173048
A racing car moving towards a cliff, sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If $v$ is the velocity of sound, then the velocity of the car is :
1 $v / 2$
2 $\mathrm{v} / \sqrt{2}$
3 $\mathrm{v} / 4$
4 $v / 3$
Explanation:
D Given that, $\mathrm{n}^{\prime}=2 \mathrm{n}$ If $v_{\mathrm{s}}$ be the velocity of car then frequency of reflected sound heard by driver, $\mathrm{n}^{\prime}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $2 \mathrm{n}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{car}}}{\mathrm{v}-\mathrm{v}_{\mathrm{car}}}\right)$ $2 \mathrm{v}-2 \mathrm{v}_{\mathrm{car}}=\mathrm{v}+\mathrm{v}_{\mathrm{car}}$ $2 \mathrm{v}-\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}_{\mathrm{car}}=\frac{\mathrm{v}}{3}$
Karnataka CET-2002
WAVES
173049
A bus is moving with a velocity of $5 \mathrm{~ms}^{-1}$ towards a huge wall. The driver sounds a horn of frequency $165 \mathrm{~Hz}$. If the speed of sound in air is $335 \mathrm{~m} / \mathrm{s}$, the number of beats heard per second by a passenger on the bus will be :
1 3
2 4
3 6
4 5
Explanation:
D Given that, $\mathrm{v}_{\mathrm{s}}=5 \mathrm{~m} / \mathrm{s}, \mathrm{v}=335 \mathrm{~m} / \mathrm{s}$ Frequency $(\mathrm{n})=165 \mathrm{~Hz}$ As the observer is moving with the source then apparent frequency of reflected wave is $n^{\prime}=n\left(\frac{v+v_{s}}{v-v_{s}}\right)$ $\mathrm{n}^{\prime}=165 \times\left(\frac{335+5}{335-5}\right)$ $\mathrm{n}^{\prime}=170$ $\therefore$ No. of beats $=\mathrm{n}^{\prime}-\mathrm{n}=170-165=5$
Karnataka CET-2001
WAVES
173052
What should be the velocity of a sound source moving towards a stationary observer so that the apparent frequency is double the actual frequency. Velocity of sound is $\mathrm{V}$.
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $\mathrm{V} / 2$
4 $\mathrm{V} / 4$
Explanation:
C Given that, The apparent frequency is double the actual frequency, $\mathrm{f}^{\prime}=2 \mathrm{f}$ Velocity of sound $=(\mathrm{V})$ By using formula, $\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right) \mathrm{n}$ $2 \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right)$ $2\left(\mathrm{~V}-\mathrm{V}_{\mathrm{s}}\right)=\mathrm{V}$ $2 \mathrm{~V}-2 \mathrm{~V}_{\mathrm{s}}=\mathrm{V}$ $\mathrm{V}=2 \mathrm{~V}_{\mathrm{s}}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{V}}{2}$
J and K CET- 1999
WAVES
173053
Sonar fitted in a submarine operates at 30 kHz. An enemy submarine approaches it with a speed of $360 \mathrm{~km} / \mathrm{h}$. The speed of sound in sea water is $1450 \mathrm{~m} / \mathrm{s}$. The frequency of sound heard after reflection from enemy submarine is
173055
An observer standing near the sea-coast counts 48 waves per min. If the wavelength of the wave is $10 \mathrm{~m}$, the velocity of the waves will be
1 $8 \mathrm{~m} / \mathrm{s}$
2 $12 \mathrm{~m} / \mathrm{s}$
3 $16 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Total number of waves, $\lambda=48$ Wavelength of each wave $=10 \mathrm{~m}$ Time taken, $\mathrm{t}=60$ second $\operatorname{Velocity}(\mathrm{v}) =\frac{\lambda}{\mathrm{T}}$ $\mathrm{v} =\frac{48}{60} \mathrm{~Hz}$ $\mathrm{v} =\frac{4}{5}$ $\mathrm{v} =0.8 \mathrm{~Hz}$ $\therefore \quad$ Wavelength velocity $=\mathrm{v} \cdot \lambda$ $=(0.8) \times(10)$ $=8 \mathrm{~m} / \mathrm{s}$
173048
