⇒Superposition of Waves

Correct Option is (C)

$\mathrm{m}=\frac{50}{10}=5\mathrm{kg}$ $(\because \mathrm{W}=\mathrm{mg})$

Tension in the mid-point of the wire is:

$\mathrm{T}=\frac{\mathrm{m}}{2}\mathrm{g}=\frac{5}{2}\times 10=25\mathrm{N}$

$\therefore$ Speed of the wave at mid-point of the wire is:

$\begin{array}{rl}\mathrm{v}& =\sqrt{\frac{\mathrm{T}}{\mu }}=\sqrt{\frac{25}{\left(\frac{5}{20}\right)}}(\because \mu =\frac{\mathrm{m}}{\mathrm{L}})\\ \therefore \mathrm{v}& =10\mathrm{m}/\mathrm{s}\end{array}$

Correct Option is (C)

The frequency of a vibrating wire $f=\frac{1}{2l}\sqrt{\frac{T}{m}}$

$\therefore \mathrm{f}\propto \frac{1}{l}\Rightarrow \mathrm{fl}=$ constant

Let the frequency of the fork be $f$ and the initial and final frequencies of the wire be $f_1$ and $f_2$.

The number of beats heard before decreasing the length is $f-f_1=3$ .... (i)

The number of beats after decreasing the length is ${\mathrm{f}}_2-\mathrm{f}=3$ .... (ii)

$\therefore {\mathrm{f}}_1{l}_1={\mathrm{f}}_2{l}_2$

$\therefore (\mathrm{f}-3)50=(\mathrm{f}+3)49$ ... from (i) and (ii)

$50\mathrm{f}-49\mathrm{f}=147+150$

$\therefore \mathrm{f}=297\mathrm{Hz}$

Correct Option is (C)

$\begin{array}{rl}& l_2=2l_1,{\mathrm{R}}_2=2{\mathrm{R}}_1,{\mathrm{T}}_1={\mathrm{T}}_2\\ & \mathrm{n}=\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\\ & \text{ Where, }\mathrm{m}=\text{ mass per unit length }=\frac{\left(\pi {\mathrm{R}}^{2}l\right)\rho }{l}\\ & \therefore \mathrm{m}\propto {\mathrm{R}}^2\\ & \therefore \frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\frac{l_1}{l_2}\times \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\\ & \therefore \frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\frac{l_1}{2l_1}\times \frac{{\mathrm{R}}_1}{2{\mathrm{R}}_1}\\ & \therefore {\mathrm{n}}_2=\frac{{\mathrm{n}}_1}{4}=\frac{\mathrm{n}}{4}\end{array}$

Correct Option is (D)

The given equation is $y=A\sin 2(\omega t-kx)$

$\therefore$ Velocity of the particle, $v=\frac{dy}{dt}$

$\begin{array}{ll}& =2\mathrm{A}\omega \cos 2(\omega \mathrm{t}-\mathrm{kx})\\ \therefore & \text{ Maximum velocity }=2\mathrm{A}\omega \\ & \text{ Velocity of the wave }=\frac{\omega }{\mathrm{k}}\end{array}$

$\begin{array}{rl}& \text{ Given }2\mathrm{A}\omega =\frac{\omega }{\mathrm{k}}\\ & \therefore \mathrm{A}=\frac{1}{2\mathrm{k}}=\frac{\lambda }{(2\pi {)}^2}=\frac{\lambda }{4\pi }\end{array}$

Correct Option is (C)

The formula for frequency of a sonometer is $\mathrm{f}=\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\pi \rho {\mathrm{D}}^2}}$

Here $l$ is length, T is tension, D is diameter and $\rho$ is density.

The frequency of both the wires is same.

The frequency of the wire $A$ is $f_A=\frac{1}{2l}\sqrt{\frac{T}{\pi {\rho }_1{D}^2}}$

The frequency of the wire $B$ is $f_B=\frac{1}{2l}\sqrt{\frac{2T}{\pi {\rho }_2(2D{)}^2}}$

Equating both the frequencies

$\begin{array}{rl}& \frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\pi \rho {\mathrm{D}}^2}}=\frac{1}{2l}\sqrt{\frac{2\mathrm{T}}{\pi {\rho }_2(2\mathrm{D}{)}^2}}\\ & \sqrt{\frac{1}{{\rho }_1}}=\sqrt{\frac{1}{2{\rho }_2}}\\ & \frac{1}{{\rho }_1}=\frac{1}{2{\rho }_2}\\ \therefore & {\rho }_2=\frac{{\rho }_1}{2}\end{array}$