NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
173015
A source of sound emitting a tone of frequency $200 \mathrm{~Hz}$ moves towards an observer with a velocity $v$ equal to the velocity of sound. If the observer also moves away from the source with the same velocity $v$, the apparent frequency heard by the observer is
1 $0 \mathrm{~Hz}$
2 $100 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
D Given that, source and observer moving with the same velocity of sound in the same direction, their relative velocity is zero. $\mathrm{v}_{0}=\mathrm{v}_{\mathrm{s}}=\mathrm{v}$ So, there is no Doppler effect and thus apparent frequency heard by observer is same as original frequency Hence, $\mathrm{f}^{\prime}=\mathrm{f}=200 \mathrm{~Hz}$
JIPMER - 2015
WAVES
172980
When source of sound moves towards a stationary observer, the wavelength of sound received by him
1 decreases while frequency increase
2 remains the same whereas frequency increases
3 increases and frequency also increases
4 decreases while frequency remains the same
Explanation:
A The frequency rises as the sources of sound draws closer to the observer. However the sound velocity remains constant. $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ Hence, frequency increases and wavelength $(\lambda)$ decreases.
MHT-CET 2018
WAVES
172982
A source emits a sound of frequency of $400 \mathrm{~Hz}$, but the listener hears it to be $390 \mathrm{~Hz}$. Then
1 the listener is moving towards the source
2 the source is moving towards the listener
3 the listener is moving away from the source
4 the listener is moving has a defective ear.
Explanation:
C Given that, Sound frequency $(f)=400 \mathrm{~Hz}$ Apparent frequency $\left(f^{\prime}\right)=390 \mathrm{~Hz}$ $\because f^{\prime}Apparent frequency is lesser then the actual frequency, hence the relative separation between source and listener should be increases. So, the distance between the source and listener increases or the listener is moving away from source.
COMEDK 2018
WAVES
172988
A reflector is moving with $20 \mathrm{~ms}^{-1}$ towards a stationary source of sound. If the source is producing sound waves of $160 \mathrm{~Hz}$, then the wavelength of the reflected wave is (Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )
1 $\frac{17}{8} \mathrm{~m}$
2 $\frac{17}{11} \mathrm{~m}$
3 $\frac{17}{9} \mathrm{~m}$
4 $\frac{17}{16} \mathrm{~m}$
Explanation:
C Given, $v=340 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{\mathrm{s}}=20 \mathrm{~m} / \mathrm{s}, \mathrm{f}=160 \mathrm{~Hz}$ Wavelength of the reflected wave is- $\lambda^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \lambda=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \frac{\mathrm{v}}{\mathrm{f}}$ $\lambda^{\prime}=\left(\frac{340-20}{340+20}\right) \frac{340}{160}=\frac{320}{360} \times \frac{340}{160}$ $\lambda^{\prime}=\frac{17}{9} \mathrm{~m} \quad\left\{\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right\}$
173015
A source of sound emitting a tone of frequency $200 \mathrm{~Hz}$ moves towards an observer with a velocity $v$ equal to the velocity of sound. If the observer also moves away from the source with the same velocity $v$, the apparent frequency heard by the observer is
1 $0 \mathrm{~Hz}$
2 $100 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
D Given that, source and observer moving with the same velocity of sound in the same direction, their relative velocity is zero. $\mathrm{v}_{0}=\mathrm{v}_{\mathrm{s}}=\mathrm{v}$ So, there is no Doppler effect and thus apparent frequency heard by observer is same as original frequency Hence, $\mathrm{f}^{\prime}=\mathrm{f}=200 \mathrm{~Hz}$
JIPMER - 2015
WAVES
172980
When source of sound moves towards a stationary observer, the wavelength of sound received by him
1 decreases while frequency increase
2 remains the same whereas frequency increases
3 increases and frequency also increases
4 decreases while frequency remains the same
Explanation:
A The frequency rises as the sources of sound draws closer to the observer. However the sound velocity remains constant. $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ Hence, frequency increases and wavelength $(\lambda)$ decreases.
MHT-CET 2018
WAVES
172982
A source emits a sound of frequency of $400 \mathrm{~Hz}$, but the listener hears it to be $390 \mathrm{~Hz}$. Then
1 the listener is moving towards the source
2 the source is moving towards the listener
3 the listener is moving away from the source
4 the listener is moving has a defective ear.
Explanation:
C Given that, Sound frequency $(f)=400 \mathrm{~Hz}$ Apparent frequency $\left(f^{\prime}\right)=390 \mathrm{~Hz}$ $\because f^{\prime}Apparent frequency is lesser then the actual frequency, hence the relative separation between source and listener should be increases. So, the distance between the source and listener increases or the listener is moving away from source.
