173064
A source of sound is travelling with a velocity $40 \mathrm{~km} / \mathrm{h}$ towards observer and emits sound of frequency $2000\mathrm{~Hz}$. If velocity of sound is $\mathbf{1 2 2 0}$ $\mathrm{km} / \mathrm{h}$, then what is the apparent frequency heard by an observer ?
1 $2210 \mathrm{~Hz}$
2 $1920 \mathrm{~Hz}$
3 $2068 \mathrm{~Hz}$
4 $2086 \mathrm{~Hz}$
Explanation:
C Given that, Velocity of sound $(\mathrm{v})=1220 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=338.88 \mathrm{~ms}^{-1}$ Source Velocity $\left(\mathrm{v}_{\mathrm{S}}\right)=40 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=11.1 \mathrm{~ms}^{-1}$ Frequency $(\mathrm{f})=2000 \mathrm{~Hz}$ $\mathrm{f}^{\prime} =\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\mathrm{f}^{\prime} =\left(\frac{338.88}{338.88-11.1}\right) \times 2000$ $\mathrm{f}^{\prime} =2067.72 = 2068 \mathrm{~Hz}$
UP CPMT-2011
WAVES
173066
A source and observer are approaching each other with $50 \mathrm{~ms}^{-1}$ velocity. What will be original frequency if the observer receives 400 cycle/s
173067
A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. Ratio of observed and real frequencies will be
1 $\frac{9}{10}$
2 $\frac{11}{10}$
3 $\frac{10}{11}$
4 $\frac{10}{9}$
Explanation:
D Given, A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. We know that, $f^{\prime}=f_{0}\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observer }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observe }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\because \mathrm{v}_{\text {observer }}=0, \text { because observer is stationary }$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}-0}{\mathrm{v}-\frac{\mathrm{v}}{10}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}}{10 \mathrm{v}-\mathrm{v}}\right.$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{10 \mathrm{v}}{9 \mathrm{v}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\frac{10}{9}$
AP EAMCET (18.09.2020) Shift-II
WAVES
173071
If siren emitting sound of frequency $500 \mathrm{~Hz}$ is going away from a stationary listener with a speed of $50 \mathrm{~m} / \mathrm{s}$, the frequency of sound heard directly from the siren is
1 $286.5 \mathrm{~Hz}$
2 $481 \mathrm{~Hz}$
3 $434.2 \mathrm{~Hz}$
4 $580 \mathrm{~Hz}$
Explanation:
C Given data, Frequency of siren, $\mathrm{f}_{0}=500 \mathrm{~Hz}, \mathrm{v}_{0}=0$ The speed of siren, $\mathrm{v}_{\mathrm{sr}}=50 \mathrm{~m} / \mathrm{s}$ Let, $\mathrm{v}_{\text {sound }}=\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ According to the Doppler's principal, $\mathrm{f}^{\prime}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \mathrm{f}_{0}$ $\mathrm{f}^{\prime}=500\left(\frac{330+0}{330+50}\right)=500\left(\frac{330}{380}\right)$ $\mathrm{f}^{\prime}=434.2 \mathrm{~Hz}$
173064
A source of sound is travelling with a velocity $40 \mathrm{~km} / \mathrm{h}$ towards observer and emits sound of frequency $2000\mathrm{~Hz}$. If velocity of sound is $\mathbf{1 2 2 0}$ $\mathrm{km} / \mathrm{h}$, then what is the apparent frequency heard by an observer ?
