173122
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
1 decrease
2 become zero
3 increase
4 not changes
Explanation:
C We know $\text { Frequency }(n)=\frac{1}{2 \operatorname{Lr}} \sqrt{\frac{T}{\pi \rho}}$ $\therefore$ If $\mathrm{L}$ and $\mathrm{r}$ decrease, the frequency will increase. Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.
MHT-CET 2020
WAVES
173127
The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency
1 will not change.
2 may increase or decrease.
3 will decrease.
4 will increase.
Explanation:
A Frequency depends upon the speed of the sound and length of the column. The frequency is independent of the material in the resonance. Hence option (a) is correct - resonating frequency will be not change.
MHT-CET 2020
WAVES
173128
In resonance tube experiment, the first and second resonating length of air column is $0.2 \mathrm{~m}$ and $0.62 \mathrm{~m}$ respectively. The inner diameter of the tube is
1 $3.33 \mathrm{~cm}$
2 $0.2 \mathrm{~m}$
3 $0.4 \mathrm{~m}$
4 $0.33 \mathrm{~cm}$
Explanation:
A Given, $\begin{aligned} l_{1}=0.2 \mathrm{~m} \text { and } l_{2}=0.62 \mathrm{~m} \\ \text { So, } \quad \text { end correction }(\mathrm{e}) =\frac{l_{2}-3 l_{1}}{2} \\ = \frac{0.62-0.60}{2}=0.01 \mathrm{~m}=1 \mathrm{~cm}\end{aligned}$ $\because \quad \mathrm{e} =0.3 \mathrm{D}$ $\mathrm{D} =\frac{\mathrm{e}}{0.3}=\frac{1}{0.3}=\frac{10}{3}$ The inner diameter of the tube, $\mathrm{D}=3.33 \mathrm{~cm}$
MHT-CET 2020
WAVES
173137
In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction ?
1 $1.05 \mathrm{~cm}$
2 $115.5 \mathrm{~cm}$
3 $92.5 \mathrm{~cm}$
4 $113.5 \mathrm{~cm}$
Explanation:
A Let, $l_{1}$ be the depth of first resonance of the pipe $\left(l_{1}\right)=22.7 \mathrm{~cm}$ and $l_{2}$ be the depth of second resonance of the pipe $\left(l_{2}\right)=70.2 \mathrm{~cm}$ Now for end correction $\mathrm{x}$, $\frac{l_{2}+\mathrm{x}}{l_{1}+\mathrm{x}}=\frac{3 \lambda / 4}{\lambda / 4}=3$ $\mathrm{x}=\frac{l_{2}-3 l_{1}}{2}=\frac{70.2-3 \times 22.7}{2}$ $\mathrm{x}=1.05 \mathrm{~cm}$
MHT-CET 2009
WAVES
173141
A car moving with a velocity of $36 \mathrm{~km} / \mathrm{hr}$ crosses a siren of frequency $500 \mathrm{~Hz}$. The apparent frequency of siren after passing it will be
1 $520 \mathrm{~Hz}$
2 $485 \mathrm{~Hz}$
3 $540 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
B Given, velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=36 \mathrm{~km} / \mathrm{hr}$ $=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{sec}, \mathrm{f}=500 \mathrm{~Hz}$ Velocity of sound frequency $\left(\mathrm{v}_{\mathrm{s}}\right)=330 \mathrm{~m} / \mathrm{sec}$ $\therefore$ Apparent frequency $\mathrm{v}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{s}}} \times f$ $=\frac{330-10}{330} \times 500=484.84=485 \mathrm{~Hz}$
173122
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
1 decrease
2 become zero
3 increase
4 not changes
Explanation:
C We know $\text { Frequency }(n)=\frac{1}{2 \operatorname{Lr}} \sqrt{\frac{T}{\pi \rho}}$ $\therefore$ If $\mathrm{L}$ and $\mathrm{r}$ decrease, the frequency will increase. Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.
MHT-CET 2020
WAVES
173127
The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency
1 will not change.
2 may increase or decrease.
3 will decrease.
4 will increase.