A racing car moving towards a cliff, sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If $v$ is the velocity of sound, then the velocity of the car is :
1 $v / 2$
2 $\mathrm{v} / \sqrt{2}$
3 $\mathrm{v} / 4$
4 $v / 3$
Explanation:
D Given that, $\mathrm{n}^{\prime}=2 \mathrm{n}$ If $v_{\mathrm{s}}$ be the velocity of car then frequency of reflected sound heard by driver, $\mathrm{n}^{\prime}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $2 \mathrm{n}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{car}}}{\mathrm{v}-\mathrm{v}_{\mathrm{car}}}\right)$ $2 \mathrm{v}-2 \mathrm{v}_{\mathrm{car}}=\mathrm{v}+\mathrm{v}_{\mathrm{car}}$ $2 \mathrm{v}-\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}_{\mathrm{car}}=\frac{\mathrm{v}}{3}$
Karnataka CET-2002
WAVES
173049
A bus is moving with a velocity of $5 \mathrm{~ms}^{-1}$ towards a huge wall. The driver sounds a horn of frequency $165 \mathrm{~Hz}$. If the speed of sound in air is $335 \mathrm{~m} / \mathrm{s}$, the number of beats heard per second by a passenger on the bus will be :
1 3
2 4
3 6
4 5
Explanation:
D Given that, $\mathrm{v}_{\mathrm{s}}=5 \mathrm{~m} / \mathrm{s}, \mathrm{v}=335 \mathrm{~m} / \mathrm{s}$ Frequency $(\mathrm{n})=165 \mathrm{~Hz}$ As the observer is moving with the source then apparent frequency of reflected wave is $n^{\prime}=n\left(\frac{v+v_{s}}{v-v_{s}}\right)$ $\mathrm{n}^{\prime}=165 \times\left(\frac{335+5}{335-5}\right)$ $\mathrm{n}^{\prime}=170$ $\therefore$ No. of beats $=\mathrm{n}^{\prime}-\mathrm{n}=170-165=5$
Karnataka CET-2001
WAVES
173052
What should be the velocity of a sound source moving towards a stationary observer so that the apparent frequency is double the actual frequency. Velocity of sound is $\mathrm{V}$.
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $\mathrm{V} / 2$
4 $\mathrm{V} / 4$
Explanation:
C Given that, The apparent frequency is double the actual frequency, $\mathrm{f}^{\prime}=2 \mathrm{f}$ Velocity of sound $=(\mathrm{V})$ By using formula, $\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right) \mathrm{n}$ $2 \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right)$ $2\left(\mathrm{~V}-\mathrm{V}_{\mathrm{s}}\right)=\mathrm{V}$ $2 \mathrm{~V}-2 \mathrm{~V}_{\mathrm{s}}=\mathrm{V}$ $\mathrm{V}=2 \mathrm{~V}_{\mathrm{s}}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{V}}{2}$
J and K CET- 1999
WAVES
173053
Sonar fitted in a submarine operates at 30 kHz. An enemy submarine approaches it with a speed of $360 \mathrm{~km} / \mathrm{h}$. The speed of sound in sea water is $1450 \mathrm{~m} / \mathrm{s}$. The frequency of sound heard after reflection from enemy submarine is
173055
An observer standing near the sea-coast counts 48 waves per min. If the wavelength of the wave is $10 \mathrm{~m}$, the velocity of the waves will be
1 $8 \mathrm{~m} / \mathrm{s}$
2 $12 \mathrm{~m} / \mathrm{s}$
3 $16 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Total number of waves, $\lambda=48$ Wavelength of each wave $=10 \mathrm{~m}$ Time taken, $\mathrm{t}=60$ second $\operatorname{Velocity}(\mathrm{v}) =\frac{\lambda}{\mathrm{T}}$ $\mathrm{v} =\frac{48}{60} \mathrm{~Hz}$ $\mathrm{v} =\frac{4}{5}$ $\mathrm{v} =0.8 \mathrm{~Hz}$ $\therefore \quad$ Wavelength velocity $=\mathrm{v} \cdot \lambda$ $=(0.8) \times(10)$ $=8 \mathrm{~m} / \mathrm{s}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
