⇒Superposition of Waves

Correct Option is (A)

Fundamental frequency of the first wire is

$\mathrm{n}=\frac{1}{2l_1}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{2l_1}\sqrt{\frac{\mathrm{T}}{\pi {\mathrm{r}}_1^2\rho }}=\frac{1}{2l_1{\mathrm{r}}_1}\sqrt{\frac{\mathrm{T}}{\pi \rho }}$

The first overtone $n_1=2n=\frac{1}{l_1{r}_1}\sqrt{\frac{T}{\pi \rho }}$

Similarly, the second overtone of the second wire will be,

${\mathrm{n}}_2=\frac{3}{2l_2{\mathrm{r}}_2}\sqrt{\frac{\mathrm{T}}{\pi \rho }}$

Given that ${\mathrm{n}}_1={\mathrm{n}}_2$

$\begin{array}{rl}& \therefore \frac{1}{l_1{\mathrm{r}}_1}\sqrt{\frac{\mathrm{T}}{\pi \rho }}=\frac{3}{2l_2{\mathrm{r}}_2}\sqrt{\frac{\mathrm{T}}{\pi \rho }}\\ & \therefore 3l_1{\mathrm{r}}_1=2l_2{\mathrm{r}}_2\\ & \frac{l_1}{l_2}=\frac{2r_2}{3r_1}\\ & =\frac{2r_2}{3\left(2r_2\right)}\dots (\because r_1=2r_2)\\ & =\frac{1}{3}\end{array}$

Correct Option is (A)

Let velocity of pulse at lower end be $v_1$ and at top be $v_2$

$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}(\because \lambda =\frac{\mathrm{v}}{\mathrm{n}}\text{ and }\mathrm{n}=\text{ constant })$

Velocity of transverse wave on a string is

$v=\sqrt{\frac{T}{m}}$

where, $\mathrm{m}$ is linear density.

In this case, $\mathrm{v}\propto \sqrt{\mathrm{T}}$

$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}=\sqrt{\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}}=\sqrt{\frac{({\mathrm{m}}_2+{\mathrm{m}}_1)}{{\mathrm{m}}_2}}$

Where, $T_2$ is tension at upper end of rope and $T_1$ is tension at lower end of rope.

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{m_2}{m_2+m_1}}$

Correct Option is (A)

Let the initial length and tension be $l$ and $T$ respectively.

After shortening,

The new length $l_{\text{new }}=l-\frac{40}{100}l=\frac{3}{5}l$

After increase in tension,

the new tension ${\mathrm{T}}_{\text{new }}=\mathrm{T}+\frac{44}{100}\mathrm{T}=\frac{144\mathrm{T}}{100}$

Fundamental frequency of a vibrating string is given by

$\begin{array}{rl}\mathrm{n}& =\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\\ \therefore {\mathrm{n}}_1& =\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\\ {\mathrm{n}}_2& =\frac{1}{2l}\sqrt{\frac{{\mathrm{T}}_{\text{new }}}{\mathrm{m}}}\\ \therefore \frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}& =\frac{l_{\text{new }}}{l}\times \frac{\sqrt{\mathrm{T}}}{\sqrt{{\mathrm{T}}_{\text{new }}}}\\ & =\frac{\frac{3}{5}l}{l}\times \sqrt{\frac{100\mathrm{T}}{144\mathrm{T}}}\\ & =\frac{3}{5}\times \frac{10}{12}=\frac{1}{2}\\ \therefore \frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}& =\frac{2}{1}\end{array}$

Correct Option is (C)

The fundamental frequency of a string is given by

Given: $l=l_1+l_2+l_3$ ..... (i)

$\mathrm{n}=\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

$\Rightarrow \mathrm{n}\propto \frac{1}{l}$ or $\mathrm{n}l=\mathrm{k}$

$\therefore l_1=\frac{\mathrm{k}}{{\mathrm{n}}_1},l_2=\frac{\mathrm{k}}{{\mathrm{n}}_2}\text{ and }{l}_3=\frac{\mathrm{k}}{\mathrm{n}3}$ .... (ii)

$\therefore$ Original length $l=\frac{\mathrm{k}}{\mathrm{n}}$ .... (iii)

Putting eq (ii) and (iii) into eq (i)

$\begin{array}{rl}& \frac{\mathrm{k}}{\mathrm{n}}=\frac{\mathrm{k}}{{\mathrm{n}}_1}+\frac{\mathrm{k}}{{\mathrm{n}}_2}+\frac{\mathrm{k}}{{\mathrm{n}}_3}\\ \therefore \frac{1}{\mathrm{n}}& =\frac{1}{{\mathrm{n}}_1}+\frac{1}{{\mathrm{n}}_2}+\frac{1}{{\mathrm{n}}_3}\end{array}$

Correct Option is (A)

The standard equation of a wave can be written as

$\mathrm{y}=\mathrm{A}\sin (\mathrm{kx}+\omega \mathrm{t})$

Speed of wave $V=\frac{\omega }{k}=\frac{60}{3}=20\mathrm{m}/\mathrm{s}$

$\begin{array}{rl}& \text{ Also, }\mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\\ & \therefore \mathrm{m}=\frac{\mathrm{T}}{{\mathrm{V}}^2}=\frac{0.4}{400}={10}^{-3}\mathrm{kg}{\mathrm{m}}^{-1}\end{array}$