Energy and Intensity of Waves Questions in English

(A) The intensity $I$ of sound from a point source in a non-absorbing medium is inversely proportional to the square of the distance $r$ from the source,given by $I \propto \frac{1}{r^2}$.
Given distances are $r_P = 2 \ m$ and $r_Q = 3 \ m$.
The ratio of intensities at $P$ and $Q$ is $\frac{I_P}{I_Q} = \frac{r_Q^2}{r_P^2}$.
Substituting the values,we get $\frac{I_P}{I_Q} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
Thus,the ratio is $9:4$.

(C) For a spherical wave originating from a point source,the intensity $I$ is given by $I = \frac{P}{4\pi r^2}$,where $P$ is the power of the source.
Since intensity $I$ is proportional to the square of the amplitude $A$ $(I \propto A^2)$,we have $A^2 \propto \frac{1}{r^2}$.
This implies that the amplitude $A$ is inversely proportional to the distance $r$,i.e.,$A \propto \frac{1}{r}$.
Given that at distance $r$,the amplitude is $A_1 = A$.
At distance $r_2 = 2r$,the new amplitude $A_2$ will be:
$\frac{A_2}{A_1} = \frac{r_1}{r_2} = \frac{r}{2r} = \frac{1}{2}$.
Therefore,$A_2 = \frac{A}{2}$.

(C) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the equations for the two particles:
${y_1} = 0.06 \sin 2\pi (1.04t + {\phi _1})$
${y_2} = 0.03 \sin 2\pi (1.04t + {\phi _2})$
The amplitudes are ${a_1} = 0.06$ and ${a_2} = 0.03$.
The ratio of the intensities is given by:
$\frac{{{I_1}}}{{{I_2}}} = \frac{{a_1^2}}{{a_2^2}} = \left( \frac{{0.06}}{{0.03}} \right)^2 = (2)^2 = \frac{4}{1}$.
Therefore,the ratio of the intensities is $4:1$.

(D) The correct answer is $(d)$.
For a plane wave,the wavefronts are parallel planes,and the energy is distributed uniformly,so the intensity remains constant as the wave propagates.
For a spherical wave,the energy emitted by the source spreads over an increasing surface area $A = 4\pi r^2$. Thus,the intensity $I = P / A = P / (4\pi r^2)$ is inversely proportional to the square of the distance $r$ from the source.
The total power $P$ (often referred to as total intensity in this context) passing through any spherical surface centered at the source remains constant,as energy is conserved.

(D) wave is defined as a disturbance that travels through a medium,transporting energy and momentum from one point to another without the net transport of matter.
In a longitudinal wave,particles of the medium oscillate back and forth about their mean positions in the direction of wave propagation.
While the particles themselves do not travel with the wave (there is no net mass transfer),the wave carries energy and linear momentum through the medium.
Therefore,the correct quantities transmitted are energy and linear momentum.

(D) The intensity $I$ of a progressive plane wave in a medium of density $\rho$ is given by the formula:
$I = 2\pi^2 n^2 a^2 \rho v$
where $n$ is the frequency,$a$ is the amplitude,and $v$ is the wave velocity.
From this expression,it is clear that:
$1. I \propto a^2$ (Directly proportional to the square of amplitude)
$2. I \propto v$ (Directly proportional to the velocity)
$3. I \propto n^2$ (Directly proportional to the square of frequency)
Therefore,all the given statements are correct.
Hence,the correct option is $D$.

(B) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$ $(I \propto a^2)$.
Given the wave equations:
$y_1 = 5\sin 2\pi (75t - 0.25x) \implies a_1 = 5$
$y_2 = 10\sin 2\pi (150t - 0.50x) \implies a_2 = 10$
The ratio of intensities is given by $\frac{I_1}{I_2} = \frac{a_1^2}{a_2^2}$.
Substituting the values: $\frac{I_1}{I_2} = \frac{5^2}{10^2} = \frac{25}{100} = \frac{1}{4}$.
Thus,the intensity ratio is $1:4$.

(B) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the ratio of intensities is $\frac{I_1}{I_2} = \frac{1}{16}$.
Using the relation $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2$,we have $\left( \frac{a_1}{a_2} \right)^2 = \frac{1}{16}$.
Taking the square root on both sides,we get $\frac{a_1}{a_2} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Therefore,the ratio of their amplitudes is $1:4$.