173048
A racing car moving towards a cliff, sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If $v$ is the velocity of sound, then the velocity of the car is :
1 $v / 2$
2 $\mathrm{v} / \sqrt{2}$
3 $\mathrm{v} / 4$
4 $v / 3$
Explanation:
D Given that, $\mathrm{n}^{\prime}=2 \mathrm{n}$ If $v_{\mathrm{s}}$ be the velocity of car then frequency of reflected sound heard by driver, $\mathrm{n}^{\prime}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $2 \mathrm{n}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{car}}}{\mathrm{v}-\mathrm{v}_{\mathrm{car}}}\right)$ $2 \mathrm{v}-2 \mathrm{v}_{\mathrm{car}}=\mathrm{v}+\mathrm{v}_{\mathrm{car}}$ $2 \mathrm{v}-\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}_{\mathrm{car}}=\frac{\mathrm{v}}{3}$
Karnataka CET-2002
WAVES
173049
A bus is moving with a velocity of $5 \mathrm{~ms}^{-1}$ towards a huge wall. The driver sounds a horn of frequency $165 \mathrm{~Hz}$. If the speed of sound in air is $335 \mathrm{~m} / \mathrm{s}$, the number of beats heard per second by a passenger on the bus will be :
1 3
2 4
3 6
4 5
Explanation:
D Given that, $\mathrm{v}_{\mathrm{s}}=5 \mathrm{~m} / \mathrm{s}, \mathrm{v}=335 \mathrm{~m} / \mathrm{s}$ Frequency $(\mathrm{n})=165 \mathrm{~Hz}$ As the observer is moving with the source then apparent frequency of reflected wave is $n^{\prime}=n\left(\frac{v+v_{s}}{v-v_{s}}\right)$ $\mathrm{n}^{\prime}=165 \times\left(\frac{335+5}{335-5}\right)$ $\mathrm{n}^{\prime}=170$ $\therefore$ No. of beats $=\mathrm{n}^{\prime}-\mathrm{n}=170-165=5$
Karnataka CET-2001
WAVES
173052
What should be the velocity of a sound source moving towards a stationary observer so that the apparent frequency is double the actual frequency. Velocity of sound is $\mathrm{V}$.
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $\mathrm{V} / 2$
4 $\mathrm{V} / 4$
Explanation:
C Given that, The apparent frequency is double the actual frequency, $\mathrm{f}^{\prime}=2 \mathrm{f}$ Velocity of sound $=(\mathrm{V})$ By using formula, $\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right) \mathrm{n}$ $2 \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right)$ $2\left(\mathrm{~V}-\mathrm{V}_{\mathrm{s}}\right)=\mathrm{V}$ $2 \mathrm{~V}-2 \mathrm{~V}_{\mathrm{s}}=\mathrm{V}$ $\mathrm{V}=2 \mathrm{~V}_{\mathrm{s}}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{V}}{2}$
J and K CET- 1999
WAVES
173053
Sonar fitted in a submarine operates at 30 kHz. An enemy submarine approaches it with a speed of $360 \mathrm{~km} / \mathrm{h}$. The speed of sound in sea water is $1450 \mathrm{~m} / \mathrm{s}$. The frequency of sound heard after reflection from enemy submarine is
173055
An observer standing near the sea-coast counts 48 waves per min. If the wavelength of the wave is $10 \mathrm{~m}$, the velocity of the waves will be
1 $8 \mathrm{~m} / \mathrm{s}$
2 $12 \mathrm{~m} / \mathrm{s}$
3 $16 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Total number of waves, $\lambda=48$ Wavelength of each wave $=10 \mathrm{~m}$ Time taken, $\mathrm{t}=60$ second $\operatorname{Velocity}(\mathrm{v}) =\frac{\lambda}{\mathrm{T}}$ $\mathrm{v} =\frac{48}{60} \mathrm{~Hz}$ $\mathrm{v} =\frac{4}{5}$ $\mathrm{v} =0.8 \mathrm{~Hz}$ $\therefore \quad$ Wavelength velocity $=\mathrm{v} \cdot \lambda$ $=(0.8) \times(10)$ $=8 \mathrm{~m} / \mathrm{s}$
173048
