In an experiment with a closed organ pipe, it is filled with water by ((1/5))th of its volume. The frequency of the…

Year	2026	2025	2024	Total
Questions	7	10	5	22

Q43 jee_main_2026_21_jan_morning Wave on a String

Two strings (A, B) having linear densities

μA = 2 \times 10⁻⁴ kg/m

and

μB = 4 \times 10⁻⁴ kg/m

and lengths

LA = 2.5 m

and

LB = 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

t₁

and

t₂

, respectively, to reach the joint. The ratio

t₁/t₂

is :

A. 1.08
B. 1.90
C. 1.67
D. 1.18

Solution

Related Formula

v = \sqrt((T)/(μ))

t = (L)/(v)

Core Logic

Given $LA = 2.5 m$ $L_{A} = 2.5\text{ m}$, $LB = 1.5 m$ $L_{B} = 1.5\text{ m}$, $T = 500 N$ $T = 500\text{ N}$. Velocity in string A:

vA = \sqrt((T)/(μA)) = 5002 \times 10⁻⁴ = \sqrt(2500000) = 5 \sqrt(10) \times 10² m/s

Velocity in string B:

vB = \sqrt((T)/(μB)) = 5004 \times 10⁻⁴ = \sqrt(1250000) = 5 \sqrt(5) \times 10² m/s

Step 1: Calculate Ratio of Times

Time taken to reach joint:

t₁ = LAvA = 2.55 \sqrt(10) \times 10²

t₂ = LBvB = 1.55 \sqrt(5) \times 10²

Ratio:

t₁t₂ = 2.55\sqrt(10) \times 5\sqrt(5)1.5 = (2.5)/(1.5) \times \sqrt(5)\sqrt(10) = (5)/(3) \times 1\sqrt(2)

t₁t₂ = (1.666)/(1.414) \approx 1.18

Pattern Recognition

Time ratio is proportional to $L\sqrt(T/μ) = L \sqrt(μ/T)$ . Thus $t₁/t₂ = (L₁/L₂) \sqrt(μ₁/μ₂)$ directly.

Chapter Mix

Class 11 Physics: Waves

Q46 jee_main_2026_22_january_morning Interference of Sound Waves

Two loudspeakers

L₁

and

L₂

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 AB 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

)

Waves and Sound diagram for Q46 - JEE Main 2026 January Morning — Two loudspeakers setup with path difference for sound interference.

Numerical Answer. Answer: 600 to 600

Solution

Related Formula

Δ x = nλ, v = fλ

Core Logic

Solution geometry diagram for Q46 - JEE Main 2026 Morning — Two loudspeakers setup with path difference for sound interference.

Path difference at point B:

L₁ B = \sqrt(20² + 40²) = 20\sqrt(5) m = 44.6 m

L₂ B = \sqrt(40² + 30²) = 50 m

Δ x = 50 - 44.6 = 5.4 m

Δ x = nλ 5.4 = 10 \times λ λ = 0.54 m

f = (v)/(λ) = (324)/(0.54) = 600 Hz

Pattern Recognition

Sees: Loudspeaker interference cycles along displacement line. Shortcut: Compute path difference at endpoint B, relate to total interference cycles to find wavelength, then frequency. Check: Numerical answer is 600 Hz. ✓

Chapter Mix

Class 11 Physics: Waves and Sound

Q33 jee_main_2026_22_january_evening Organ Pipe Harmonics

In an open organ pipe

v₃

and

v₆

are

3rd

and

6th

harmonic frequencies, respectively. If

v₆ - v₃ = 2200

Hz then length of the pipe is ____ mm. (Take velocity of sound in air is 330 m/s.)

A. $275$ $275$
B. $225$ $225$
C. $200$ $200$
D. $250$ $250$

Solution

Related Formula

fₙ = n ((v)/(2L))

Core Logic

For an open organ pipe, harmonic frequency $vₙ = n \cdot f₀ = n ((v)/(2L))$ .

Given $v₆ - v₃ = 2200$ $v_6 - v_3 = 2200$ Hz:

(6v)/(2L) - (3v)/(2L) = 2200

(3v)/(2L) = 2200

Substituting speed of sound $v = 330 ~m/s$ $v = 330 \mathrm{~m/s}$:

(3 \times 330)/(2L) = 2200

L = (990)/(4400) = 0.225 ~m = 225 ~mm

Step 1: Final Conclusion

The length of the pipe is $225 ~mm$ $225 \mathrm{~mm}$.

Pattern Recognition

Open organ pipe: Frequency difference $Δ f = (6-3)f₀ = 3 f₀ = 2200 f₀ = 2200/3$ . Since $f₀ = v/(2L)$ $f_0 = v/(2L)$, solve directly for $L$ $L$.

Chapter Mix

Class 11 Physics: Waves

Q50 jee_main_2026_23_january_evening Speed of Sound

The velocity of sound in air is doubled when the temperature is raised from

0°

C to

α°

C. The value of

α

is ____.

Numerical Answer. Answer: 819 to 819

Solution

Related Formula

V = \sqrt((γ RT)/(M))

(V₁)/(V₂) = \sqrt((T₁)/(T₂))

Core Logic

The speed of sound in an ideal gas is directly proportional to the square root of its absolute temperature in Kelvin. Initial state: $T₁ = 0°C = 273 K$ , $V₁ = V₀$ $V_1 = V_0$ Final state: $T₂ = α°C = (α + 273) K$ , $V₂ = 2V₀$ $V_2 = 2V_0$

Step 1: Apply Ratio Equation

(V₀)/(2V₀) = \sqrt((273)/(T₂))

(1)/(4) = (273)/(T₂)

T₂ = 4 \times 273 = 1092 K

Step 2: Convert to Celsius

T₂ = α + 273 = 1092

α = 1092 - 273 = 819°C

Pattern Recognition

If velocity doubles, absolute temperature must quadruple ( $2² = 4$ $2^2 = 4$). $4 \times 273 = 1092 K$ $4 \times 273 = 1092 \, \mathrm{K}$, which translates perfectly to $819°C$ $819^{\circ}\mathrm{C}$.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases

Q34 jee_main_2026_24_january_evening Organ Pipes and Resonance

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 ____.

A. $4$ $4$
B. $2$ $2$
C. $1$ $1$
D. $3$ $3$

Solution

Related Formula

$fn, closed = n v4Lclosed$ (where $n$ $n$ is an odd integer) $fn, open = n v2Lₒₚₑₙ$ (where $n$ $n$ is any integer)

Core Logic

We are given that the fifth harmonic of the closed organ pipe equals the first harmonic of the open pipe:

f5, closed = f1, open

5v4Lclosed = v2Lₒₚₑₙ

Step 1: Simplify Ratio

Isolating the length ratio:

LclosedLₒₚₑₙ = (5)/(4) \times 2 = (10)/(4) = (5)/(2)

Equating this to $5/x$ $5/x$ gives $x = 2$ $x = 2$.

Pattern Recognition

For a closed pipe, the fundamental is $v/4L$ $v/4L$. For an open pipe, it's $v/2L$ $v/2L$. If an odd harmonic $n$ $n$ of a closed pipe matches the $m$ $m$-th harmonic of an open pipe, $Lc / Lₒ = n/2m$ $L_c / L_o = n/2m$.

Chapter Mix

Class 11 Physics: Waves

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate New Frequency

Step 2: Determine Percentage Shift

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

More Waves Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate Ratio of Times

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Ratio Equation

Step 2: Convert to Celsius

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Ratio

Pattern Recognition

Chapter Mix

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