(A) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Let the amplitudes be $a_1$ and $a_2$. Given $\frac{a_1}{a_2} = \frac{2}{1}$.
The maximum intensity $I_{\max}$ is proportional to $(a_1 + a_2)^2$ and the minimum intensity $I_{\min}$ is proportional to $(a_1 - a_2)^2$.
Therefore,the ratio of maximum to minimum intensity is given by:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \left( \frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1} \right)^2$.
Substituting the given ratio $\frac{a_1}{a_2} = 2$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = \frac{9}{1}$.
Thus,the ratio is $9:1$.

(C) The intensity $I$ of a wave is proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
For two waves with amplitudes $a_1$ and $a_2$,the ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \left( \frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1} \right)^2$.
Given the ratio of amplitudes $\frac{a_1}{a_2} = \frac{4}{3}$.
Substituting the values into the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{4}{3} + 1}{\frac{4}{3} - 1} \right)^2 = \left( \frac{\frac{7}{3}}{\frac{1}{3}} \right)^2 = (7)^2 = \frac{49}{1}$.
Thus,the ratio is $49:1$.

(A) The intensity $I$ of a spherical wave at a distance $r$ from a source of power $P$ is given by the formula: $I = \frac{P}{A} = \frac{P}{4\pi r^2}$.
Given: Power $P = 4 W$,distance $r = 200 m$.
Substituting the values: $I = \frac{4}{4\pi \times (200)^2}$.
$I = \frac{1}{\pi \times 40000} = \frac{1}{125663.7} W/m^2$.
$I \approx 7.957 \times 10^{-6} W/m^2$.
Rounding to the nearest provided option,we get $I \approx 8 \times 10^{-6} W/m^2$.

(A) The intensity $I$ of a sound wave is directly proportional to the square of the pressure amplitude $P_0$,given by the relation $I \propto P_0^2$.
If the pressure amplitude is tripled,the new pressure amplitude $P_0' = 3P_0$.
The new intensity $I'$ will be $I' \propto (P_0')^2 = (3P_0)^2 = 9P_0^2$.
Therefore,$I' = 9I$.
The intensity of the sound increases by a factor of $9$.

(C) The intensity of a sound wave is given by the formula $I = 2{\pi ^2}{a^2}{n^2}\rho v$,where $a$ is the amplitude and $n$ is the frequency.
From this,we see that $I \propto {a^2}{n^2}$.
Let the initial amplitude be $a_1$ and frequency be $n_1$. Let the final amplitude be $a_2 = 2a_1$ and frequency be $n_2 = n_1/4$.
The ratio of intensities is $\frac{I_2}{I_1} = \left( \frac{a_2}{a_1} \right)^2 \times \left( \frac{n_2}{n_1} \right)^2$.
Substituting the values: $\frac{I_2}{I_1} = (2)^2 \times (1/4)^2 = 4 \times (1/16) = 1/4$.
Therefore,$I_2 = I_1/4$,which means the intensity is decreased by a factor of $4$.

(D) The energy density $(u)$ of a wave is directly proportional to the square of its amplitude $(A)$.
Mathematically,$u \propto A^2$.
Given the ratio of amplitudes $A_1 : A_2 = 5 : 2$.
The ratio of their energy densities is given by $\frac{u_1}{u_2} = \left( \frac{A_1}{A_2} \right)^2$.
Substituting the values,$\frac{u_1}{u_2} = \left( \frac{5}{2} \right)^2 = \frac{25}{4}$.
Thus,the ratio of their energy densities is $25 : 4$.

(D) The energy $E$ of a sound wave is given by the relation $E \propto a^2 n^2$,where $a$ is the amplitude and $n$ is the frequency.
Since the energies of the two sounds are equal,we have $E_A = E_B$.
Therefore,$a_A^2 n_A^2 = a_B^2 n_B^2$.
Given that the frequency of $B$ is one-eighth the frequency of $A$,we have $n_B = \frac{1}{8} n_A$,or $\frac{n_A}{n_B} = 8$.
Rearranging the energy equation: $\frac{a_B^2}{a_A^2} = \frac{n_A^2}{n_B^2}$.
Taking the square root on both sides: $\frac{a_B}{a_A} = \frac{n_A}{n_B} = 8$.
Thus,$a_B = 8 a_A$.
The amplitude of the note of $B$ is eight times that of $A$.

