173136
In a resonance tube the first resonance with a tuning fork occurs at $16 \mathrm{~cm}$ and second at 49 $\mathrm{cm}$. If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, the frequency of tuning fork is
1 $500 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $330 \mathrm{~Hz}$
4 $165 \mathrm{~Hz}$
Explanation:
A Given, First resonance occurs at, $l_{1}=16 \mathrm{~cm}=0.16 \mathrm{~m}$ Second resonance occurs at, $l_{2}=49 \mathrm{~cm}=0.49 \mathrm{~m}$ Velocity of sound, $\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\frac{\lambda}{2}=\left(l_{2}-l_{1}\right)$ From equations (i) and (ii), we get - $\frac{\mathrm{v}}{\mathrm{f}}=2\left(l_{2}-l_{1}\right)$ $\mathrm{f}=\frac{\mathrm{v}}{2\left(l_{2}-l_{1}\right)}$ $\mathrm{f}=\frac{330}{2(0.49-0.16)}=\frac{330}{0.66}$ $\mathrm{f}=500 \mathrm{~Hz}$
MHT-CET 2011
WAVES
173138
A light points fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. If eight oscillations are counted when the plate falls through $10 \mathrm{~cm}$, the frequency of the tuning fork is :
1 $280 \mathrm{~Hz}$
2 $360 \mathrm{~Hz}$
3 $56 \mathrm{~Hz}$
4 $560 \mathrm{~Hz}$
Explanation:
C Time taken by plate to fall through $10 \mathrm{~cm}$, $\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 0.1}{9.8}}$ $=\frac{1}{7}$ Thus, tuning fork Completes ' 8 ' oscillations in $1 / 7$ seconds $\text { Frequency } =\frac{\text { number of oscillations }}{\text { time }}$ $=\frac{8}{\mathrm{t}}=\frac{8}{1 / 7}$ $=8 \times 7$ $=56 \mathrm{~Hz}$
Karnataka CET-2002
WAVES
173139
If $L_{1}$ and $L_{2}$ are the lengths of the first and second resonating air columns in a resonance tube, then the wavelength of the note produced is
B If $L_{1}$ and $L_{2}$ are the lengths of first and second resonances, then we know that $\mathrm{L}_{1}+\mathrm{e}=\frac{\lambda}{4} \quad \text {............(i) (first resonance) }$ And, $\mathrm{L}_{2}+\mathrm{e}=\frac{3 \lambda}{4}$ (ii) (second resonance) From equation (i) and (ii), $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{\lambda}{2}$ $\lambda=2\left(\mathrm{~L}_{2}-\mathrm{L}_{1}\right)$
J and K CET- 2008
WAVES
173140
A segment of wire vibrates with fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg}$ weight. Then tension at which the fundamental frequency of the wire becomes $900 \mathrm{~Hz}$ is
1 $36 \mathrm{~kg}$ weight
2 $27 \mathrm{~kg}$ weight
3 $18 \mathrm{~kg}$ weight
4 $72 \mathrm{~kg}$ weight
Explanation:
A Given that, $\mathrm{f}_{1}=450 \mathrm{~Hz}$ and $\mathrm{f}_{2}=900 \mathrm{~Hz}$ $\mathrm{T}_{1}=9 \mathrm{~kg}$ and $\mathrm{T}_{2}=$ ? Fundamental frequency, $(\mathrm{f})=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ $f \propto \sqrt{T}$ $\frac{f_{1}}{f_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}$ $\therefore \quad \frac{450}{900} =\sqrt{\frac{9}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{9}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =36 \text { kg weight }$
EAMCET - 2007
WAVES
173144
A body vibrating with a certain frequency sends waves of wavelength $15 \mathrm{~cm}$ in a medium $A$ and $20 \mathrm{~cm}$ in medium $B$. If $v$ of waves in $A$ is $120 \mathrm{~ms}^{-1}$. That in $B$ will be