A racing car moving towards a cliff, sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If $v$ is the velocity of sound, then the velocity of the car is :
1 $v / 2$
2 $\mathrm{v} / \sqrt{2}$
3 $\mathrm{v} / 4$
4 $v / 3$
Explanation:
D Given that, $\mathrm{n}^{\prime}=2 \mathrm{n}$ If $v_{\mathrm{s}}$ be the velocity of car then frequency of reflected sound heard by driver, $\mathrm{n}^{\prime}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $2 \mathrm{n}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{car}}}{\mathrm{v}-\mathrm{v}_{\mathrm{car}}}\right)$ $2 \mathrm{v}-2 \mathrm{v}_{\mathrm{car}}=\mathrm{v}+\mathrm{v}_{\mathrm{car}}$ $2 \mathrm{v}-\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}_{\mathrm{car}}=\frac{\mathrm{v}}{3}$
Karnataka CET-2002
WAVES
173049
A bus is moving with a velocity of $5 \mathrm{~ms}^{-1}$ towards a huge wall. The driver sounds a horn of frequency $165 \mathrm{~Hz}$. If the speed of sound in air is $335 \mathrm{~m} / \mathrm{s}$, the number of beats heard per second by a passenger on the bus will be :
1 3
2 4
3 6
4 5
Explanation:
D Given that, $\mathrm{v}_{\mathrm{s}}=5 \mathrm{~m} / \mathrm{s}, \mathrm{v}=335 \mathrm{~m} / \mathrm{s}$ Frequency $(\mathrm{n})=165 \mathrm{~Hz}$ As the observer is moving with the source then apparent frequency of reflected wave is $n^{\prime}=n\left(\frac{v+v_{s}}{v-v_{s}}\right)$ $\mathrm{n}^{\prime}=165 \times\left(\frac{335+5}{335-5}\right)$ $\mathrm{n}^{\prime}=170$ $\therefore$ No. of beats $=\mathrm{n}^{\prime}-\mathrm{n}=170-165=5$
Karnataka CET-2001
WAVES
173052
What should be the velocity of a sound source moving towards a stationary observer so that the apparent frequency is double the actual frequency. Velocity of sound is $\mathrm{V}$.
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $\mathrm{V} / 2$
4 $\mathrm{V} / 4$
Explanation:
C Given that, The apparent frequency is double the actual frequency, $\mathrm{f}^{\prime}=2 \mathrm{f}$ Velocity of sound $=(\mathrm{V})$ By using formula, $\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right) \mathrm{n}$ $2 \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right)$ $2\left(\mathrm{~V}-\mathrm{V}_{\mathrm{s}}\right)=\mathrm{V}$ $2 \mathrm{~V}-2 \mathrm{~V}_{\mathrm{s}}=\mathrm{V}$ $\mathrm{V}=2 \mathrm{~V}_{\mathrm{s}}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{V}}{2}$
J and K CET- 1999
WAVES
173053
Sonar fitted in a submarine operates at 30 kHz. An enemy submarine approaches it with a speed of $360 \mathrm{~km} / \mathrm{h}$. The speed of sound in sea water is $1450 \mathrm{~m} / \mathrm{s}$. The frequency of sound heard after reflection from enemy submarine is
173055
An observer standing near the sea-coast counts 48 waves per min. If the wavelength of the wave is $10 \mathrm{~m}$, the velocity of the waves will be
1 $8 \mathrm{~m} / \mathrm{s}$
2 $12 \mathrm{~m} / \mathrm{s}$
3 $16 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Total number of waves, $\lambda=48$ Wavelength of each wave $=10 \mathrm{~m}$ Time taken, $\mathrm{t}=60$ second $\operatorname{Velocity}(\mathrm{v}) =\frac{\lambda}{\mathrm{T}}$ $\mathrm{v} =\frac{48}{60} \mathrm{~Hz}$ $\mathrm{v} =\frac{4}{5}$ $\mathrm{v} =0.8 \mathrm{~Hz}$ $\therefore \quad$ Wavelength velocity $=\mathrm{v} \cdot \lambda$ $=(0.8) \times(10)$ $=8 \mathrm{~m} / \mathrm{s}$
173048
