C Given, Two tuning forks of frequencies $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ which produces $n$ beat/sec then - We know that, Beat frequency $=$ number of beat $/ \mathrm{sec}$ $\mathrm{n}=\mathrm{n}_{2}-\mathrm{n}_{1} \text { or } \mathrm{n}_{1}-\mathrm{n}_{2}$ $\therefore \quad \mathrm{n}_{1}=\mathrm{n}_{2} \pm \mathrm{n}$
WB JEE 2008
WAVES
172783
There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next. The first one is octave of the last. What is the frequency of $18^{\text {th }}$ tuning fork?
1 $100 \mathrm{~Hz}$
2 $99 \mathrm{~Hz}$
3 $96 \mathrm{~Hz}$
4 $103 \mathrm{~Hz}$
Explanation:
B Let the frequency of first tuning fork is $\mathbf{f}$. The frequencies of other tuning fork are $(f-3),(f-2 \times 3), \ldots \ldots-\ldots(f-17 \times 3), \ldots .,(f-25 \times 3)$. According to question, $f=2(f-25 \times 3)=2 f-25 \times 6$ or $f=25 \times 6=150 \mathrm{~Hz}$ The frequency of the $18^{\text {th }}$ turning fork $f=150-51$ $f=99 \mathrm{~Hz}$
JIPMER-2005
WAVES
172784
The ratio of the velocity of sound in oxygen to that in hydrogen at same temperature and pressure is approximately :
1 $16: 1$
2 $1: 16$
3 $4: 1$
4 $1: 4$
Explanation:
D Oxygen and hydrogen both are diatomic gases. According to Laplace, the speed of sound in gas is given by $\mathrm{v}=\frac{\sqrt{\gamma \mathrm{RT}}}{\mathrm{M}}$ Where, $\mathrm{v}=$ speed of sound $\gamma=\text { adiabatic constant for gas }$ $\mathrm{R}=\text { Universal gas constant }$ $\mathrm{T}=\text { temperature of gas }$ $\mathrm{M}=\text { molecule weight of gas }$ The given parameters in above question are - Hydrogen $\quad \Rightarrow \mathrm{M}_{\mathrm{H}_{2}}=2$ $\text { Oxygen } \Rightarrow \mathrm{M}_{\mathrm{O}_{2}}=32$ $\therefore \quad \mathrm{v} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$ $\therefore \quad \mathrm{v}_{\mathrm{O}_{2}}: \mathrm{v}_{\mathrm{H}_{2}}=1: 4$
UPSEE - 2005
WAVES
172785
Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is
1 1
2 2
3 4
4 8
Explanation:
C Given that, The two wavelengths are $\left(\lambda_{1}\right)=99 \mathrm{~cm}=0.99 \mathrm{~m}$, and $\left(\lambda_{2}\right)$ $=100 \mathrm{~cm}=1 \mathrm{~m}$, velocity $(\mathrm{v})=396 \mathrm{~m} / \mathrm{s}$ Now, frequency $\mathrm{f}_{1}=\frac{\mathrm{v}}{\lambda_{1}}=\frac{396}{99 \times 10^{-2}}=400 \mathrm{~Hz} .$ $\mathrm{f}_{2}=\frac{\mathrm{v}}{\lambda_{2}}=\frac{396}{1}=396 \mathrm{~Hz}$ Number of beats produced by the waves $=\mathrm{f}_{1} \sim \mathrm{f}_{2}$ $=400-396$ $=4 \text { beats }$
C Given, Two tuning forks of frequencies $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ which produces $n$ beat/sec then - We know that, Beat frequency $=$ number of beat $/ \mathrm{sec}$ $\mathrm{n}=\mathrm{n}_{2}-\mathrm{n}_{1} \text { or } \mathrm{n}_{1}-\mathrm{n}_{2}$ $\therefore \quad \mathrm{n}_{1}=\mathrm{n}_{2} \pm \mathrm{n}$
WB JEE 2008
WAVES
172783
There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next. The first one is octave of the last. What is the frequency of $18^{\text {th }}$ tuning fork?