1 $2210 \mathrm{~Hz}$
2 $1920 \mathrm{~Hz}$
3 $2068 \mathrm{~Hz}$
4 $2086 \mathrm{~Hz}$
Explanation:
C Given that, Velocity of sound $(\mathrm{v})=1220 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=338.88 \mathrm{~ms}^{-1}$ Source Velocity $\left(\mathrm{v}_{\mathrm{S}}\right)=40 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=11.1 \mathrm{~ms}^{-1}$ Frequency $(\mathrm{f})=2000 \mathrm{~Hz}$ $\mathrm{f}^{\prime} =\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\mathrm{f}^{\prime} =\left(\frac{338.88}{338.88-11.1}\right) \times 2000$ $\mathrm{f}^{\prime} =2067.72 = 2068 \mathrm{~Hz}$
UP CPMT-2011
WAVES
173066
A source and observer are approaching each other with $50 \mathrm{~ms}^{-1}$ velocity. What will be original frequency if the observer receives 400 cycle/s
173067
A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. Ratio of observed and real frequencies will be
1 $\frac{9}{10}$
2 $\frac{11}{10}$
3 $\frac{10}{11}$
4 $\frac{10}{9}$
Explanation:
D Given, A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. We know that, $f^{\prime}=f_{0}\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observer }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observe }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\because \mathrm{v}_{\text {observer }}=0, \text { because observer is stationary }$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}-0}{\mathrm{v}-\frac{\mathrm{v}}{10}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}}{10 \mathrm{v}-\mathrm{v}}\right.$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{10 \mathrm{v}}{9 \mathrm{v}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\frac{10}{9}$
AP EAMCET (18.09.2020) Shift-II
WAVES
173071
If siren emitting sound of frequency $500 \mathrm{~Hz}$ is going away from a stationary listener with a speed of $50 \mathrm{~m} / \mathrm{s}$, the frequency of sound heard directly from the siren is
1 $286.5 \mathrm{~Hz}$
2 $481 \mathrm{~Hz}$
3 $434.2 \mathrm{~Hz}$
4 $580 \mathrm{~Hz}$
Explanation:
C Given data, Frequency of siren, $\mathrm{f}_{0}=500 \mathrm{~Hz}, \mathrm{v}_{0}=0$ The speed of siren, $\mathrm{v}_{\mathrm{sr}}=50 \mathrm{~m} / \mathrm{s}$ Let, $\mathrm{v}_{\text {sound }}=\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ According to the Doppler's principal, $\mathrm{f}^{\prime}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \mathrm{f}_{0}$ $\mathrm{f}^{\prime}=500\left(\frac{330+0}{330+50}\right)=500\left(\frac{330}{380}\right)$ $\mathrm{f}^{\prime}=434.2 \mathrm{~Hz}$
173064
A source of sound is travelling with a velocity $40 \mathrm{~km} / \mathrm{h}$ towards observer and emits sound of frequency $2000\mathrm{~Hz}$. If velocity of sound is $\mathbf{1 2 2 0}$ $\mathrm{km} / \mathrm{h}$, then what is the apparent frequency heard by an observer ?
1 $2210 \mathrm{~Hz}$
2 $1920 \mathrm{~Hz}$
3 $2068 \mathrm{~Hz}$
4 $2086 \mathrm{~Hz}$
Explanation:
C Given that, Velocity of sound $(\mathrm{v})=1220 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=338.88 \mathrm{~ms}^{-1}$ Source Velocity $\left(\mathrm{v}_{\mathrm{S}}\right)=40 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=11.1 \mathrm{~ms}^{-1}$ Frequency $(\mathrm{f})=2000 \mathrm{~Hz}$ $\mathrm{f}^{\prime} =\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\mathrm{f}^{\prime} =\left(\frac{338.88}{338.88-11.1}\right) \times 2000$ $\mathrm{f}^{\prime} =2067.72 = 2068 \mathrm{~Hz}$
UP CPMT-2011
WAVES
173066
A source and observer are approaching each other with $50 \mathrm{~ms}^{-1}$ velocity. What will be original frequency if the observer receives 400 cycle/s
173067
A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. Ratio of observed and real frequencies will be
1 $\frac{9}{10}$
2 $\frac{11}{10}$
3 $\frac{10}{11}$
4 $\frac{10}{9}$
Explanation:
D Given, A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. We know that, $f^{\prime}=f_{0}\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observer }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observe }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\because \mathrm{v}_{\text {observer }}=0, \text { because observer is stationary }$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}-0}{\mathrm{v}-\frac{\mathrm{v}}{10}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}}{10 \mathrm{v}-\mathrm{v}}\right.$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{10 \mathrm{v}}{9 \mathrm{v}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\frac{10}{9}$
AP EAMCET (18.09.2020) Shift-II
WAVES
173071
If siren emitting sound of frequency $500 \mathrm{~Hz}$ is going away from a stationary listener with a speed of $50 \mathrm{~m} / \mathrm{s}$, the frequency of sound heard directly from the siren is
1 $286.5 \mathrm{~Hz}$
2 $481 \mathrm{~Hz}$
3 $434.2 \mathrm{~Hz}$
4 $580 \mathrm{~Hz}$
Explanation:
C Given data, Frequency of siren, $\mathrm{f}_{0}=500 \mathrm{~Hz}, \mathrm{v}_{0}=0$ The speed of siren, $\mathrm{v}_{\mathrm{sr}}=50 \mathrm{~m} / \mathrm{s}$ Let, $\mathrm{v}_{\text {sound }}=\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ According to the Doppler's principal, $\mathrm{f}^{\prime}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \mathrm{f}_{0}$ $\mathrm{f}^{\prime}=500\left(\frac{330+0}{330+50}\right)=500\left(\frac{330}{380}\right)$ $\mathrm{f}^{\prime}=434.2 \mathrm{~Hz}$
173064
A source of sound is travelling with a velocity $40 \mathrm{~km} / \mathrm{h}$ towards observer and emits sound of frequency $2000\mathrm{~Hz}$. If velocity of sound is $\mathbf{1 2 2 0}$ $\mathrm{km} / \mathrm{h}$, then what is the apparent frequency heard by an observer ?
