172764
In a gas, two waves of wavelengths $1 \mathrm{~m}$ and $1.01 \mathrm{~m}$ are superposed and produce 10 beats in 3 s. The velocity of sound in the medium is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $336.7 \mathrm{~m} / \mathrm{s}$
3 $360.2 \mathrm{~m} / \mathrm{s}$
4 $270 \mathrm{~m} / \mathrm{s}$
Explanation:
B We know that wavelength. $\mathrm{v}=\mathrm{f} \lambda$ Where, $\mathrm{v}$ is the velocity and $\lambda$ is $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ Number of beats $=$ difference in frequencies $\therefore \quad \mathrm{f}_{1} -\mathrm{f}_{2}=\mathrm{v}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{1}{1}-\frac{1}{1.01}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{0.01}{1.01}\right)$ $\mathrm{v} =\left(\frac{10 \times 1.01}{3 \times 0.01}\right)$ $\mathrm{v} =336.7 \mathrm{~m} / \mathrm{s}$
Manipal UGET-2017
WAVES
172766
The equations of two sound waves are given by, $y_{1}=3 \sin (100 \pi t)$ and $y_{2}=4 \sin (150 \pi t)$. The ratio of intensities of sound produced in the medium is
1 $1: 4$
2 $2: 3$
3 $3: 4$
4 $9: 16$
Explanation:
D Given, $y_{1}=3 \sin (100 \pi t)$ $y_{2}=4 \sin (150 \pi t)$ Comparing both equations with standard equation of wave, $y=A \sin (2 \pi n t)$ We get, $A_{1}=3, n_{1}=50 \text { and } A_{2}=4, n_{2}=75$ We know that, $I =A^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{A_{1}}{A_{2}}\right)^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{3}{4}\right)^{2}$ $I_{1}: I_{2} =9: 16$
COMEDK 2017
WAVES
172768
When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 \mathrm{~Hz}$. When one of the prongs of the fork $A$ is filled and sounded with $B$, the beat frequency increases, then the frequency of the fork $A$ is :
1 $379 \mathrm{~Hz}$
2 $380 \mathrm{~Hz}$
3 $389 \mathrm{~Hz}$
4 $388 \mathrm{~Hz}$
Explanation:
D Given that, Frequency of fork $B\left(f_{B}\right)=384 \mathrm{~Hz}$ We know that beats, $b=\left|f_{A}-f_{B}\right|$ As on filling the fork $\mathrm{A}$, beat frequency increases. Thus fork A has greater frequency than fork B as on filling, frequency of fork increases. $\mathrm{b}=\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}$ $4=\mathrm{f}_{\mathrm{A}}-384$ $\mathrm{f}_{\mathrm{A}}=384+4$ $\mathrm{f}_{\mathrm{A}}=388 \mathrm{~Hz} .$
Karnataka CET-2015
WAVES
172769
Two waves each of loudness $L$ superimpose to produce beats. The maximum loudness of beats will be
1 $4 \mathrm{~L}$
2 $\mathrm{L}$
3 $2 \mathrm{~L}$
4 $5 \mathrm{~L}$
Explanation:
A Loudness L $\propto$ square of amplitude (A) $\mathrm{L} \propto \mathrm{A}^{2}$ $\mathrm{L}=\mathrm{kA}{ }^{2}$ On superposition, Resultant amplitude $=2 \mathrm{~A}$ Hence, resultant loudness, $4 \mathrm{~A}^{2}=4 \mathrm{~L}$
172764
In a gas, two waves of wavelengths $1 \mathrm{~m}$ and $1.01 \mathrm{~m}$ are superposed and produce 10 beats in 3 s. The velocity of sound in the medium is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $336.7 \mathrm{~m} / \mathrm{s}$
3 $360.2 \mathrm{~m} / \mathrm{s}$
4 $270 \mathrm{~m} / \mathrm{s}$
Explanation:
B We know that wavelength. $\mathrm{v}=\mathrm{f} \lambda$ Where, $\mathrm{v}$ is the velocity and $\lambda$ is $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ Number of beats $=$ difference in frequencies $\therefore \quad \mathrm{f}_{1} -\mathrm{f}_{2}=\mathrm{v}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{1}{1}-\frac{1}{1.01}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{0.01}{1.01}\right)$ $\mathrm{v} =\left(\frac{10 \times 1.01}{3 \times 0.01}\right)$ $\mathrm{v} =336.7 \mathrm{~m} / \mathrm{s}$
Manipal UGET-2017
WAVES
172766
