JEE Physics 2025 Question with Solution

Given: Two closed air columns have lengths $l_1 = 100 \, \text{cm} = 1 \, \text{m}$ and $l_2 = 120 \, \text{cm} = 1.2 \, \text{m}$. The beat frequency is $15 \, \text{Hz}$.

Find: The velocity of sound $v$ in air.

For an air column closed at one end, the fundamental frequency is

$$f = \frac{v}{4l}$$

So, for the first air column,

$$f_1 = \frac{v}{4l_1} = \frac{v}{4 \times 1} = \frac{v}{4}$$

For the second air column,

$$f_2 = \frac{v}{4l_2} = \frac{v}{4 \times 1.2} = \frac{v}{4.8}$$

The beat frequency is the absolute difference of the two frequencies:

$$|f_1 - f_2| = 15$$

Substituting the expressions for $f_1$ and $f_2$,

$$15 = \left| \frac{v}{4} - \frac{v}{4.8} \right|$$ $$15 = v \left| \frac{1}{4} - \frac{1}{4.8} \right|$$ $$15 = v \left| \frac{4.8 - 4}{4 \times 4.8} \right|$$ $$15 = v \left( \frac{0.8}{19.2} \right)$$ $$15 = v \left( \frac{1}{24} \right)$$ $$v = 15 \times 24 = 360 \, \text{m/s}$$

Therefore, the velocity of sound in the air column is $360 \, \text{m/s}$. The correct option is D.

Given: The two closed air columns have lengths $100 \, \text{cm}$ and $120 \, \text{cm}$, and they produce $15$ beats per second.

Find: The velocity of sound $v$.

Using the fundamental frequency formula for a closed pipe,

$$f = \frac{v}{4L}$$

If lengths are kept in centimetres, then

$$f_1 = \frac{v}{4 \times 100} = \frac{v}{400}$$ $$f_2 = \frac{v}{4 \times 120} = \frac{v}{480}$$

The beat frequency is

$$|f_1 - f_2| = 15$$

So,

$$\left| \frac{v}{400} - \frac{v}{480} \right| = 15$$

Taking the difference,

$$\frac{6v}{2400} - \frac{5v}{2400} = 15$$ $$\frac{v}{2400} = 15$$ $$v = 15 \times 2400 = 36000 \, \text{cm/s}$$

Converting to SI units,

$$36000 \, \text{cm/s} = 360 \, \text{m/s}$$

Therefore, the velocity of sound is $360 \, \text{m/s}$. The correct option is D.