(C) The intensity of a spherical sound wave is inversely proportional to the square of the distance from the source: $I \propto \frac{1}{r^2}$.
Given: $I_1 = 1 \times 10^{-2} \ \mu W/m^2$ at $r_1 = 2 \ m$.
We need to find $I_2$ at $r_2 = 10 \ m$.
Using the relation $\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$:
$\frac{I_2}{1 \times 10^{-2}} = \frac{2^2}{10^2} = \frac{4}{100} = 0.04$.
Therefore,$I_2 = 0.04 \times 10^{-2} \ \mu W/m^2 = 4 \times 10^{-4} \ \mu W/m^2$.

(B) The intensity of the sound wave decreases by $10\%$ for every $1 \ m$ distance covered.
This means that after $1 \ m$,the intensity becomes $90\%$ of the initial intensity,i.e.,$0.9 \ I_0$.
After passing through $3 \ m$,the intensity $I_3$ will be:
$I_3 = I_0 \times (0.9) \times (0.9) \times (0.9)$
$I_3 = I_0 \times (0.9)^3$
$I_3 = I_0 \times 0.729$
Given the initial intensity $I_0 = 100 \ dB$,the final intensity is:
$I_3 = 100 \times 0.729 = 72.9 \ dB$.

(D) The intensity $I$ of a point source at a distance $r$ is given by the inverse square law: $I \propto \frac{1}{r^2}$.
Taking the logarithmic derivative,we get: $\frac{\Delta I}{I} = -2 \frac{\Delta r}{r}$.
Given that the separation $r$ is increased by $2\%$,we have $\frac{\Delta r}{r} = 0.02$.
Substituting this value: $\frac{\Delta I}{I} = -2 \times 0.02 = -0.04$.
This corresponds to a decrease of $4\%$.
Therefore,the intensity decreases by $4\%$.

(B) The intensity $I$ of a spherical sound wave follows the inverse square law: $I \propto \frac{1}{r^2}$.
Given $r_1 = 200 \ cm$ and $r_2 = 400 \ cm$,the ratio of intensities is $\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{200}{400}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,$I_1 = 4I_2$.
The intensity level $L$ in decibels is given by $L = 10 \log_{10}\left(\frac{I}{I_0}\right)$.
The difference in intensity levels is $L_1 - L_2 = 10 \log_{10}\left(\frac{I_1}{I_2}\right)$.
Substituting the ratio: $L_1 - L_2 = 10 \log_{10}(4) = 10 \times 0.602 = 6.02 \ dB \approx 6 \ dB$.
Therefore,$L_2 = L_1 - 6 \ dB = 80 \ dB - 6 \ dB = 74 \ dB$.

(B) For the interference of two waves,the resultant intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
If the phase difference $\phi$ varies randomly with time,the average value of the cosine term is $(\cos \phi)_{av} = 0$.
Therefore,the average resultant intensity becomes $I = I_1 + I_2$.
For $n$ identical sources,each with intensity $I_0$,the total average intensity is the sum of the individual intensities: $I = I_0 + I_0 + \dots + I_0$ ($n$ times).
Thus,$I = n\,I_0$.
Given $n = 10$,the resultant intensity is $I = 10\,I_0$.

(C) The sound intensity level in decibels is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
At $r_1 = 1 \ m$,$\beta_1 = 40 \ dB$. Thus,$40 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \implies \frac{I_1}{I_0} = 10^4$.
At distance $r_2$,$\beta_2 = 20 \ dB$. Thus,$20 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \implies \frac{I_2}{I_0} = 10^2$.
Dividing the two equations: $\frac{I_1}{I_2} = \frac{10^4}{10^2} = 10^2 = 100$.
Since intensity $I \propto \frac{1}{r^2}$,we have $\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$.
Substituting the values: $100 = \frac{r_2^2}{(1)^2} \implies r_2^2 = 100 \implies r_2 = 10 \ m$.

