Correct Answer is Option (C)

Speed of transverse wave $=\sqrt{\frac{T}{\mu }}$ $=\sqrt{\frac{70}{7\times {10}^{-3}}}=100\mathrm{m}/\mathrm{s}$

Correct Answer is 90

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f=\frac{1}{2L}\cdot v$

where $f$ is the fundamental frequency, $L$ is the length of the string, and $v$ is the speed of the transverse waves.

First, we are given the mass of the string ($m=18g$) and the linear mass density ($\mu =20g/m$). We can find the length of the string by dividing the mass by the linear mass density:

$L=\frac{m}{\mu }=\frac{18g}{20g/m}=0.9m$

Now we can plug in the values for the fundamental frequency ($f=50Hz$) and the length of the string ($L=0.9m$) into the formula:

$50Hz=\frac{1}{2(0.9m)}\cdot v$

To isolate $v$, we multiply both sides by $2(0.9m)$:

$v=50Hz\cdot 2(0.9m)=90{\mathrm{ms}}^{-1}$

The speed of the transverse waves produced in the string is $90{\mathrm{ms}}^{-1}$.

Alternate Method:

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

where $f_1$ is the fundamental frequency, $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the linear mass density of the string.

We're given that the fundamental frequency $f_1=50,\text{Hz}$, the mass of the string $m=18,\text{g}$, and the linear mass density $\mu =20,\text{g/m}$. To find the speed of the transverse waves, we need to find the tension $T$ and the length $L$ of the string.

First, let's find the length $L$ of the string using the mass and linear mass density:

$L=\frac{m}{\mu }=\frac{18,\text{g}}{20,\text{g/m}}=0.9\text{m}$

Now, we can rearrange the formula for the fundamental frequency to solve for the tension $T$:

$T=\mu {\left(\frac{2L{f}_1}{1}\right)}^2$

Substitute the known values:

$T=20,\text{g/m}\cdot {\left(\frac{2\cdot 0.9\text{m}\cdot 50\text{Hz}}{1}\right)}^2$

$T=20,\text{g/m}\cdot (90\text{m/s}{)}^2$

$T=20,\text{g/m}\cdot 8100{\text{m}}^2/{\text{s}}^2$

$T=162000,\text{g m}/{\text{s}}^2$

Now, we can find the speed of the transverse waves $v$ using the formula:

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

Substitute the known values:

$v=\sqrt{\frac{162000}{20}}$

$v=\sqrt{8100}=90\text{m/s}$

The speed of the transverse waves produced in the string is $90\text{m/s}$.

Correct Answer is 500

We can use the fact that the ratio of frequencies is equal to the square root of the ratio of tensions:

$\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$

In the first case, the mass attached to the string is $180\mathrm{g}$ and the frequency is $30\mathrm{Hz}$, so we have:

$\frac{f_2}{30\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$

In the second case, the frequency is $50\mathrm{Hz}$, so we have:

$\frac{50\mathrm{Hz}}{30\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$

Simplifying, we get:

$\frac{5}{3}=\sqrt{\frac{T_2}{T_1}}$

Squaring both sides, we get:

$\frac{25}{9}=\frac{T_2}{T_1}$

Since the tension in the string is proportional to the mass attached to it, we can write:

$\frac{m}{180\mathrm{g}}=\frac{T_2}{T_1}=\frac{25}{9}$

Solving for $m$, we get:

$m=\frac{25}{9}(180\mathrm{g})=\boxed{{\displaystyle 500\mathrm{g}}}$

Therefore, the mass attached to the string in the second case is $500\mathrm{g}$.

Correct Answer is 60

The fundamental frequency (also known as the first harmonic) of a vibrating string is given by the formula:

$f=\frac{v}{2L}$

where:

• (f) is the frequency,
• (v) is the speed of the wave in the string, and
• (L) is the length of the string.

In this case, the speed of the wave in the string stays the same because it depends on the properties of the string and the tension in it, which we can assume to be constant.

We can write the equation for the fundamental frequency of the original string and the shorter string:

$f_1=\frac{v}{2L_1}$

$f_2=\frac{v}{2L_2}$

where:

• $(f_1=120\text{Hz})$ and $(L_1=90\text{cm})$ for the original string, and
• $(f_2=180\text{Hz})$ and $(L_2)$ is what we're trying to find for the shorter string.

We can set up a ratio of these two equations:

$\frac{f_1}{f_2}=\frac{L_2}{L_1}$

Substituting in the given values, we get:

$\frac{120\text{Hz}}{180\text{Hz}}=\frac{L_2}{90\text{cm}}$

Solving for ($L_2$) gives:

$L_2=90\text{cm}\times \frac{120\text{Hz}}{180\text{Hz}}=60\text{cm}$

So, the length of the string producing a fundamental frequency of 180 Hz will be 60 cm.

Correct Answer is 3

$\frac{v}{2l}=50$ Hz

$\Rightarrow T={\left[100\times \left(\frac{30}{100}\right)\right]}^2\times \mu$

$\Rightarrow \mu =\frac{2700}{900}=3$