;\frac{8}{9} I_c \times \;\frac{1}{4}=\;\frac{2}{9} I_c$ $I_o=\;\frac{20}{9} cm$

Q15JEE Main 2025MCQ

Two strings with circular cross section and made of same material are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section $R$ is $v_1$ , and that in the other string having radius of cross section $R / 2$ is $v_2$ . Then $\;\frac{v_2}{v_1}=$

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📖 Explanation

$\;v=\sqrt{\;\frac{T}{\mu }}$ $\;\;\frac{v_1}{v_2}=\sqrt{\;\frac{R_2^2}{R_1^2}}=\;\frac{R_2}{R_1}=\;\frac{1}{2}$ $\;\;\frac{v_2}{v_1}=2$

Q16JEE Main 2025MCQ

In an experiment with a closed organ pipe, it is filled with water by $(\;\frac{1}{5})$ th of its volume. The frequency of the fundamental note will change by

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📖 Explanation

The fundamental frequency of a closed organ pipe of length $L$ is $f_1 = \frac{v}{4L}$ . When filled with water to $\frac{1}{5}$ th of its volume, the remaining air column length is $L' = L - \frac{L}{5} = \frac{4L}{5}$ . The new fundamental frequency becomes $f_2 = \frac{v}{4L'} = \frac{v}{4(4L/5)} = \frac{5}{4} \left(\frac{v}{4L}\right) = \frac{5}{4} f_1$ . The change in frequency is $\Delta f = f_2 - f_1 = \frac{5}{4}f_1 - f_1 = \frac{1}{4}f_1$ . Thus, the percentage change is $\frac{\Delta f}{f_1} \times 100\% = \frac{1}{4} \times 100\% = 25\%$ . Since the frequency increases, the change is $+25\%$ .

Q17JEE Main 2025MCQ

In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is

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📖 Explanation

$\;f_0=\;\frac{v}{4 L}$ $\;\Delta f=\;\frac{v}{4} \{\;\frac{1}{1}-\;\frac{1}{1.2}\}$ $\;\;\frac{15 \times 4 \times 1.2}{0.2}=v$ $\;360=v$

Q18JEE Main 2025MCQ

The equation of a transverse wave travelling along a string is $y(x, t)=4.0 sin \; \$ [20 × 10^{-3} x+600 f]$ {\text{mm}} $, where$ x $is in mm and$ t$ is in second. The velocity of the wave is :

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📖 Explanation

$\;k=20 \times 10^{-3} mm^{-1}=20 m^{-1}$ $\;w=600 s^{-1}$ $\;v=\;\frac{w}{k}=\;\frac{600}{20}=30 m / s$ and $x & t$ carry same signTherefore $v =-30 m / s$

Q19JEE Main 2024MCQ

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 {\text{cm}}$ , the length of the closed pipe will be : [31-Jan-2024 Shift 1]

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📖 Explanation

$\;\frac{\lambda }{4}= {\text{L}}_1\;\;\;\;\;\;2 (\;\frac{\lambda }{2}) =\lambda$ ${\text{v}}= {\text{f}} \lambda \;\;\;\;\;\;\;\; {\text{f}}_2=\;{2 {\text{v}}}/{2 {\text{L}}_2}$ ${\text{v}}= {\text{f}}_1 (4 {\text{L}}_1) \;\;\;\;\;\;\;\; {\text{f}}_2=\;{ {\text{v}}}/{ {\text{L}}_2}$ ${\text{f}}_1=\;{ {\text{v}}}/{4 {\text{L}}_1}\;$ ${\text{f}}_1= {\text{f}}_2\;$ $\;{ {\text{v}}}/{4 {\text{L}}_1}=\;{ {\text{v}}}/{ {\text{L}}_2}\;$ $\Rightarrow {\text{L}}_2=4 {\text{L}}_1\;$ $60=4 \times {\text{L}}_1\;$ ${\text{L}}_1=15 {\text{cm}}\;$

Q20JEE Main 2024NAT

In a closed organ pipe, the frequency of fundamental note is $30 {\text{Hz}}$ . A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $110 {\text{Hz}}$ . If the organ pipe has a cross-sectional area of $2 {\text{cm}}^2$ , the amount of water poured in the organ tube is _______g. (Take speed of sound in air is $330 {\text{m}} / {\ s}$ ) [30-Jan-2024 Shift 1]

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📖 Explanation

$\;\;\frac{V}{4 ℓ_1}=30 \Rightarrow ℓ_1=\;\frac{11}{4} m$ $\;\;\frac{V}{4 ℓ_2}=110 \Rightarrow ℓ_2=\;\frac{3}{4} m$ $\;\Delta ℓ=2 m,$ Change in volume $=A \Delta ℓ=400 {\text{cm}}^3$ ${M}={4 0 0} {g} ; ( \because \rho =1 {\text{g}} / {\text{cm}}^3)$

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