(C) The energy density $(E)$ of a sound wave is given by the formula: $E = 2{\pi ^2}\rho {n^2}{A^2}$.
Here,$\rho$ is the density of the medium,$n$ is the frequency,and $A$ is the amplitude.
The maximum speed of the particles of the medium is given by: ${v_{\max }} = \omega A = 2\pi nA$.
Squaring the expression for ${v_{\max }}$,we get: ${({v_{\max }})^2} = 4{\pi ^2}{n^2}{A^2}$.
Comparing the two expressions,we can see that $E = \frac{1}{2}\rho {({v_{\max }})^2}$.
This implies that $E \propto {({v_{\max }})^2}$.
Since $E$ is proportional to the square of ${v_{\max }}$,the graph between $E$ and ${v_{\max }}$ will be a parabola opening upwards,which is represented by Graph $C$.

(D) The intensity $I$ of a sound wave is proportional to the square of its amplitude $a$ and the square of its angular frequency $\omega$,i.e.,$I \propto a^2 \omega^2$.
From the given graph:
For wave $A$,the amplitude $a_A = 2$ units and the time period $T_A = 2$ units (as it completes one cycle in $2$ units of time).
For wave $B$,the amplitude $a_B = 1$ unit and the time period $T_B = 1$ unit (as it completes one cycle in $1$ unit of time).
Since angular frequency $\omega = 2\pi/T$,we have $\omega_A / \omega_B = T_B / T_A = 1/2$.
Now,the ratio of intensities is:
$\frac{I_A}{I_B} = \left( \frac{a_A}{a_B} \right)^2 \times \left( \frac{\omega_A}{\omega_B} \right)^2$
$\frac{I_A}{I_B} = \left( \frac{2}{1} \right)^2 \times \left( \frac{1}{2} \right)^2 = 4 \times \frac{1}{4} = 1$
Thus,the ratio $I_A/I_B$ is $1:1$.

(A) wave is defined as a disturbance that travels through a medium or vacuum,transferring energy from one point to another without the net transport of matter.
Therefore,the correct option is $A$.

(D) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the ratio of intensities is $\frac{I_1}{I_2} = \frac{1}{25}$.
Since $\frac{I_1}{I_2} = \frac{a_1^2}{a_2^2}$,we have $\frac{a_1^2}{a_2^2} = \frac{1}{25}$.
Taking the square root on both sides,we get $\frac{a_1}{a_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$.
Therefore,the ratio of their amplitudes is $1:5$.

(A) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the ratio of intensities is $\frac{I_1}{I_2} = \frac{4}{1}$.
Since $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2$,we have $\left( \frac{a_1}{a_2} \right)^2 = \frac{4}{1}$.
Taking the square root on both sides,we get $\frac{a_1}{a_2} = \sqrt{\frac{4}{1}} = \frac{2}{1}$.
Therefore,the ratio of the amplitudes is $2:1$.

(C) The intensity $I$ of a wave is proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
For two waves with amplitudes $a_1$ and $a_2$,the ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \left( \frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1} \right)^2$
Given the ratio of amplitudes $\frac{a_1}{a_2} = \frac{4}{3}$,we substitute this into the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{4}{3} + 1}{\frac{4}{3} - 1} \right)^2 = \left( \frac{\frac{7}{3}}{\frac{1}{3}} \right)^2 = (7)^2 = \frac{49}{1}$
Thus,the ratio of maximum and minimum intensity is $49:1$.

(D) The intensity $I$ of a point source is inversely proportional to the square of the distance $r$ from the source: $I \propto \frac{1}{r^2}$.
Given that the distance $r$ increases by $2\%$,the new distance $r_2 = r_1 + 0.02r_1 = 1.02r_1$.
The ratio of intensities is given by $\frac{I_2}{I_1} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{1.02r_1} \right)^2 = (1.02)^{-2}$.
Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$,where $x = 0.02$ and $n = -2$:
$I_2 = I_1(1 + 0.02)^{-2} \approx I_1(1 - 2 \times 0.02) = I_1(1 - 0.04)$.
$I_2 = I_1 - 0.04I_1$.
This represents a decrease of $4\%$ in intensity.