1 $196 \mathrm{~ms}^{-1}$
2 $160 \mathrm{~ms}^{-1}$
3 $360 \mathrm{~ms}^{-1}$
4 $260 \mathrm{~ms}^{-1}$
Explanation:
B Given that, Wave in A medium $\left(\mathrm{v}_{\mathrm{A}}\right)=120 \mathrm{~ms}^{-1}$ Wavelength in A medium $\left(\lambda_{A}\right)=15 \mathrm{~cm}$ Wavelength in $B$ medium $\left(\lambda_{B}\right)=20 \mathrm{~cm}$ Wave in $B$ medium $\left(v_{B}\right)=$ ? We know that, $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}$ $\frac{120}{\mathrm{v}_{\mathrm{B}}}=\frac{15}{20}$ $\mathrm{v}_{\mathrm{B}}=\frac{20 \times 120}{15}=160 \mathrm{~ms}^{-1}$
173136
In a resonance tube the first resonance with a tuning fork occurs at $16 \mathrm{~cm}$ and second at 49 $\mathrm{cm}$. If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, the frequency of tuning fork is
1 $500 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $330 \mathrm{~Hz}$
4 $165 \mathrm{~Hz}$
Explanation:
A Given, First resonance occurs at, $l_{1}=16 \mathrm{~cm}=0.16 \mathrm{~m}$ Second resonance occurs at, $l_{2}=49 \mathrm{~cm}=0.49 \mathrm{~m}$ Velocity of sound, $\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\frac{\lambda}{2}=\left(l_{2}-l_{1}\right)$ From equations (i) and (ii), we get - $\frac{\mathrm{v}}{\mathrm{f}}=2\left(l_{2}-l_{1}\right)$ $\mathrm{f}=\frac{\mathrm{v}}{2\left(l_{2}-l_{1}\right)}$ $\mathrm{f}=\frac{330}{2(0.49-0.16)}=\frac{330}{0.66}$ $\mathrm{f}=500 \mathrm{~Hz}$
MHT-CET 2011
WAVES
173138
A light points fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. If eight oscillations are counted when the plate falls through $10 \mathrm{~cm}$, the frequency of the tuning fork is :
1 $280 \mathrm{~Hz}$
2 $360 \mathrm{~Hz}$
3 $56 \mathrm{~Hz}$
4 $560 \mathrm{~Hz}$
Explanation:
C Time taken by plate to fall through $10 \mathrm{~cm}$, $\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 0.1}{9.8}}$ $=\frac{1}{7}$ Thus, tuning fork Completes ' 8 ' oscillations in $1 / 7$ seconds $\text { Frequency } =\frac{\text { number of oscillations }}{\text { time }}$ $=\frac{8}{\mathrm{t}}=\frac{8}{1 / 7}$ $=8 \times 7$ $=56 \mathrm{~Hz}$
Karnataka CET-2002
WAVES
173139
If $L_{1}$ and $L_{2}$ are the lengths of the first and second resonating air columns in a resonance tube, then the wavelength of the note produced is
B If $L_{1}$ and $L_{2}$ are the lengths of first and second resonances, then we know that $\mathrm{L}_{1}+\mathrm{e}=\frac{\lambda}{4} \quad \text {............(i) (first resonance) }$ And, $\mathrm{L}_{2}+\mathrm{e}=\frac{3 \lambda}{4}$ (ii) (second resonance) From equation (i) and (ii), $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{\lambda}{2}$ $\lambda=2\left(\mathrm{~L}_{2}-\mathrm{L}_{1}\right)$
J and K CET- 2008
WAVES
173140
A segment of wire vibrates with fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg}$ weight. Then tension at which the fundamental frequency of the wire becomes $900 \mathrm{~Hz}$ is
1 $36 \mathrm{~kg}$ weight
2 $27 \mathrm{~kg}$ weight
3 $18 \mathrm{~kg}$ weight
4 $72 \mathrm{~kg}$ weight
Explanation:
