QJEEAdv21Ph55ae: JEE Advanced 2021 Physics - Class 11 JEEAdv21Ph55ae Solution: A source, approaching with speed u towards the open end of a…

Natural frequency of closed pipe,

f = (2n + 1)f₀

f₀ is fundamental frequency

n = 0, 1, 2 .......


Frequency of source received by pipe,

$$f^{\prime }=f_s\left[\frac{v-0}{v-u}\right]$$

For resonance,

$$f^{\prime }=f$$

$$f_s\left[\frac{v}{v-u}\right]=(2n+1)f_0$$

If u = 0.8v, f_s = f₀

$$f^{\prime }=\frac{v}{0.2v}{f}_0=5f_0$$

For n = 2 pipe can be in resonance

Hence, option (a) is correct.

If u = 0.8v, f_s = 2f₀

$$f^{\prime }=\frac{v}{0.2v}\times 2f_0=10f_0$$

If u = 0.8v, f_s = 0.5f₀

$$f^{\prime }=\frac{v}{0.2v}\times 0.5f_0=2.5f_0$$

Not possible.

If u = 0.5v, f_s = 1.5f₀

$$f^{\prime }=\frac{v}{0.5v}\times 1.5f_0=3f_0$$

For n = 1 f = 3f₀

Pipe can be in resonance.

Hence, option (d) is correct.