(A) The intensity $I$ of a wave is defined as the power $P$ per unit area $A$.
For a circular wave spreading in two dimensions on the surface of water,the area $A$ is proportional to the circumference,which is $2\pi r$.
Thus,the intensity $I$ is given by $I = \frac{P}{2\pi r}$,which implies $I \propto \frac{1}{r}$.
Since the intensity $I$ is proportional to the square of the amplitude $A_{amp}$ (i.e.,$I \propto A_{amp}^2$),we have $A_{amp}^2 \propto \frac{1}{r}$.
Therefore,the amplitude $A_{amp}$ is proportional to $r^{-1/2}$.

(C) In a progressive wave,energy is transported from one point to another through the medium as the wave propagates.
In a stationary (standing) wave,the energy is confined within the medium between the nodes and does not propagate or transport across the medium.
Therefore,the property that distinguishes them is the transport of energy.

(C) For a spherical wave originating from a point source,the intensity $I$ is inversely proportional to the square of the distance $r$,i.e.,$I \propto 1/r^2$.
Since the intensity $I$ is also proportional to the square of the amplitude $A$ $(I \propto A^2)$,we can write $A^2 \propto 1/r^2$.
Taking the square root on both sides,we get $A \propto 1/r$.
Therefore,the ratio of amplitudes at two different distances is given by $A_1/A_2 = r_2/r_1$.
Given $r_1 = r$,$A_1 = A$,and $r_2 = 2r$,we substitute these values:
$A/A_2 = (2r)/r = 2$.
Solving for $A_2$,we get $A_2 = A/2$.

(B) The energy per unit time intercepted by a surface is given by the power $P$,which is proportional to the intensity $I$ and the area $S$.
Since intensity $I \propto A^2$,where $A$ is the amplitude,the power intercepted is $P \propto A^2 S$.
Let the initial power be $P_1 = E = k A^2 S$,where $k$ is a constant.
For the new conditions,the new area is $S' = \frac{1}{2} S$ and the new amplitude is $A' = 2A$.
The new power $P_2$ is given by $P_2 = k (A')^2 S' = k (2A)^2 (\frac{1}{2} S)$.
$P_2 = k (4A^2) (\frac{1}{2} S) = 2 k A^2 S$.
Since $E = k A^2 S$,we have $P_2 = 2E$.

(A) When a wave travels in a medium,the displacement and velocity of each particle vary with time.
For any elemental length of the string,both its kinetic energy $(K.E.)$ and potential energy $(P.E.)$ change periodically with time.
However,the total energy (sum of $K.E.$ and $P.E.$) of an elemental length remains constant as the wave propagates.
Therefore,the oscillation energy,which is the total energy of the element,remains constant.

(C) The intensity $I$ of a wave is given by the formula $I = 2\pi^2 f^2 A^2 \rho v$,where $f$ is frequency,$A$ is amplitude,$\rho$ is density,and $v$ is wave speed.
For wave $1$: $A_1 = 2a$ and $\omega_1 = \omega$ (so $f_1 = \omega / 2\pi$).
Intensity $I_1 \propto f_1^2 A_1^2 = (\omega/2\pi)^2 (2a)^2 = 4 \cdot (\omega/2\pi)^2 a^2$.
For wave $2$: $A_2 = a$ and $\omega_2 = 2\omega$ (so $f_2 = 2\omega / 2\pi$).
Intensity $I_2 \propto f_2^2 A_2^2 = (2\omega/2\pi)^2 (a)^2 = 4 \cdot (\omega/2\pi)^2 a^2$.
Since $I_1 = I_2$,Statement-$1$ is True.
Statement-$2$ is False because intensity is proportional to the square of both amplitude $AND$ frequency $(I \propto A^2 f^2)$. Therefore,Statement-$1$ is True and Statement-$2$ is False.

(B) For a long linear source of sound,the intensity $I$ at a distance $r$ is given by $I = \frac{P}{2 \pi r L}$.
The sound level $\beta$ in decibels is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right) = 10 \log_{10} \left( \frac{P}{2 \pi L I_0 r} \right)$.
Let $\beta_A$ be the sound level at distance $r$ and $\beta_B$ be the sound level at distance $10r$.
$\beta_A - \beta_B = 10 \log_{10} \left( \frac{P}{2 \pi L I_0 r} \right) - 10 \log_{10} \left( \frac{P}{2 \pi L I_0 (10r)} \right)$.
$\beta_A - \beta_B = 10 \log_{10} \left( \frac{P / (2 \pi L I_0 r)}{P / (2 \pi L I_0 10r)} \right) = 10 \log_{10} (10) = 10 \ dB$.
Given $\beta_A = 80 \ dB$,we have $80 \ dB - \beta_B = 10 \ dB$.
Therefore,$\beta_B = 80 \ dB - 10 \ dB = 70 \ dB$.

