172744
The number of beats heard per second, by three sound sources of equal intensities and frequencies of 300,301 and $302 \mathrm{~Hz}$, is
1 4
2 3
3 2
4 1
Explanation:
D Given, $\mathrm{f}_{1}=300 \mathrm{~Hz}, \mathrm{f}_{2}=301 \mathrm{~Hz}, \mathrm{f}_{3}=302 \mathrm{~Hz}$ Beat formula, $f_{b}=\left|f_{2}-f_{1}\right|$ When first and second source are sounded together then beat frequency, $\mathrm{f}_{\mathrm{b}}=(301-300)=1$ When second and third source is sounded together then beat frequency $f_{b}=(302-301)=1$
UPSEE 2019
WAVES
172746
Two identical wires with equal tension $T$ have fundamental frequency $900 \mathrm{~Hz}$. The tension in one of the wires is increased by $\Delta \mathrm{T}$ in order to increase the beats by 9 beats/sec. The magnitude of $\frac{\Delta T}{T}$ is
1 0.01
2 0.02
3 0.03
4 0.04
Explanation:
B Given, $\Delta \mathrm{f}=9, \mathrm{f}=900 \mathrm{~Hz}$ Frequency $(\mathrm{f})=\frac{\mathrm{V}}{2 \mathrm{~L}}$ Where, $v=\sqrt{\frac{T}{\mu}}$ $\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Thus fractional change in frequency $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{9}{900}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=2 \times 0.01$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=0.02$
TS EAMCET 08.05.2019
WAVES
172747
A stone is dropped into a lake by a person form a $500 \mathrm{~m}$ high tower. He would hear the sound after approximately
1 $10 \mathrm{sec}$
2 $11.5 \mathrm{sec}$
3 $14 \mathrm{sec}$
4 $21 \mathrm{sec}$
Explanation:
B Given that, Height of the tower $(\mathrm{s})=500 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Time taken by stone $(\mathrm{t})=\sqrt{\frac{2 \mathrm{~s}}{\mathrm{~g}}}$ $t=\sqrt{\frac{2 \times 500}{10}}$ \(t=\sqrt{100}\) \(\mathrm{t}=10 \mathrm{~s}\) \(\text { Time taken by sound }\left(t^{\prime}\right)=\frac{\mathrm{s}}{\mathrm{v}}\) \(\mathrm{t}^{\prime}=\frac{500}{330}\) \(\mathrm{t}^{\prime}=1.51\) \(\mathrm{t}^{\prime} = 1.5 \mathrm{sec}\) \(\therefore \quad \text { Total time } \quad(\mathrm{T})=\mathrm{t}+\mathrm{t}^{\prime}\) \(\mathrm{T}=10+1.5=11.5 \mathrm{sec}\)
SRMJEE-2019
WAVES
172748
A tuning fork arrangement produces 4 beats/ second with one fork of frequency $288 \mathrm{~Hz}$. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of this unknown fork is ..... Hz.
1 286
2 292
3 294
4 288
Explanation:
B Let the frequency of the unknown fork be $f$. As it produces, 4 beats with $288 \mathrm{~Hz}$. $|\mathrm{f}-288|=4$ $\mathrm{f}-288= \pm 4$ $\mathrm{f}=292 \text { or } 284 \mathrm{~Hz}$ Waxing of fork decreases its frequency. Hence, beat frequency also drops, the frequency of unknown fork has to be, $\mathrm{f}=292 \mathrm{~Hz}$.