COMEDK 2018
WAVES
172988
A reflector is moving with $20 \mathrm{~ms}^{-1}$ towards a stationary source of sound. If the source is producing sound waves of $160 \mathrm{~Hz}$, then the wavelength of the reflected wave is (Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )
1 $\frac{17}{8} \mathrm{~m}$
2 $\frac{17}{11} \mathrm{~m}$
3 $\frac{17}{9} \mathrm{~m}$
4 $\frac{17}{16} \mathrm{~m}$
Explanation:
C Given, $v=340 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{\mathrm{s}}=20 \mathrm{~m} / \mathrm{s}, \mathrm{f}=160 \mathrm{~Hz}$ Wavelength of the reflected wave is- $\lambda^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \lambda=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \frac{\mathrm{v}}{\mathrm{f}}$ $\lambda^{\prime}=\left(\frac{340-20}{340+20}\right) \frac{340}{160}=\frac{320}{360} \times \frac{340}{160}$ $\lambda^{\prime}=\frac{17}{9} \mathrm{~m} \quad\left\{\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right\}$
173015
A source of sound emitting a tone of frequency $200 \mathrm{~Hz}$ moves towards an observer with a velocity $v$ equal to the velocity of sound. If the observer also moves away from the source with the same velocity $v$, the apparent frequency heard by the observer is
1 $0 \mathrm{~Hz}$
2 $100 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
D Given that, source and observer moving with the same velocity of sound in the same direction, their relative velocity is zero. $\mathrm{v}_{0}=\mathrm{v}_{\mathrm{s}}=\mathrm{v}$ So, there is no Doppler effect and thus apparent frequency heard by observer is same as original frequency Hence, $\mathrm{f}^{\prime}=\mathrm{f}=200 \mathrm{~Hz}$
JIPMER - 2015
WAVES
172980
When source of sound moves towards a stationary observer, the wavelength of sound received by him
1 decreases while frequency increase
2 remains the same whereas frequency increases
3 increases and frequency also increases
4 decreases while frequency remains the same
Explanation:
A The frequency rises as the sources of sound draws closer to the observer. However the sound velocity remains constant. $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ Hence, frequency increases and wavelength $(\lambda)$ decreases.
MHT-CET 2018
WAVES
172982
A source emits a sound of frequency of $400 \mathrm{~Hz}$, but the listener hears it to be $390 \mathrm{~Hz}$. Then
1 the listener is moving towards the source
2 the source is moving towards the listener
3 the listener is moving away from the source
4 the listener is moving has a defective ear.
Explanation:
C Given that, Sound frequency $(f)=400 \mathrm{~Hz}$ Apparent frequency $\left(f^{\prime}\right)=390 \mathrm{~Hz}$ $\because f^{\prime}Apparent frequency is lesser then the actual frequency, hence the relative separation between source and listener should be increases. So, the distance between the source and listener increases or the listener is moving away from source.
COMEDK 2018
WAVES
172988
A reflector is moving with $20 \mathrm{~ms}^{-1}$ towards a stationary source of sound. If the source is producing sound waves of $160 \mathrm{~Hz}$, then the wavelength of the reflected wave is (Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )
1 $\frac{17}{8} \mathrm{~m}$
2 $\frac{17}{11} \mathrm{~m}$
3 $\frac{17}{9} \mathrm{~m}$
4 $\frac{17}{16} \mathrm{~m}$
Explanation:
C Given, $v=340 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{\mathrm{s}}=20 \mathrm{~m} / \mathrm{s}, \mathrm{f}=160 \mathrm{~Hz}$ Wavelength of the reflected wave is- $\lambda^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \lambda=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \frac{\mathrm{v}}{\mathrm{f}}$ $\lambda^{\prime}=\left(\frac{340-20}{340+20}\right) \frac{340}{160}=\frac{320}{360} \times \frac{340}{160}$ $\lambda^{\prime}=\frac{17}{9} \mathrm{~m} \quad\left\{\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right\}$
173015
A source of sound emitting a tone of frequency $200 \mathrm{~Hz}$ moves towards an observer with a velocity $v$ equal to the velocity of sound. If the observer also moves away from the source with the same velocity $v$, the apparent frequency heard by the observer is
1 $0 \mathrm{~Hz}$
2 $100 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
D Given that, source and observer moving with the same velocity of sound in the same direction, their relative velocity is zero. $\mathrm{v}_{0}=\mathrm{v}_{\mathrm{s}}=\mathrm{v}$ So, there is no Doppler effect and thus apparent frequency heard by observer is same as original frequency Hence, $\mathrm{f}^{\prime}=\mathrm{f}=200 \mathrm{~Hz}$
JIPMER - 2015
WAVES
172980
When source of sound moves towards a stationary observer, the wavelength of sound received by him
1 decreases while frequency increase
2 remains the same whereas frequency increases
3 increases and frequency also increases
4 decreases while frequency remains the same
Explanation:
A The frequency rises as the sources of sound draws closer to the observer. However the sound velocity remains constant. $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ Hence, frequency increases and wavelength $(\lambda)$ decreases.
MHT-CET 2018
WAVES
172982
A source emits a sound of frequency of $400 \mathrm{~Hz}$, but the listener hears it to be $390 \mathrm{~Hz}$. Then
1 the listener is moving towards the source
2 the source is moving towards the listener
3 the listener is moving away from the source
4 the listener is moving has a defective ear.
Explanation:
C Given that, Sound frequency $(f)=400 \mathrm{~Hz}$ Apparent frequency $\left(f^{\prime}\right)=390 \mathrm{~Hz}$ $\because f^{\prime}Apparent frequency is lesser then the actual frequency, hence the relative separation between source and listener should be increases. So, the distance between the source and listener increases or the listener is moving away from source.
COMEDK 2018
WAVES
172988
A reflector is moving with $20 \mathrm{~ms}^{-1}$ towards a stationary source of sound. If the source is producing sound waves of $160 \mathrm{~Hz}$, then the wavelength of the reflected wave is (Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )
1 $\frac{17}{8} \mathrm{~m}$
2 $\frac{17}{11} \mathrm{~m}$
3 $\frac{17}{9} \mathrm{~m}$
4 $\frac{17}{16} \mathrm{~m}$
Explanation:
C Given, $v=340 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{\mathrm{s}}=20 \mathrm{~m} / \mathrm{s}, \mathrm{f}=160 \mathrm{~Hz}$ Wavelength of the reflected wave is- $\lambda^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \lambda=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \frac{\mathrm{v}}{\mathrm{f}}$ $\lambda^{\prime}=\left(\frac{340-20}{340+20}\right) \frac{340}{160}=\frac{320}{360} \times \frac{340}{160}$ $\lambda^{\prime}=\frac{17}{9} \mathrm{~m} \quad\left\{\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right\}$