(C) For a point source emitting sound in all directions,the intensity $I$ at a distance $r$ is given by $I = \frac{P}{4\pi r^2}$,where $P$ is the power of the source. Thus,$I \propto \frac{1}{r^2}$.
Since the intensity $I$ is proportional to the square of the amplitude $A$ $(I \propto A^2)$,we have $A^2 \propto \frac{1}{r^2}$,which implies $A \propto \frac{1}{r}$.
Therefore,the ratio of the amplitudes at points $P$ and $Q$ is $\frac{A_P}{A_Q} = \frac{r_Q}{r_P}$.
Given $r_P = 9 \ m$ and $r_Q = 25 \ m$,the ratio is $\frac{A_P}{A_Q} = \frac{25}{9}$.

(B) The power $P$ transmitted by a transverse wave in a string is given by $P = F v_p$,where $F$ is the transverse component of the tension force and $v_p$ is the transverse velocity of the particle at point $P$.
For a wave traveling in the $x$-direction,the transverse force is $F = -T \frac{\partial y}{\partial x}$.
Since $\frac{\partial y}{\partial x} = -\frac{v_p}{v}$,where $v$ is the wave speed,we have $F = T \frac{v_p}{v}$.
Thus,the power is $P = (T \frac{v_p}{v}) v_p = \frac{T}{v} v_p^2$.
Given $P = 40 \ mW = 40 \times 10^{-3} \ W$,$T = 20 \ N$,and $v = 20 \ m/s$.
Substituting the values: $40 \times 10^{-3} = \frac{20}{20} v_p^2$.
$40 \times 10^{-3} = v_p^2$.
$v_p = \sqrt{0.04} \ m/s = 0.2 \ m/s$.
Converting to $cm/s$: $v_p = 0.2 \times 100 \ cm/s = 20 \ cm/s$.

(B) The mass of the string is $m = 100 \, g = 0.1 \, kg$.
The amplitude of the wave is $A = 2 \, mm = 2 \times 10^{-3} \, m$.
The angular frequency is $\omega = 5000 \, rad/s$.
The total energy $E$ of a wave in a string is given by the formula $E = \frac{1}{2} m \omega^2 A^2$.
Substituting the values:
$E = \frac{1}{2} \times (0.1) \times (5000)^2 \times (2 \times 10^{-3})^2$
$E = 0.05 \times (25 \times 10^6) \times (4 \times 10^{-6})$
$E = 0.05 \times 100 = 5 \, J$.

(A) The intensity $I$ of a sound wave is given by the relation $I = \frac{1}{2} \rho v \omega^2 A^2$,where $\rho$ is the density of the medium,$v$ is the wave velocity,$\omega$ is the angular frequency,and $A$ is the displacement amplitude.
Since $\rho$ and $v$ are constant for all waves traveling in the same medium,the intensity is proportional to the square of the product of angular frequency and amplitude: $I \propto (\omega A)^2$.
Let us calculate the value of $(\omega A)$ for each option:
$A) \omega A = 500 \times 10 \times 10^{-4} = 0.5$
$B) \omega A = 2000 \times 2 \times 10^{-4} = 0.4$
$C) \omega A = 115 \times 2 \times 10^{-4} = 0.023$
$D) \omega A = 200 \times 20 \times 10^{-4} = 0.4$
Comparing the values,option $A$ has the highest value of $(\omega A)$,and therefore,the highest intensity.