172744
The number of beats heard per second, by three sound sources of equal intensities and frequencies of 300,301 and $302 \mathrm{~Hz}$, is
1 4
2 3
3 2
4 1
Explanation:
D Given, $\mathrm{f}_{1}=300 \mathrm{~Hz}, \mathrm{f}_{2}=301 \mathrm{~Hz}, \mathrm{f}_{3}=302 \mathrm{~Hz}$ Beat formula, $f_{b}=\left|f_{2}-f_{1}\right|$ When first and second source are sounded together then beat frequency, $\mathrm{f}_{\mathrm{b}}=(301-300)=1$ When second and third source is sounded together then beat frequency $f_{b}=(302-301)=1$
UPSEE 2019
WAVES
172746
Two identical wires with equal tension $T$ have fundamental frequency $900 \mathrm{~Hz}$. The tension in one of the wires is increased by $\Delta \mathrm{T}$ in order to increase the beats by 9 beats/sec. The magnitude of $\frac{\Delta T}{T}$ is
1 0.01
2 0.02
3 0.03
4 0.04
Explanation:
B Given, $\Delta \mathrm{f}=9, \mathrm{f}=900 \mathrm{~Hz}$ Frequency $(\mathrm{f})=\frac{\mathrm{V}}{2 \mathrm{~L}}$ Where, $v=\sqrt{\frac{T}{\mu}}$ $\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Thus fractional change in frequency $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{9}{900}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=2 \times 0.01$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=0.02$
TS EAMCET 08.05.2019
WAVES
172747
A stone is dropped into a lake by a person form a $500 \mathrm{~m}$ high tower. He would hear the sound after approximately
1 $10 \mathrm{sec}$
2 $11.5 \mathrm{sec}$
3 $14 \mathrm{sec}$
4 $21 \mathrm{sec}$
Explanation:
B Given that, Height of the tower $(\mathrm{s})=500 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Time taken by stone $(\mathrm{t})=\sqrt{\frac{2 \mathrm{~s}}{\mathrm{~g}}}$ $t=\sqrt{\frac{2 \times 500}{10}}$ \(t=\sqrt{100}\) \(\mathrm{t}=10 \mathrm{~s}\) \(\text { Time taken by sound }\left(t^{\prime}\right)=\frac{\mathrm{s}}{\mathrm{v}}\) \(\mathrm{t}^{\prime}=\frac{500}{330}\) \(\mathrm{t}^{\prime}=1.51\) \(\mathrm{t}^{\prime} = 1.5 \mathrm{sec}\) \(\therefore \quad \text { Total time } \quad(\mathrm{T})=\mathrm{t}+\mathrm{t}^{\prime}\) \(\mathrm{T}=10+1.5=11.5 \mathrm{sec}\)
SRMJEE-2019
WAVES
172748
A tuning fork arrangement produces 4 beats/ second with one fork of frequency $288 \mathrm{~Hz}$. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of this unknown fork is ..... Hz.
1 286
2 292
3 294
4 288
Explanation:
B Let the frequency of the unknown fork be $f$. As it produces, 4 beats with $288 \mathrm{~Hz}$. $|\mathrm{f}-288|=4$ $\mathrm{f}-288= \pm 4$ $\mathrm{f}=292 \text { or } 284 \mathrm{~Hz}$ Waxing of fork decreases its frequency. Hence, beat frequency also drops, the frequency of unknown fork has to be, $\mathrm{f}=292 \mathrm{~Hz}$.
172744
The number of beats heard per second, by three sound sources of equal intensities and frequencies of 300,301 and $302 \mathrm{~Hz}$, is
1 4
2 3
3 2
4 1
Explanation:
D Given, $\mathrm{f}_{1}=300 \mathrm{~Hz}, \mathrm{f}_{2}=301 \mathrm{~Hz}, \mathrm{f}_{3}=302 \mathrm{~Hz}$ Beat formula, $f_{b}=\left|f_{2}-f_{1}\right|$ When first and second source are sounded together then beat frequency, $\mathrm{f}_{\mathrm{b}}=(301-300)=1$ When second and third source is sounded together then beat frequency $f_{b}=(302-301)=1$
UPSEE 2019
WAVES
172746
Two identical wires with equal tension $T$ have fundamental frequency $900 \mathrm{~Hz}$. The tension in one of the wires is increased by $\Delta \mathrm{T}$ in order to increase the beats by 9 beats/sec. The magnitude of $\frac{\Delta T}{T}$ is
1 0.01
2 0.02
3 0.03
4 0.04
Explanation:
B Given, $\Delta \mathrm{f}=9, \mathrm{f}=900 \mathrm{~Hz}$ Frequency $(\mathrm{f})=\frac{\mathrm{V}}{2 \mathrm{~L}}$ Where, $v=\sqrt{\frac{T}{\mu}}$ $\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Thus fractional change in frequency $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{9}{900}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=2 \times 0.01$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=0.02$
TS EAMCET 08.05.2019
WAVES
172747
A stone is dropped into a lake by a person form a $500 \mathrm{~m}$ high tower. He would hear the sound after approximately
1 $10 \mathrm{sec}$
2 $11.5 \mathrm{sec}$
3 $14 \mathrm{sec}$
4 $21 \mathrm{sec}$
Explanation:
B Given that, Height of the tower $(\mathrm{s})=500 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Time taken by stone $(\mathrm{t})=\sqrt{\frac{2 \mathrm{~s}}{\mathrm{~g}}}$ $t=\sqrt{\frac{2 \times 500}{10}}$ \(t=\sqrt{100}\) \(\mathrm{t}=10 \mathrm{~s}\) \(\text { Time taken by sound }\left(t^{\prime}\right)=\frac{\mathrm{s}}{\mathrm{v}}\) \(\mathrm{t}^{\prime}=\frac{500}{330}\) \(\mathrm{t}^{\prime}=1.51\) \(\mathrm{t}^{\prime} = 1.5 \mathrm{sec}\) \(\therefore \quad \text { Total time } \quad(\mathrm{T})=\mathrm{t}+\mathrm{t}^{\prime}\) \(\mathrm{T}=10+1.5=11.5 \mathrm{sec}\)
SRMJEE-2019
WAVES
172748
A tuning fork arrangement produces 4 beats/ second with one fork of frequency $288 \mathrm{~Hz}$. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of this unknown fork is ..... Hz.