1 $100 \mathrm{~Hz}$
2 $99 \mathrm{~Hz}$
3 $96 \mathrm{~Hz}$
4 $103 \mathrm{~Hz}$
Explanation:
B Let the frequency of first tuning fork is $\mathbf{f}$. The frequencies of other tuning fork are $(f-3),(f-2 \times 3), \ldots \ldots-\ldots(f-17 \times 3), \ldots .,(f-25 \times 3)$. According to question, $f=2(f-25 \times 3)=2 f-25 \times 6$ or $f=25 \times 6=150 \mathrm{~Hz}$ The frequency of the $18^{\text {th }}$ turning fork $f=150-51$ $f=99 \mathrm{~Hz}$
JIPMER-2005
WAVES
172784
The ratio of the velocity of sound in oxygen to that in hydrogen at same temperature and pressure is approximately :
1 $16: 1$
2 $1: 16$
3 $4: 1$
4 $1: 4$
Explanation:
D Oxygen and hydrogen both are diatomic gases. According to Laplace, the speed of sound in gas is given by $\mathrm{v}=\frac{\sqrt{\gamma \mathrm{RT}}}{\mathrm{M}}$ Where, $\mathrm{v}=$ speed of sound $\gamma=\text { adiabatic constant for gas }$ $\mathrm{R}=\text { Universal gas constant }$ $\mathrm{T}=\text { temperature of gas }$ $\mathrm{M}=\text { molecule weight of gas }$ The given parameters in above question are - Hydrogen $\quad \Rightarrow \mathrm{M}_{\mathrm{H}_{2}}=2$ $\text { Oxygen } \Rightarrow \mathrm{M}_{\mathrm{O}_{2}}=32$ $\therefore \quad \mathrm{v} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$ $\therefore \quad \mathrm{v}_{\mathrm{O}_{2}}: \mathrm{v}_{\mathrm{H}_{2}}=1: 4$
UPSEE - 2005
WAVES
172785
Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is
1 1
2 2
3 4
4 8
Explanation:
C Given that, The two wavelengths are $\left(\lambda_{1}\right)=99 \mathrm{~cm}=0.99 \mathrm{~m}$, and $\left(\lambda_{2}\right)$ $=100 \mathrm{~cm}=1 \mathrm{~m}$, velocity $(\mathrm{v})=396 \mathrm{~m} / \mathrm{s}$ Now, frequency $\mathrm{f}_{1}=\frac{\mathrm{v}}{\lambda_{1}}=\frac{396}{99 \times 10^{-2}}=400 \mathrm{~Hz} .$ $\mathrm{f}_{2}=\frac{\mathrm{v}}{\lambda_{2}}=\frac{396}{1}=396 \mathrm{~Hz}$ Number of beats produced by the waves $=\mathrm{f}_{1} \sim \mathrm{f}_{2}$ $=400-396$ $=4 \text { beats }$
C Given, Two tuning forks of frequencies $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ which produces $n$ beat/sec then - We know that, Beat frequency $=$ number of beat $/ \mathrm{sec}$ $\mathrm{n}=\mathrm{n}_{2}-\mathrm{n}_{1} \text { or } \mathrm{n}_{1}-\mathrm{n}_{2}$ $\therefore \quad \mathrm{n}_{1}=\mathrm{n}_{2} \pm \mathrm{n}$
WB JEE 2008
WAVES
172783
There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next. The first one is octave of the last. What is the frequency of $18^{\text {th }}$ tuning fork?
1 $100 \mathrm{~Hz}$
2 $99 \mathrm{~Hz}$
3 $96 \mathrm{~Hz}$
4 $103 \mathrm{~Hz}$
Explanation:
B Let the frequency of first tuning fork is $\mathbf{f}$. The frequencies of other tuning fork are $(f-3),(f-2 \times 3), \ldots \ldots-\ldots(f-17 \times 3), \ldots .,(f-25 \times 3)$. According to question, $f=2(f-25 \times 3)=2 f-25 \times 6$ or $f=25 \times 6=150 \mathrm{~Hz}$ The frequency of the $18^{\text {th }}$ turning fork $f=150-51$ $f=99 \mathrm{~Hz}$
JIPMER-2005
WAVES
172784
The ratio of the velocity of sound in oxygen to that in hydrogen at same temperature and pressure is approximately :
1 $16: 1$
2 $1: 16$
3 $4: 1$
4 $1: 4$
Explanation:
D Oxygen and hydrogen both are diatomic gases. According to Laplace, the speed of sound in gas is given by $\mathrm{v}=\frac{\sqrt{\gamma \mathrm{RT}}}{\mathrm{M}}$ Where, $\mathrm{v}=$ speed of sound $\gamma=\text { adiabatic constant for gas }$ $\mathrm{R}=\text { Universal gas constant }$ $\mathrm{T}=\text { temperature of gas }$ $\mathrm{M}=\text { molecule weight of gas }$ The given parameters in above question are - Hydrogen $\quad \Rightarrow \mathrm{M}_{\mathrm{H}_{2}}=2$ $\text { Oxygen } \Rightarrow \mathrm{M}_{\mathrm{O}_{2}}=32$ $\therefore \quad \mathrm{v} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$ $\therefore \quad \mathrm{v}_{\mathrm{O}_{2}}: \mathrm{v}_{\mathrm{H}_{2}}=1: 4$