A Given that, $\mathrm{f}_{1}=450 \mathrm{~Hz}$ and $\mathrm{f}_{2}=900 \mathrm{~Hz}$ $\mathrm{T}_{1}=9 \mathrm{~kg}$ and $\mathrm{T}_{2}=$ ? Fundamental frequency, $(\mathrm{f})=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ $f \propto \sqrt{T}$ $\frac{f_{1}}{f_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}$ $\therefore \quad \frac{450}{900} =\sqrt{\frac{9}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{9}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =36 \text { kg weight }$
EAMCET - 2007
WAVES
173144
A body vibrating with a certain frequency sends waves of wavelength $15 \mathrm{~cm}$ in a medium $A$ and $20 \mathrm{~cm}$ in medium $B$. If $v$ of waves in $A$ is $120 \mathrm{~ms}^{-1}$. That in $B$ will be
1 $196 \mathrm{~ms}^{-1}$
2 $160 \mathrm{~ms}^{-1}$
3 $360 \mathrm{~ms}^{-1}$
4 $260 \mathrm{~ms}^{-1}$
Explanation:
B Given that, Wave in A medium $\left(\mathrm{v}_{\mathrm{A}}\right)=120 \mathrm{~ms}^{-1}$ Wavelength in A medium $\left(\lambda_{A}\right)=15 \mathrm{~cm}$ Wavelength in $B$ medium $\left(\lambda_{B}\right)=20 \mathrm{~cm}$ Wave in $B$ medium $\left(v_{B}\right)=$ ? We know that, $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}$ $\frac{120}{\mathrm{v}_{\mathrm{B}}}=\frac{15}{20}$ $\mathrm{v}_{\mathrm{B}}=\frac{20 \times 120}{15}=160 \mathrm{~ms}^{-1}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
173136
In a resonance tube the first resonance with a tuning fork occurs at $16 \mathrm{~cm}$ and second at 49 $\mathrm{cm}$. If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, the frequency of tuning fork is
1 $500 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $330 \mathrm{~Hz}$
4 $165 \mathrm{~Hz}$
Explanation:
A Given, First resonance occurs at, $l_{1}=16 \mathrm{~cm}=0.16 \mathrm{~m}$ Second resonance occurs at, $l_{2}=49 \mathrm{~cm}=0.49 \mathrm{~m}$ Velocity of sound, $\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\frac{\lambda}{2}=\left(l_{2}-l_{1}\right)$ From equations (i) and (ii), we get - $\frac{\mathrm{v}}{\mathrm{f}}=2\left(l_{2}-l_{1}\right)$ $\mathrm{f}=\frac{\mathrm{v}}{2\left(l_{2}-l_{1}\right)}$ $\mathrm{f}=\frac{330}{2(0.49-0.16)}=\frac{330}{0.66}$ $\mathrm{f}=500 \mathrm{~Hz}$
MHT-CET 2011
WAVES
173138
A light points fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. If eight oscillations are counted when the plate falls through $10 \mathrm{~cm}$, the frequency of the tuning fork is :
1 $280 \mathrm{~Hz}$
2 $360 \mathrm{~Hz}$
3 $56 \mathrm{~Hz}$
4 $560 \mathrm{~Hz}$
Explanation:
C Time taken by plate to fall through $10 \mathrm{~cm}$, $\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 0.1}{9.8}}$ $=\frac{1}{7}$ Thus, tuning fork Completes ' 8 ' oscillations in $1 / 7$ seconds $\text { Frequency } =\frac{\text { number of oscillations }}{\text { time }}$ $=\frac{8}{\mathrm{t}}=\frac{8}{1 / 7}$ $=8 \times 7$ $=56 \mathrm{~Hz}$
Karnataka CET-2002
WAVES
173139
If $L_{1}$ and $L_{2}$ are the lengths of the first and second resonating air columns in a resonance tube, then the wavelength of the note produced is
