$\mathrm{n}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
The tuning fork produces $5$ beats per second with lengths $20 \mathrm{\,cm}$ and $21 \mathrm{\,cm}$. If $\mathrm{n}$ be frequency of fork, then
$\mathrm{n}+5=\frac{1}{2 \times 0.20} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ .........$(i)$
$\mathrm{n}-5=\frac{1}{2 \times 0.21} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ .........$(ii)$
Dividing eqn. $(i)$ by $(ii)$ we get:
hence, $\quad \frac{n+5}{n-5}=\frac{0.21}{0.20}$
or $0.2 n+1=0.21 n-1.05$
or $1+1.05=0.21 n-0.2 n^{\prime}$
or $0.05=0.01\, n$
or $\mathrm{n}=205 \mathrm{\,Hz}$

${z_1} = a\cos (kx - \omega \,t)$.....$(A)$
${z_2} = a\cos (kx + \omega \,t)$.....$(B)$
${z_3} = a\cos (ky - \omega \,t)$..... $(C)$
Assume that the sound of the whistle is composed of components varying in frequency from $f_1=800 \mathrm{~Hz}$ to $f_2=1120 \mathrm{~Hz}$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $320 \mathrm{~Hz}$. The speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$.
$1.$ The speed of sound of the whistle is
$(A)$ $340 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$
$(B)$ $360 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$
$(C)$ $310 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $360 \mathrm{~m} / \mathrm{s}$ for passengers in $B$
$(D)$ $340 \mathrm{~m} / \mathrm{s}$ for passengers in both the trains
$2.$ The distribution of the sound intensity of the whistle as observed by the passengers in train $\mathrm{A}$ is best represented by
$Image$
$3.$ The spread of frequency as observed by the passengers in train $B$ is
$(A)$ $310 \mathrm{~Hz}$ $(B)$ $330 \mathrm{~Hz}$ $(C)$ $350 \mathrm{~Hz}$ $(D)$ $290 \mathrm{~Hz}$
Give the answer question $1,2$ and $3.$