1 286
2 292
3 294
4 288
Explanation:
B Let the frequency of the unknown fork be $f$. As it produces, 4 beats with $288 \mathrm{~Hz}$. $|\mathrm{f}-288|=4$ $\mathrm{f}-288= \pm 4$ $\mathrm{f}=292 \text { or } 284 \mathrm{~Hz}$ Waxing of fork decreases its frequency. Hence, beat frequency also drops, the frequency of unknown fork has to be, $\mathrm{f}=292 \mathrm{~Hz}$.
172744
The number of beats heard per second, by three sound sources of equal intensities and frequencies of 300,301 and $302 \mathrm{~Hz}$, is
1 4
2 3
3 2
4 1
Explanation:
D Given, $\mathrm{f}_{1}=300 \mathrm{~Hz}, \mathrm{f}_{2}=301 \mathrm{~Hz}, \mathrm{f}_{3}=302 \mathrm{~Hz}$ Beat formula, $f_{b}=\left|f_{2}-f_{1}\right|$ When first and second source are sounded together then beat frequency, $\mathrm{f}_{\mathrm{b}}=(301-300)=1$ When second and third source is sounded together then beat frequency $f_{b}=(302-301)=1$
UPSEE 2019
WAVES
172746
Two identical wires with equal tension $T$ have fundamental frequency $900 \mathrm{~Hz}$. The tension in one of the wires is increased by $\Delta \mathrm{T}$ in order to increase the beats by 9 beats/sec. The magnitude of $\frac{\Delta T}{T}$ is
1 0.01
2 0.02
3 0.03
4 0.04
Explanation:
B Given, $\Delta \mathrm{f}=9, \mathrm{f}=900 \mathrm{~Hz}$ Frequency $(\mathrm{f})=\frac{\mathrm{V}}{2 \mathrm{~L}}$ Where, $v=\sqrt{\frac{T}{\mu}}$ $\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Thus fractional change in frequency $\frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{9}{900}=\frac{1}{2} \frac{\Delta \mathrm{T}}{\mathrm{T}}$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=2 \times 0.01$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=0.02$
TS EAMCET 08.05.2019
WAVES
172747
A stone is dropped into a lake by a person form a $500 \mathrm{~m}$ high tower. He would hear the sound after approximately
1 $10 \mathrm{sec}$
2 $11.5 \mathrm{sec}$
3 $14 \mathrm{sec}$
4 $21 \mathrm{sec}$
Explanation:
B Given that, Height of the tower $(\mathrm{s})=500 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Time taken by stone $(\mathrm{t})=\sqrt{\frac{2 \mathrm{~s}}{\mathrm{~g}}}$ $t=\sqrt{\frac{2 \times 500}{10}}$ \(t=\sqrt{100}\) \(\mathrm{t}=10 \mathrm{~s}\) \(\text { Time taken by sound }\left(t^{\prime}\right)=\frac{\mathrm{s}}{\mathrm{v}}\) \(\mathrm{t}^{\prime}=\frac{500}{330}\) \(\mathrm{t}^{\prime}=1.51\) \(\mathrm{t}^{\prime} = 1.5 \mathrm{sec}\) \(\therefore \quad \text { Total time } \quad(\mathrm{T})=\mathrm{t}+\mathrm{t}^{\prime}\) \(\mathrm{T}=10+1.5=11.5 \mathrm{sec}\)
SRMJEE-2019
WAVES
172748
A tuning fork arrangement produces 4 beats/ second with one fork of frequency $288 \mathrm{~Hz}$. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of this unknown fork is ..... Hz.
1 286
2 292
3 294
4 288
Explanation:
B Let the frequency of the unknown fork be $f$. As it produces, 4 beats with $288 \mathrm{~Hz}$. $|\mathrm{f}-288|=4$ $\mathrm{f}-288= \pm 4$ $\mathrm{f}=292 \text { or } 284 \mathrm{~Hz}$ Waxing of fork decreases its frequency. Hence, beat frequency also drops, the frequency of unknown fork has to be, $\mathrm{f}=292 \mathrm{~Hz}$.
