JEE Advanced 2021 — Paper 2 — Question 2

Natural frequency of closed pipe $\mathrm{f}=(2 \mathrm{n}+1) \mathrm{f}_{0}$ $\mathrm{f}_{0}$

is fundamental frequency $\mathrm{n}=0,1,2$, $\qquad$

frequency of source received by pipe

$f^{\prime}=f_{s}\left[\frac{v-0}{v-u}\right]$ L For resonance $\mathrm{f}^{\prime}=\mathrm{f}$

$f_{s}\left[\frac{v}{v-u}\right]=(2 n+1) f_{0}$ If $u=0.8 v \quad f_{s}=f_{0}$

$\mathrm{f}^{\prime}=\frac{\mathrm{v}}{0.2 \mathrm{v}} \mathrm{f}_{0}=5 \mathrm{f}_{0}$

for $n=2$ pipe can be in resonance Hence, option (A) is correct. If $u=0.8 v \quad \mathrm{f}_{\mathrm{s}}=2 \mathrm{f}_{0}$

$\mathrm{f}^{\prime}=\frac{\mathrm{v}}{0.2 \mathrm{v}} \times 2 \mathrm{f}_{0}=10 \mathrm{f}_{0}$ If $u=0.8 v, f_{S}=0.5 f_{0}$ $\mathrm{f}^{\prime}=\frac{\mathrm{v}}{0.2 \mathrm{v}} \times 0.5 \mathrm{f}_{0}=2.5 \mathrm{f}_{0}$

Not possible If $u=0.5 \mathrm{v}, \mathrm{f}_{\mathrm{S}}=1.5 \mathrm{f}_{0}$

$\mathrm{f}^{\prime}=\frac{\mathrm{v}}{0.5 \mathrm{v}} \times 1.5 \mathrm{f}_{0}=3 \mathrm{f}_{0}$

for $\mathrm{n}=1 \mathrm{f}=3 \mathrm{f}_{0}$ pipe can be in resonance Hence, option (D) is correct.