(A) The intensity $I$ of a travelling wave is given by $I = \frac{1}{2} \rho \omega^2 A^2 v$,where $\rho$ is the density of the medium,$\omega$ is the angular frequency,$A$ is the amplitude,and $v$ is the wave speed.
For the incident and reflected waves,they travel in the same medium,so $\rho$,$\omega$,and $v$ are identical for both. Thus,the ratio of intensities is proportional to the square of the ratio of amplitudes: $\frac{I_i}{I_r} = \frac{\frac{1}{2} \rho \omega^2 a_i^2 v}{\frac{1}{2} \rho \omega^2 a_r^2 v} = \left( \frac{a_i}{a_r} \right)^2$.
Since the transmitted wave travels into a different medium,the density $\rho$ and wave speed $v$ change,meaning the relationship between $I_t$ and $a_t$ involves different constants compared to the incident wave. Therefore,only the ratio involving the incident and reflected waves in the same medium holds as $\frac{I_i}{I_r} = \left( \frac{a_i}{a_r} \right)^2$.

(D) Let the initial intensity of sound be $I_0$.
After passing through one slab,the intensity decreases by $30\%$,so the remaining intensity is $I_1 = I_0 - 0.30 I_0 = 0.70 I_0$.
When passing through a second identical slab,the intensity again decreases by $30\%$ of the incident intensity on that slab.
So,the final intensity $I_2 = 0.70 I_1 = 0.70 \times (0.70 I_0) = 0.49 I_0$.
The total decrease in intensity is $I_0 - I_2 = I_0 - 0.49 I_0 = 0.51 I_0$.
Therefore,the intensity decreases by $51\%$.

(A) From the graph,for the first wave (solid line),the amplitude $A_1 = 1$ and the time period $T_1 = 2$ units.
For the second wave (dotted line),the amplitude $A_2 = 2$ and the time period $T_2 = 4$ units.
The intensity $I$ of a wave is given by $I = 2 \pi^2 \rho v f^2 A^2$,where $f = 1/T$.
Thus,$I \propto (A/T)^2$.
Therefore,the ratio of intensities is $\frac{I_1}{I_2} = \left( \frac{A_1}{A_2} \cdot \frac{T_2}{T_1} \right)^2$.
Substituting the values: $\frac{I_1}{I_2} = \left( \frac{1}{2} \cdot \frac{4}{2} \right)^2 = (1)^2 = 1 : 1$.

(A, C) plane wave does not have any radial dependence,so it travels with constant intensity.
For a spherical wave,the power $P$ emitted by the source is distributed over the surface area of a sphere of radius $r$,which is $4 \pi r^2$.
Thus,the intensity $I$ is given by $I = \frac{P}{4 \pi r^2}$.
This shows that the intensity of a spherical wave decreases as the inverse square of the distance from the source $(I \propto \frac{1}{r^2})$.
Additionally,the total power passing through any spherical surface centered at the source remains constant,but the intensity (power per unit area) decreases with distance.
Therefore,options $A$ and $C$ are correct.

(A) When a stone is thrown into still water, it creates circular ripples. The energy $E$ of the wave is conserved as it spreads outwards.
The energy of a circular crest of width $dr$ and amplitude $h$ is proportional to its mass and the square of its amplitude. The mass of the crest is proportional to its circumference $(2 \pi r)$ and its width $(dr)$.
Thus, the energy $E$ is given by:
$E \propto (2 \pi r) \cdot dr \cdot h^2$
Since the total energy $E$ remains constant as the wave propagates outwards, we have:
$r \cdot h^2 \approx \text{constant}$
Therefore, $h^2 \propto \frac{1}{r}$, which implies $h \propto r^{-1/2}$.
Thus, the amplitude of the wave varies as $r^{-1/2}$.

(C) wave is a disturbance that travels through a medium or space,transferring energy from one point to another without the net transport of matter (mass).
While the particles of the medium oscillate about their equilibrium positions,they do not travel along with the wave.
Velocity is a property of the wave's propagation,not something that is 'carried' by the wave.
Therefore,the fundamental quantity transported by waves is energy.

(A) Yes,waves carry both energy and momentum.
When a wave propagates through a medium,the particles of the medium oscillate about their mean positions.
Due to the elasticity and inertia of the medium,these particles transfer energy and momentum to their neighboring particles.
This process continues,allowing the wave to transport both energy and momentum from one point to another without the net transfer of matter.