UPSEE - 2005
WAVES
172785
Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is
1 1
2 2
3 4
4 8
Explanation:
C Given that, The two wavelengths are $\left(\lambda_{1}\right)=99 \mathrm{~cm}=0.99 \mathrm{~m}$, and $\left(\lambda_{2}\right)$ $=100 \mathrm{~cm}=1 \mathrm{~m}$, velocity $(\mathrm{v})=396 \mathrm{~m} / \mathrm{s}$ Now, frequency $\mathrm{f}_{1}=\frac{\mathrm{v}}{\lambda_{1}}=\frac{396}{99 \times 10^{-2}}=400 \mathrm{~Hz} .$ $\mathrm{f}_{2}=\frac{\mathrm{v}}{\lambda_{2}}=\frac{396}{1}=396 \mathrm{~Hz}$ Number of beats produced by the waves $=\mathrm{f}_{1} \sim \mathrm{f}_{2}$ $=400-396$ $=4 \text { beats }$
C Given, Two tuning forks of frequencies $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ which produces $n$ beat/sec then - We know that, Beat frequency $=$ number of beat $/ \mathrm{sec}$ $\mathrm{n}=\mathrm{n}_{2}-\mathrm{n}_{1} \text { or } \mathrm{n}_{1}-\mathrm{n}_{2}$ $\therefore \quad \mathrm{n}_{1}=\mathrm{n}_{2} \pm \mathrm{n}$
WB JEE 2008
WAVES
172783
There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next. The first one is octave of the last. What is the frequency of $18^{\text {th }}$ tuning fork?
1 $100 \mathrm{~Hz}$
2 $99 \mathrm{~Hz}$
3 $96 \mathrm{~Hz}$
4 $103 \mathrm{~Hz}$
Explanation:
B Let the frequency of first tuning fork is $\mathbf{f}$. The frequencies of other tuning fork are $(f-3),(f-2 \times 3), \ldots \ldots-\ldots(f-17 \times 3), \ldots .,(f-25 \times 3)$. According to question, $f=2(f-25 \times 3)=2 f-25 \times 6$ or $f=25 \times 6=150 \mathrm{~Hz}$ The frequency of the $18^{\text {th }}$ turning fork $f=150-51$ $f=99 \mathrm{~Hz}$
JIPMER-2005
WAVES
172784
The ratio of the velocity of sound in oxygen to that in hydrogen at same temperature and pressure is approximately :
1 $16: 1$
2 $1: 16$
3 $4: 1$
4 $1: 4$
Explanation:
D Oxygen and hydrogen both are diatomic gases. According to Laplace, the speed of sound in gas is given by $\mathrm{v}=\frac{\sqrt{\gamma \mathrm{RT}}}{\mathrm{M}}$ Where, $\mathrm{v}=$ speed of sound $\gamma=\text { adiabatic constant for gas }$ $\mathrm{R}=\text { Universal gas constant }$ $\mathrm{T}=\text { temperature of gas }$ $\mathrm{M}=\text { molecule weight of gas }$ The given parameters in above question are - Hydrogen $\quad \Rightarrow \mathrm{M}_{\mathrm{H}_{2}}=2$ $\text { Oxygen } \Rightarrow \mathrm{M}_{\mathrm{O}_{2}}=32$ $\therefore \quad \mathrm{v} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\frac{\mathrm{v}_{\mathrm{O}_{2}}}{\mathrm{v}_{\mathrm{H}_{2}}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$ $\therefore \quad \mathrm{v}_{\mathrm{O}_{2}}: \mathrm{v}_{\mathrm{H}_{2}}=1: 4$
UPSEE - 2005
WAVES
172785
Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is
1 1
2 2
3 4
4 8
Explanation:
C Given that, The two wavelengths are $\left(\lambda_{1}\right)=99 \mathrm{~cm}=0.99 \mathrm{~m}$, and $\left(\lambda_{2}\right)$ $=100 \mathrm{~cm}=1 \mathrm{~m}$, velocity $(\mathrm{v})=396 \mathrm{~m} / \mathrm{s}$ Now, frequency $\mathrm{f}_{1}=\frac{\mathrm{v}}{\lambda_{1}}=\frac{396}{99 \times 10^{-2}}=400 \mathrm{~Hz} .$ $\mathrm{f}_{2}=\frac{\mathrm{v}}{\lambda_{2}}=\frac{396}{1}=396 \mathrm{~Hz}$ Number of beats produced by the waves $=\mathrm{f}_{1} \sim \mathrm{f}_{2}$ $=400-396$ $=4 \text { beats }$