B If $L_{1}$ and $L_{2}$ are the lengths of first and second resonances, then we know that $\mathrm{L}_{1}+\mathrm{e}=\frac{\lambda}{4} \quad \text {............(i) (first resonance) }$ And, $\mathrm{L}_{2}+\mathrm{e}=\frac{3 \lambda}{4}$ (ii) (second resonance) From equation (i) and (ii), $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{\lambda}{2}$ $\lambda=2\left(\mathrm{~L}_{2}-\mathrm{L}_{1}\right)$
J and K CET- 2008
WAVES
173140
A segment of wire vibrates with fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg}$ weight. Then tension at which the fundamental frequency of the wire becomes $900 \mathrm{~Hz}$ is
1 $36 \mathrm{~kg}$ weight
2 $27 \mathrm{~kg}$ weight
3 $18 \mathrm{~kg}$ weight
4 $72 \mathrm{~kg}$ weight
Explanation:
A Given that, $\mathrm{f}_{1}=450 \mathrm{~Hz}$ and $\mathrm{f}_{2}=900 \mathrm{~Hz}$ $\mathrm{T}_{1}=9 \mathrm{~kg}$ and $\mathrm{T}_{2}=$ ? Fundamental frequency, $(\mathrm{f})=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ $f \propto \sqrt{T}$ $\frac{f_{1}}{f_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}$ $\therefore \quad \frac{450}{900} =\sqrt{\frac{9}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{9}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =36 \text { kg weight }$
EAMCET - 2007
WAVES
173144
A body vibrating with a certain frequency sends waves of wavelength $15 \mathrm{~cm}$ in a medium $A$ and $20 \mathrm{~cm}$ in medium $B$. If $v$ of waves in $A$ is $120 \mathrm{~ms}^{-1}$. That in $B$ will be
1 $196 \mathrm{~ms}^{-1}$
2 $160 \mathrm{~ms}^{-1}$
3 $360 \mathrm{~ms}^{-1}$
4 $260 \mathrm{~ms}^{-1}$
Explanation:
B Given that, Wave in A medium $\left(\mathrm{v}_{\mathrm{A}}\right)=120 \mathrm{~ms}^{-1}$ Wavelength in A medium $\left(\lambda_{A}\right)=15 \mathrm{~cm}$ Wavelength in $B$ medium $\left(\lambda_{B}\right)=20 \mathrm{~cm}$ Wave in $B$ medium $\left(v_{B}\right)=$ ? We know that, $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}$ $\frac{120}{\mathrm{v}_{\mathrm{B}}}=\frac{15}{20}$ $\mathrm{v}_{\mathrm{B}}=\frac{20 \times 120}{15}=160 \mathrm{~ms}^{-1}$
173136
In a resonance tube the first resonance with a tuning fork occurs at $16 \mathrm{~cm}$ and second at 49 $\mathrm{cm}$. If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, the frequency of tuning fork is
1 $500 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $330 \mathrm{~Hz}$
4 $165 \mathrm{~Hz}$
Explanation:
A Given, First resonance occurs at, $l_{1}=16 \mathrm{~cm}=0.16 \mathrm{~m}$ Second resonance occurs at, $l_{2}=49 \mathrm{~cm}=0.49 \mathrm{~m}$ Velocity of sound, $\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\frac{\lambda}{2}=\left(l_{2}-l_{1}\right)$ From equations (i) and (ii), we get - $\frac{\mathrm{v}}{\mathrm{f}}=2\left(l_{2}-l_{1}\right)$ $\mathrm{f}=\frac{\mathrm{v}}{2\left(l_{2}-l_{1}\right)}$ $\mathrm{f}=\frac{330}{2(0.49-0.16)}=\frac{330}{0.66}$ $\mathrm{f}=500 \mathrm{~Hz}$
MHT-CET 2011
WAVES
173138
A light points fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. If eight oscillations are counted when the plate falls through $10 \mathrm{~cm}$, the frequency of the tuning fork is :
1 $280 \mathrm{~Hz}$
2 $360 \mathrm{~Hz}$
3 $56 \mathrm{~Hz}$
4 $560 \mathrm{~Hz}$
Explanation:
C Time taken by plate to fall through $10 \mathrm{~cm}$, $\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 0.1}{9.8}}$ $=\frac{1}{7}$ Thus, tuning fork Completes ' 8 ' oscillations in $1 / 7$ seconds $\text { Frequency } =\frac{\text { number of oscillations }}{\text { time }}$ $=\frac{8}{\mathrm{t}}=\frac{8}{1 / 7}$ $=8 \times 7$ $=56 \mathrm{~Hz}$
Karnataka CET-2002
WAVES
173139
If $L_{1}$ and $L_{2}$ are the lengths of the first and second resonating air columns in a resonance tube, then the wavelength of the note produced is
B If $L_{1}$ and $L_{2}$ are the lengths of first and second resonances, then we know that $\mathrm{L}_{1}+\mathrm{e}=\frac{\lambda}{4} \quad \text {............(i) (first resonance) }$ And, $\mathrm{L}_{2}+\mathrm{e}=\frac{3 \lambda}{4}$ (ii) (second resonance) From equation (i) and (ii), $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{\lambda}{2}$ $\lambda=2\left(\mathrm{~L}_{2}-\mathrm{L}_{1}\right)$
J and K CET- 2008
WAVES
173140
A segment of wire vibrates with fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg}$ weight. Then tension at which the fundamental frequency of the wire becomes $900 \mathrm{~Hz}$ is
1 $36 \mathrm{~kg}$ weight
2 $27 \mathrm{~kg}$ weight
3 $18 \mathrm{~kg}$ weight
4 $72 \mathrm{~kg}$ weight
Explanation:
A Given that, $\mathrm{f}_{1}=450 \mathrm{~Hz}$ and $\mathrm{f}_{2}=900 \mathrm{~Hz}$ $\mathrm{T}_{1}=9 \mathrm{~kg}$ and $\mathrm{T}_{2}=$ ? Fundamental frequency, $(\mathrm{f})=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ $f \propto \sqrt{T}$ $\frac{f_{1}}{f_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}$ $\therefore \quad \frac{450}{900} =\sqrt{\frac{9}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{9}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =36 \text { kg weight }$
EAMCET - 2007
WAVES
173144
A body vibrating with a certain frequency sends waves of wavelength $15 \mathrm{~cm}$ in a medium $A$ and $20 \mathrm{~cm}$ in medium $B$. If $v$ of waves in $A$ is $120 \mathrm{~ms}^{-1}$. That in $B$ will be
1 $196 \mathrm{~ms}^{-1}$
2 $160 \mathrm{~ms}^{-1}$
3 $360 \mathrm{~ms}^{-1}$
4 $260 \mathrm{~ms}^{-1}$
Explanation:
B Given that, Wave in A medium $\left(\mathrm{v}_{\mathrm{A}}\right)=120 \mathrm{~ms}^{-1}$ Wavelength in A medium $\left(\lambda_{A}\right)=15 \mathrm{~cm}$ Wavelength in $B$ medium $\left(\lambda_{B}\right)=20 \mathrm{~cm}$ Wave in $B$ medium $\left(v_{B}\right)=$ ? We know that, $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}$ $\frac{120}{\mathrm{v}_{\mathrm{B}}}=\frac{15}{20}$ $\mathrm{v}_{\mathrm{B}}=\frac{20 \times 120}{15}=160 \mathrm{~ms}^{-1}$
173136
In a resonance tube the first resonance with a tuning fork occurs at $16 \mathrm{~cm}$ and second at 49 $\mathrm{cm}$. If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, the frequency of tuning fork is
1 $500 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $330 \mathrm{~Hz}$
4 $165 \mathrm{~Hz}$
Explanation:
A Given, First resonance occurs at, $l_{1}=16 \mathrm{~cm}=0.16 \mathrm{~m}$ Second resonance occurs at, $l_{2}=49 \mathrm{~cm}=0.49 \mathrm{~m}$ Velocity of sound, $\mathrm{v}_{\mathrm{s}}=330 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{v}}{\mathrm{f}}$ $\frac{\lambda}{2}=\left(l_{2}-l_{1}\right)$ From equations (i) and (ii), we get - $\frac{\mathrm{v}}{\mathrm{f}}=2\left(l_{2}-l_{1}\right)$ $\mathrm{f}=\frac{\mathrm{v}}{2\left(l_{2}-l_{1}\right)}$ $\mathrm{f}=\frac{330}{2(0.49-0.16)}=\frac{330}{0.66}$ $\mathrm{f}=500 \mathrm{~Hz}$