(B) Given: Distance $r_{1} = 8 \,km = 8000 \,m$, Intensity level $L_{1} = 30 \,dB$.
Distance at the base $r_{2} = 80 \,m$, Intensity level $L_{2} = ?$.
The intensity level $L$ is given by $L = 10 \log_{10} \left( \frac{I}{I_{0}} \right)$.
Since the sound source acts as a point source, intensity $I = \frac{P}{4 \pi r^{2}}$, which implies $I \propto \frac{1}{r^{2}}$.
The difference in intensity levels is given by:
$L_{2} - L_{1} = 10 \log_{10} \left( \frac{I_{2}}{I_{1}} \right) = 10 \log_{10} \left( \frac{r_{1}}{r_{2}} \right)^{2} = 20 \log_{10} \left( \frac{r_{1}}{r_{2}} \right)$.
Substituting the values:
$L_{2} - 30 = 20 \log_{10} \left( \frac{8000}{80} \right)$
$L_{2} - 30 = 20 \log_{10} (100)$
$L_{2} - 30 = 20 \times 2 = 40$
$L_{2} = 40 + 30 = 70 \,dB$.
Thus, the intensity at the base of the tower is $70 \,dB$.

(B) The sound intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,where $P$ is the power of the source. Thus,$I \propto \frac{1}{r^2}$.
The sound intensity level $\beta$ in decibels $(dB)$ is defined as $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I_0$ is the reference intensity.
Let $\beta_1 = 10 \,dB$ at $r_1 = 4 \,m$ and $\beta_2$ be the level at $r_2 = 2 \,m$.
$\beta_2 - \beta_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right) = 10 \log_{10} \left( \frac{r_1^2}{r_2^2} \right) = 20 \log_{10} \left( \frac{r_1}{r_2} \right)$.
Substituting the values: $\beta_2 - 10 = 20 \log_{10} \left( \frac{4}{2} \right) = 20 \log_{10}(2)$.
Using $\log_{10}(2) \approx 0.301$,we get $\beta_2 - 10 = 20 \times 0.301 = 6.02$.
Therefore,$\beta_2 = 10 + 6.02 = 16.02 \,dB \approx 16 \,dB$.

(A) In a travelling wave on a string,the potential energy of an elemental portion is proportional to the square of the slope of the string at that point,i.e.,$dU_p \propto (\frac{\partial y}{\partial x})^2$.
$1$. At points $a$ and $c$,the slope of the wave is maximum (either positive or negative maximum).
$2$. At points $b$ and $d$,the string is at its extreme positions (trough and crest),where the slope is zero.
$3$. Since the potential energy is maximum where the slope is maximum,the portions $a$ and $c$ have the maximum potential energy.
$4$. However,looking at the options provided and the standard nature of this problem,the slope is maximum at the equilibrium position (where the wave crosses the $x$-axis). Thus,$a$ and $c$ are the points with maximum potential energy. Given the options,if only one is to be chosen,$a$ is the most appropriate representation of the maximum slope region.

(D) The intensity of a wave is given by $I = 2 \pi^2 \rho v f^2 A^2$,where $\rho$ is density,$v$ is wave speed,$f$ is frequency,and $A$ is amplitude.
For the given waves:
Wave $1$: $y_1 = 5 \sin(150 \pi t - 0.5 \pi x)$. Amplitude $A_1 = 5$,angular frequency $\omega_1 = 150 \pi$,wave number $k_1 = 0.5 \pi$. Wave speed $v_1 = \frac{\omega_1}{k_1} = \frac{150 \pi}{0.5 \pi} = 300 \ m/s$.
Wave $2$: $y_2 = 10 \sin(300 \pi t - 1.0 \pi x)$. Amplitude $A_2 = 10$,angular frequency $\omega_2 = 300 \pi$,wave number $k_2 = 1.0 \pi$. Wave speed $v_2 = \frac{\omega_2}{k_2} = \frac{300 \pi}{1.0 \pi} = 300 \ m/s$.
Since $v_1 = v_2$ and assuming the medium is the same $(\rho_1 = \rho_2)$,the intensity ratio is $\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 \times \left(\frac{f_1}{f_2}\right)^2 = \left(\frac{A_1}{A_2}\right)^2 \times \left(\frac{\omega_1}{\omega_2}\right)^2$.
Substituting the values: $\frac{I_1}{I_2} = \left(\frac{5}{10}\right)^2 \times \left(\frac{150 \pi}{300 \pi}\right)^2 = \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^2 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.