Answer: Option B. -> 36%
:
B
Intensity after passing through one slab
$I^{\prime }=[I-\frac{20}{100}\times I]=[I-\frac{1}{5}]=\frac{41}{5}$
So, intensity after passing through two slabs
$I"=[I-\frac{20}{100}\times I^{\prime }]=\frac{4I^{\prime }}{5}=\frac{161}{25}\therefore \%decrease=\left[\frac{(I=-\frac{161}{5})}{I}\right]\times 100=36\%$

Answer: Option B. -> 36%
:
B
Towards right wavelength gets compressed and towards left wavelength gets expanded.

Answer: Option A. -> 0
:
A
The frequency of direct and reflected sound is same.

Answer: Option A. -> 711 Hz
:
A
As the source and the observer are approaching one another, so n’ would be larger.
$f=\left(\frac{v+\frac{v}{15}}{v-\frac{v}{10}}\right)600=711Hz$

Answer: Option A. -> The minimum value of apparent frequency is 889 Hz.
:
A
This frequency–time curve corresponds to a source moving at an angle to a stationary observer.

In the region SN, the source is moving towards the observer, i.e., the apparent frequency
$n^{\prime }=n_0\left(\frac{v}{v-v_{s}cos\theta }\right)$
$n^{\prime }=n_0\left(\frac{300}{300-30cos\theta }\right)$
$When\theta =\frac{\pi }{2}.$ i.e., at N,
$n^{\prime }=n_0=1000Hz,$ i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency
$n^{\prime }=n_0\left(\frac{300}{300-30cos\theta }\right)$
$When\theta =0,i.e.,cos\theta =1,$
$n_{max}=n_0\frac{v}{v-v_s}=\frac{\left(1000Hz\right)\left(300m/s\right)}{(300m/s-30m/s)}$
$=\frac{10}{9}\times 1000Hz=1111Hz$
$n_{min}=n_0\frac{v}{v+v_s}=\frac{1000\times 300}{330}=909Hz$

Answer: Option A. -> 1.2f and λ
:
A
Given that velocity of source v_S = 0(because it is stationary). Velocity of observer $v_0=\left(\frac{1}{5}\right)v=0.2v$ (where v is the velocity of sound). Actual frequency of source is f and actual wavelength of source is λ. We know from the Doppler’s effect that the apparent frequency recorded, when the observer is moving towards the stationary source, is given by
$n^{\prime }=\left(\frac{v+v_0}{v-v_s}\right)\times n=\left(\frac{v+0.2v}{v-0}\right)\times n=\frac{1.2v}{v}\times n=1.2n=1.2f$
Since the source is stationary, therefore the apparent wavelength remains unchanged, i.e., $\lambda$

Answer: Option B. -> 680 Hz
:
B
Effective value of velocity of source is

$v_s=\frac{100}{3}cos\theta$
$=\frac{100}{3}\times \frac{3}{5}=20m/s$
$v^{\prime }=\frac{v}{v-v_s}v$
$v^{\prime }=\frac{340}{340-20}\times 640Hz=680Hz$

Answer: Option C. -> 1094 Hz
:
C
$f=\left(\frac{v+v_m}{v+v_m-v_{source}}\right)1000$
$=\left(\frac{340+20cos{60}^{\circ }}{340+20cos{60}^{\circ }-30}\right)1000$
=1094 Hz

Answer: Option B. -> 2.5 m/s
:
B
Apparent frequency due to source A is
$n^{\prime }=\frac{v-u}{v}\times n$
Apparent frequency due to source B is
$n"=\frac{v+u}{v}\times n$
$\therefore n"-n^{\prime }=\frac{2u}{v}\times n=10$
$\therefore u=\frac{10v}{2n}=\frac{10\times 340}{2\times 680}=2.5m/s$

Answer: Option B. -> 30 m/s
:
B
$v^{\prime }=\frac{v}{v-v_s}v,v"=\frac{v}{v+v_s}v\frac{v^{\prime }}{v^{\prime \prime }}=\frac{v+v_z}{v-v_s}or\frac{6}{5}=\frac{330+v}{330-v}11v_s=330\Rightarrow v_s=30m/s$