MHT-CET 2011
WAVES
173138
A light points fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. If eight oscillations are counted when the plate falls through $10 \mathrm{~cm}$, the frequency of the tuning fork is :
1 $280 \mathrm{~Hz}$
2 $360 \mathrm{~Hz}$
3 $56 \mathrm{~Hz}$
4 $560 \mathrm{~Hz}$
Explanation:
C Time taken by plate to fall through $10 \mathrm{~cm}$, $\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 0.1}{9.8}}$ $=\frac{1}{7}$ Thus, tuning fork Completes ' 8 ' oscillations in $1 / 7$ seconds $\text { Frequency } =\frac{\text { number of oscillations }}{\text { time }}$ $=\frac{8}{\mathrm{t}}=\frac{8}{1 / 7}$ $=8 \times 7$ $=56 \mathrm{~Hz}$
Karnataka CET-2002
WAVES
173139
If $L_{1}$ and $L_{2}$ are the lengths of the first and second resonating air columns in a resonance tube, then the wavelength of the note produced is
B If $L_{1}$ and $L_{2}$ are the lengths of first and second resonances, then we know that $\mathrm{L}_{1}+\mathrm{e}=\frac{\lambda}{4} \quad \text {............(i) (first resonance) }$ And, $\mathrm{L}_{2}+\mathrm{e}=\frac{3 \lambda}{4}$ (ii) (second resonance) From equation (i) and (ii), $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{\lambda}{2}$ $\lambda=2\left(\mathrm{~L}_{2}-\mathrm{L}_{1}\right)$
J and K CET- 2008
WAVES
173140
A segment of wire vibrates with fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg}$ weight. Then tension at which the fundamental frequency of the wire becomes $900 \mathrm{~Hz}$ is
1 $36 \mathrm{~kg}$ weight
2 $27 \mathrm{~kg}$ weight
3 $18 \mathrm{~kg}$ weight
4 $72 \mathrm{~kg}$ weight
Explanation:
A Given that, $\mathrm{f}_{1}=450 \mathrm{~Hz}$ and $\mathrm{f}_{2}=900 \mathrm{~Hz}$ $\mathrm{T}_{1}=9 \mathrm{~kg}$ and $\mathrm{T}_{2}=$ ? Fundamental frequency, $(\mathrm{f})=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ $f \propto \sqrt{T}$ $\frac{f_{1}}{f_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}$ $\therefore \quad \frac{450}{900} =\sqrt{\frac{9}{\mathrm{~T}_{2}}}$ $\frac{1}{4} =\frac{9}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =36 \text { kg weight }$
EAMCET - 2007
WAVES
173144
A body vibrating with a certain frequency sends waves of wavelength $15 \mathrm{~cm}$ in a medium $A$ and $20 \mathrm{~cm}$ in medium $B$. If $v$ of waves in $A$ is $120 \mathrm{~ms}^{-1}$. That in $B$ will be
1 $196 \mathrm{~ms}^{-1}$
2 $160 \mathrm{~ms}^{-1}$
3 $360 \mathrm{~ms}^{-1}$
4 $260 \mathrm{~ms}^{-1}$
Explanation:
B Given that, Wave in A medium $\left(\mathrm{v}_{\mathrm{A}}\right)=120 \mathrm{~ms}^{-1}$ Wavelength in A medium $\left(\lambda_{A}\right)=15 \mathrm{~cm}$ Wavelength in $B$ medium $\left(\lambda_{B}\right)=20 \mathrm{~cm}$ Wave in $B$ medium $\left(v_{B}\right)=$ ? We know that, $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}$ $\frac{120}{\mathrm{v}_{\mathrm{B}}}=\frac{15}{20}$ $\mathrm{v}_{\mathrm{B}}=\frac{20 \times 120}{15}=160 \mathrm{~ms}^{-1}$