⇒Superposition of Waves

Correct Option is (D)

In fundamental mode the length of the closed tube is $\frac{\lambda }{4}$

Time required to travel a distance $\frac{\lambda }{4}$ is t.

Hence time required to travel a distance $\lambda$ will be $4\mathrm{t}$.

$\therefore$ Time period $\mathrm{T}=4\mathrm{t}$;

Frequency $\mathrm{n}=\frac{1}{\mathrm{T}}=\frac{1}{4\mathrm{t}}=\frac{0.25}{\mathrm{t}}$

Correct Option is (D)

In a pipe closed at one end only odd harmonics of the fundamental are produced. If $\mathrm{f}$ is the fundamental frequency then the harmonics produced are $3\mathrm{f},5\mathrm{f},7\mathrm{f},\dots \dots$ The difference between the successive overtones is $2\mathrm{f}$.

$\begin{array}{rl}& \therefore 250\mathrm{Hz}-150\mathrm{Hz}=2\mathrm{f}\\ & \therefore 100\mathrm{Hz}=2\mathrm{f}\\ & \therefore \mathrm{f}=50\mathrm{Hz}\end{array}$

Correct Option is (A)

$\mathrm{n}=264\mathrm{Hz},\mathrm{V}=300\mathrm{m}/\mathrm{s}$

For fundamental frequency, $\mathrm{n}=\frac{\mathrm{V}}{4l}$

$l=\frac{\mathrm{V}}{4\mathrm{n}}=\frac{330}{4\times 264}=0.3125\mathrm{n}=31.25\mathrm{cm}$

For fundamental mode, $l=\frac{\lambda }{4}$

other possible lengths are $\frac{3\lambda }{4},\frac{5\lambda }{4},\dots \dots$

$\text{ i.e. }93.75\mathrm{cm},156.25\mathrm{cm},\dots \dots$

Hence, $62.50\mathrm{m}$ is not possible.