The equations of two sound waves are given by, $y_{1}=3 \sin (100 \pi t)$ and $y_{2}=4 \sin (150 \pi t)$. The ratio of intensities of sound produced in the medium is
1 $1: 4$
2 $2: 3$
3 $3: 4$
4 $9: 16$
Explanation:
D Given, $y_{1}=3 \sin (100 \pi t)$ $y_{2}=4 \sin (150 \pi t)$ Comparing both equations with standard equation of wave, $y=A \sin (2 \pi n t)$ We get, $A_{1}=3, n_{1}=50 \text { and } A_{2}=4, n_{2}=75$ We know that, $I =A^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{A_{1}}{A_{2}}\right)^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{3}{4}\right)^{2}$ $I_{1}: I_{2} =9: 16$
COMEDK 2017
WAVES
172768
When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 \mathrm{~Hz}$. When one of the prongs of the fork $A$ is filled and sounded with $B$, the beat frequency increases, then the frequency of the fork $A$ is :
1 $379 \mathrm{~Hz}$
2 $380 \mathrm{~Hz}$
3 $389 \mathrm{~Hz}$
4 $388 \mathrm{~Hz}$
Explanation:
D Given that, Frequency of fork $B\left(f_{B}\right)=384 \mathrm{~Hz}$ We know that beats, $b=\left|f_{A}-f_{B}\right|$ As on filling the fork $\mathrm{A}$, beat frequency increases. Thus fork A has greater frequency than fork B as on filling, frequency of fork increases. $\mathrm{b}=\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}$ $4=\mathrm{f}_{\mathrm{A}}-384$ $\mathrm{f}_{\mathrm{A}}=384+4$ $\mathrm{f}_{\mathrm{A}}=388 \mathrm{~Hz} .$
Karnataka CET-2015
WAVES
172769
Two waves each of loudness $L$ superimpose to produce beats. The maximum loudness of beats will be
1 $4 \mathrm{~L}$
2 $\mathrm{L}$
3 $2 \mathrm{~L}$
4 $5 \mathrm{~L}$
Explanation:
A Loudness L $\propto$ square of amplitude (A) $\mathrm{L} \propto \mathrm{A}^{2}$ $\mathrm{L}=\mathrm{kA}{ }^{2}$ On superposition, Resultant amplitude $=2 \mathrm{~A}$ Hence, resultant loudness, $4 \mathrm{~A}^{2}=4 \mathrm{~L}$
172764
In a gas, two waves of wavelengths $1 \mathrm{~m}$ and $1.01 \mathrm{~m}$ are superposed and produce 10 beats in 3 s. The velocity of sound in the medium is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $336.7 \mathrm{~m} / \mathrm{s}$
3 $360.2 \mathrm{~m} / \mathrm{s}$
4 $270 \mathrm{~m} / \mathrm{s}$
Explanation:
B We know that wavelength. $\mathrm{v}=\mathrm{f} \lambda$ Where, $\mathrm{v}$ is the velocity and $\lambda$ is $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ Number of beats $=$ difference in frequencies $\therefore \quad \mathrm{f}_{1} -\mathrm{f}_{2}=\mathrm{v}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{1}{1}-\frac{1}{1.01}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{0.01}{1.01}\right)$ $\mathrm{v} =\left(\frac{10 \times 1.01}{3 \times 0.01}\right)$ $\mathrm{v} =336.7 \mathrm{~m} / \mathrm{s}$
Manipal UGET-2017
WAVES
172766
The equations of two sound waves are given by, $y_{1}=3 \sin (100 \pi t)$ and $y_{2}=4 \sin (150 \pi t)$. The ratio of intensities of sound produced in the medium is
1 $1: 4$
2 $2: 3$
3 $3: 4$
4 $9: 16$
Explanation:
D Given, $y_{1}=3 \sin (100 \pi t)$ $y_{2}=4 \sin (150 \pi t)$ Comparing both equations with standard equation of wave, $y=A \sin (2 \pi n t)$ We get, $A_{1}=3, n_{1}=50 \text { and } A_{2}=4, n_{2}=75$ We know that, $I =A^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{A_{1}}{A_{2}}\right)^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{3}{4}\right)^{2}$ $I_{1}: I_{2} =9: 16$
COMEDK 2017
WAVES
172768
When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 \mathrm{~Hz}$. When one of the prongs of the fork $A$ is filled and sounded with $B$, the beat frequency increases, then the frequency of the fork $A$ is :
1 $379 \mathrm{~Hz}$
2 $380 \mathrm{~Hz}$
3 $389 \mathrm{~Hz}$
4 $388 \mathrm{~Hz}$
Explanation:
D Given that, Frequency of fork $B\left(f_{B}\right)=384 \mathrm{~Hz}$ We know that beats, $b=\left|f_{A}-f_{B}\right|$ As on filling the fork $\mathrm{A}$, beat frequency increases. Thus fork A has greater frequency than fork B as on filling, frequency of fork increases. $\mathrm{b}=\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}$ $4=\mathrm{f}_{\mathrm{A}}-384$ $\mathrm{f}_{\mathrm{A}}=384+4$ $\mathrm{f}_{\mathrm{A}}=388 \mathrm{~Hz} .$
Karnataka CET-2015
WAVES
172769
Two waves each of loudness $L$ superimpose to produce beats. The maximum loudness of beats will be
1 $4 \mathrm{~L}$
2 $\mathrm{L}$
3 $2 \mathrm{~L}$
4 $5 \mathrm{~L}$
Explanation:
A Loudness L $\propto$ square of amplitude (A) $\mathrm{L} \propto \mathrm{A}^{2}$ $\mathrm{L}=\mathrm{kA}{ }^{2}$ On superposition, Resultant amplitude $=2 \mathrm{~A}$ Hence, resultant loudness, $4 \mathrm{~A}^{2}=4 \mathrm{~L}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172764
In a gas, two waves of wavelengths $1 \mathrm{~m}$ and $1.01 \mathrm{~m}$ are superposed and produce 10 beats in 3 s. The velocity of sound in the medium is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $336.7 \mathrm{~m} / \mathrm{s}$
3 $360.2 \mathrm{~m} / \mathrm{s}$
4 $270 \mathrm{~m} / \mathrm{s}$
Explanation:
B We know that wavelength. $\mathrm{v}=\mathrm{f} \lambda$ Where, $\mathrm{v}$ is the velocity and $\lambda$ is $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ Number of beats $=$ difference in frequencies $\therefore \quad \mathrm{f}_{1} -\mathrm{f}_{2}=\mathrm{v}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{1}{1}-\frac{1}{1.01}\right)$ $\frac{10}{3} =\mathrm{v}\left(\frac{0.01}{1.01}\right)$ $\mathrm{v} =\left(\frac{10 \times 1.01}{3 \times 0.01}\right)$ $\mathrm{v} =336.7 \mathrm{~m} / \mathrm{s}$
Manipal UGET-2017
WAVES
172766
The equations of two sound waves are given by, $y_{1}=3 \sin (100 \pi t)$ and $y_{2}=4 \sin (150 \pi t)$. The ratio of intensities of sound produced in the medium is
1 $1: 4$
2 $2: 3$
3 $3: 4$
4 $9: 16$
Explanation:
D Given, $y_{1}=3 \sin (100 \pi t)$ $y_{2}=4 \sin (150 \pi t)$ Comparing both equations with standard equation of wave, $y=A \sin (2 \pi n t)$ We get, $A_{1}=3, n_{1}=50 \text { and } A_{2}=4, n_{2}=75$ We know that, $I =A^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{A_{1}}{A_{2}}\right)^{2}$ $\frac{I_{1}}{I_{2}} =\left(\frac{3}{4}\right)^{2}$ $I_{1}: I_{2} =9: 16$
COMEDK 2017
WAVES
172768
When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 \mathrm{~Hz}$. When one of the prongs of the fork $A$ is filled and sounded with $B$, the beat frequency increases, then the frequency of the fork $A$ is :
1 $379 \mathrm{~Hz}$
2 $380 \mathrm{~Hz}$
3 $389 \mathrm{~Hz}$
4 $388 \mathrm{~Hz}$
Explanation:
D Given that, Frequency of fork $B\left(f_{B}\right)=384 \mathrm{~Hz}$ We know that beats, $b=\left|f_{A}-f_{B}\right|$ As on filling the fork $\mathrm{A}$, beat frequency increases. Thus fork A has greater frequency than fork B as on filling, frequency of fork increases. $\mathrm{b}=\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}$ $4=\mathrm{f}_{\mathrm{A}}-384$ $\mathrm{f}_{\mathrm{A}}=384+4$ $\mathrm{f}_{\mathrm{A}}=388 \mathrm{~Hz} .$
Karnataka CET-2015
WAVES
172769
Two waves each of loudness $L$ superimpose to produce beats. The maximum loudness of beats will be
1 $4 \mathrm{~L}$
2 $\mathrm{L}$
3 $2 \mathrm{~L}$
4 $5 \mathrm{~L}$
Explanation:
A Loudness L $\propto$ square of amplitude (A) $\mathrm{L} \propto \mathrm{A}^{2}$ $\mathrm{L}=\mathrm{kA}{ }^{2}$ On superposition, Resultant amplitude $=2 \mathrm{~A}$ Hence, resultant loudness, $4 \mathrm{~A}^{2}=4 \mathrm{~L}$
