Physics Q35 – JEE-MAIN 2025

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:

Clue 1 of 3: Resonance in Pipes

Understand how the fundamental frequency of an air column closed at one end is related to its length and the speed of sound.

🧠 Full Solution Path

Step 1: Determine fundamental frequencies✦ Active

For an air column closed at one end, the fundamental frequency $f$ is given by $f = \frac{v}{4 L}$ . Convert lengths to meters: $L_{1} = 100 cm = 1.0 m$ and $L_{2} = 120 cm = 1.2 m$ .

f_{1} = \frac{v}{4 L_{1}} = \frac{v}{4 \times 1.0} = \frac{v}{4}

f_{2} = \frac{v}{4 L_{2}} = \frac{v}{4 \times 1.2} = \frac{v}{4.8}

Step 2: Apply the beat frequency formula○ Expand

The beat frequency $f_{b}$ is the absolute difference between the two frequencies. Since $L_{1} < L_{2}$ , $f_{1} > f_{2}$ . Given $f_{b} = 15 Hz$ .

f_{b} = f_{1} - f_{2}

15 = \frac{v}{4} - \frac{v}{4.8}

💡 Teacher's Secret Hint

Ensure to use the absolute difference for beat frequency.

Step 3: Solve for the velocity of sound○ Expand

Factor out $v$ and solve the equation.

15 = v (\frac{1}{4} - \frac{1}{4.8})

15 = v (\frac{1}{4} - \frac{5}{24})

15 = v (\frac{6}{24} - \frac{5}{24})

15 = v (\frac{1}{24})

v = 15 \times 24

v = 360 m/s

💡 Teacher's Secret Hint

Pay attention to unit consistency; lengths were converted to meters.

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Physics Question 35 – JEE-MAIN 2025

Clue 1 of 3: Resonance in Pipes

🧠 Full Solution Path