A racing car moving towards a cliff, sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If $v$ is the velocity of sound, then the velocity of the car is :
1 $v / 2$
2 $\mathrm{v} / \sqrt{2}$
3 $\mathrm{v} / 4$
4 $v / 3$
Explanation:
D Given that, $\mathrm{n}^{\prime}=2 \mathrm{n}$ If $v_{\mathrm{s}}$ be the velocity of car then frequency of reflected sound heard by driver, $\mathrm{n}^{\prime}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $2 \mathrm{n}=\mathrm{n}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{car}}}{\mathrm{v}-\mathrm{v}_{\mathrm{car}}}\right)$ $2 \mathrm{v}-2 \mathrm{v}_{\mathrm{car}}=\mathrm{v}+\mathrm{v}_{\mathrm{car}}$ $2 \mathrm{v}-\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}=3 \mathrm{v}_{\mathrm{car}}$ $\mathrm{v}_{\mathrm{car}}=\frac{\mathrm{v}}{3}$
Karnataka CET-2002
WAVES
173049
A bus is moving with a velocity of $5 \mathrm{~ms}^{-1}$ towards a huge wall. The driver sounds a horn of frequency $165 \mathrm{~Hz}$. If the speed of sound in air is $335 \mathrm{~m} / \mathrm{s}$, the number of beats heard per second by a passenger on the bus will be :
1 3
2 4
3 6
4 5
Explanation:
D Given that, $\mathrm{v}_{\mathrm{s}}=5 \mathrm{~m} / \mathrm{s}, \mathrm{v}=335 \mathrm{~m} / \mathrm{s}$ Frequency $(\mathrm{n})=165 \mathrm{~Hz}$ As the observer is moving with the source then apparent frequency of reflected wave is $n^{\prime}=n\left(\frac{v+v_{s}}{v-v_{s}}\right)$ $\mathrm{n}^{\prime}=165 \times\left(\frac{335+5}{335-5}\right)$ $\mathrm{n}^{\prime}=170$ $\therefore$ No. of beats $=\mathrm{n}^{\prime}-\mathrm{n}=170-165=5$
Karnataka CET-2001
WAVES
173052
What should be the velocity of a sound source moving towards a stationary observer so that the apparent frequency is double the actual frequency. Velocity of sound is $\mathrm{V}$.
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $\mathrm{V} / 2$
4 $\mathrm{V} / 4$
Explanation:
C Given that, The apparent frequency is double the actual frequency, $\mathrm{f}^{\prime}=2 \mathrm{f}$ Velocity of sound $=(\mathrm{V})$ By using formula, $\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right) \mathrm{n}$ $2 \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right)$ $2\left(\mathrm{~V}-\mathrm{V}_{\mathrm{s}}\right)=\mathrm{V}$ $2 \mathrm{~V}-2 \mathrm{~V}_{\mathrm{s}}=\mathrm{V}$ $\mathrm{V}=2 \mathrm{~V}_{\mathrm{s}}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{V}}{2}$
J and K CET- 1999
WAVES
173053
Sonar fitted in a submarine operates at 30 kHz. An enemy submarine approaches it with a speed of $360 \mathrm{~km} / \mathrm{h}$. The speed of sound in sea water is $1450 \mathrm{~m} / \mathrm{s}$. The frequency of sound heard after reflection from enemy submarine is
173055
An observer standing near the sea-coast counts 48 waves per min. If the wavelength of the wave is $10 \mathrm{~m}$, the velocity of the waves will be
1 $8 \mathrm{~m} / \mathrm{s}$
2 $12 \mathrm{~m} / \mathrm{s}$
3 $16 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Total number of waves, $\lambda=48$ Wavelength of each wave $=10 \mathrm{~m}$ Time taken, $\mathrm{t}=60$ second $\operatorname{Velocity}(\mathrm{v}) =\frac{\lambda}{\mathrm{T}}$ $\mathrm{v} =\frac{48}{60} \mathrm{~Hz}$ $\mathrm{v} =\frac{4}{5}$ $\mathrm{v} =0.8 \mathrm{~Hz}$ $\therefore \quad$ Wavelength velocity $=\mathrm{v} \cdot \lambda$ $=(0.8) \times(10)$ $=8 \mathrm{~m} / \mathrm{s}$