1 $2210 \mathrm{~Hz}$
2 $1920 \mathrm{~Hz}$
3 $2068 \mathrm{~Hz}$
4 $2086 \mathrm{~Hz}$
Explanation:
C Given that, Velocity of sound $(\mathrm{v})=1220 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=338.88 \mathrm{~ms}^{-1}$ Source Velocity $\left(\mathrm{v}_{\mathrm{S}}\right)=40 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=11.1 \mathrm{~ms}^{-1}$ Frequency $(\mathrm{f})=2000 \mathrm{~Hz}$ $\mathrm{f}^{\prime} =\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\mathrm{f}^{\prime} =\left(\frac{338.88}{338.88-11.1}\right) \times 2000$ $\mathrm{f}^{\prime} =2067.72 = 2068 \mathrm{~Hz}$
UP CPMT-2011
WAVES
173066
A source and observer are approaching each other with $50 \mathrm{~ms}^{-1}$ velocity. What will be original frequency if the observer receives 400 cycle/s
173067
A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. Ratio of observed and real frequencies will be
1 $\frac{9}{10}$
2 $\frac{11}{10}$
3 $\frac{10}{11}$
4 $\frac{10}{9}$
Explanation:
D Given, A source is approaching a stationary observer with velocity $\left(\frac{1}{10}\right)^{\text {th }}$ that of sound. We know that, $f^{\prime}=f_{0}\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observer }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {observe }}}{\mathrm{v}_{\text {sound }}-\mathrm{v}_{\text {source }}}\right]$ $\because \mathrm{v}_{\text {observer }}=0, \text { because observer is stationary }$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}-0}{\mathrm{v}-\frac{\mathrm{v}}{10}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{\mathrm{v}}{10 \mathrm{v}-\mathrm{v}}\right.$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\left[\frac{10 \mathrm{v}}{9 \mathrm{v}}\right]$ $\Rightarrow \frac{f^{\prime}}{f_{0}}=\frac{10}{9}$
AP EAMCET (18.09.2020) Shift-II
WAVES
173071
If siren emitting sound of frequency $500 \mathrm{~Hz}$ is going away from a stationary listener with a speed of $50 \mathrm{~m} / \mathrm{s}$, the frequency of sound heard directly from the siren is
1 $286.5 \mathrm{~Hz}$
2 $481 \mathrm{~Hz}$
3 $434.2 \mathrm{~Hz}$
4 $580 \mathrm{~Hz}$
Explanation:
C Given data, Frequency of siren, $\mathrm{f}_{0}=500 \mathrm{~Hz}, \mathrm{v}_{0}=0$ The speed of siren, $\mathrm{v}_{\mathrm{sr}}=50 \mathrm{~m} / \mathrm{s}$ Let, $\mathrm{v}_{\text {sound }}=\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ According to the Doppler's principal, $\mathrm{f}^{\prime}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \mathrm{f}_{0}$ $\mathrm{f}^{\prime}=500\left(\frac{330+0}{330+50}\right)=500\left(\frac{330}{380}\right)$ $\mathrm{f}^{\prime}=434.2 \mathrm{~Hz}$