Explanation:
A Frequency depends upon the speed of the sound and length of the column. The frequency is independent of the material in the resonance. Hence option (a) is correct - resonating frequency will be not change.
MHT-CET 2020
WAVES
173128
In resonance tube experiment, the first and second resonating length of air column is $0.2 \mathrm{~m}$ and $0.62 \mathrm{~m}$ respectively. The inner diameter of the tube is
1 $3.33 \mathrm{~cm}$
2 $0.2 \mathrm{~m}$
3 $0.4 \mathrm{~m}$
4 $0.33 \mathrm{~cm}$
Explanation:
A Given, $\begin{aligned} l_{1}=0.2 \mathrm{~m} \text { and } l_{2}=0.62 \mathrm{~m} \\ \text { So, } \quad \text { end correction }(\mathrm{e}) =\frac{l_{2}-3 l_{1}}{2} \\ = \frac{0.62-0.60}{2}=0.01 \mathrm{~m}=1 \mathrm{~cm}\end{aligned}$ $\because \quad \mathrm{e} =0.3 \mathrm{D}$ $\mathrm{D} =\frac{\mathrm{e}}{0.3}=\frac{1}{0.3}=\frac{10}{3}$ The inner diameter of the tube, $\mathrm{D}=3.33 \mathrm{~cm}$
MHT-CET 2020
WAVES
173137
In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction ?
1 $1.05 \mathrm{~cm}$
2 $115.5 \mathrm{~cm}$
3 $92.5 \mathrm{~cm}$
4 $113.5 \mathrm{~cm}$
Explanation:
A Let, $l_{1}$ be the depth of first resonance of the pipe $\left(l_{1}\right)=22.7 \mathrm{~cm}$ and $l_{2}$ be the depth of second resonance of the pipe $\left(l_{2}\right)=70.2 \mathrm{~cm}$ Now for end correction $\mathrm{x}$, $\frac{l_{2}+\mathrm{x}}{l_{1}+\mathrm{x}}=\frac{3 \lambda / 4}{\lambda / 4}=3$ $\mathrm{x}=\frac{l_{2}-3 l_{1}}{2}=\frac{70.2-3 \times 22.7}{2}$ $\mathrm{x}=1.05 \mathrm{~cm}$
MHT-CET 2009
WAVES
173141
A car moving with a velocity of $36 \mathrm{~km} / \mathrm{hr}$ crosses a siren of frequency $500 \mathrm{~Hz}$. The apparent frequency of siren after passing it will be
1 $520 \mathrm{~Hz}$
2 $485 \mathrm{~Hz}$
3 $540 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
B Given, velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=36 \mathrm{~km} / \mathrm{hr}$ $=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{sec}, \mathrm{f}=500 \mathrm{~Hz}$ Velocity of sound frequency $\left(\mathrm{v}_{\mathrm{s}}\right)=330 \mathrm{~m} / \mathrm{sec}$ $\therefore$ Apparent frequency $\mathrm{v}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{s}}} \times f$ $=\frac{330-10}{330} \times 500=484.84=485 \mathrm{~Hz}$
173122
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
1 decrease
2 become zero
3 increase
4 not changes
Explanation:
C We know $\text { Frequency }(n)=\frac{1}{2 \operatorname{Lr}} \sqrt{\frac{T}{\pi \rho}}$ $\therefore$ If $\mathrm{L}$ and $\mathrm{r}$ decrease, the frequency will increase. Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.
MHT-CET 2020
WAVES
173127
The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency
1 will not change.
2 may increase or decrease.
3 will decrease.
4 will increase.
Explanation:
A Frequency depends upon the speed of the sound and length of the column. The frequency is independent of the material in the resonance. Hence option (a) is correct - resonating frequency will be not change.
MHT-CET 2020
WAVES
173128
In resonance tube experiment, the first and second resonating length of air column is $0.2 \mathrm{~m}$ and $0.62 \mathrm{~m}$ respectively. The inner diameter of the tube is
1 $3.33 \mathrm{~cm}$
2 $0.2 \mathrm{~m}$
3 $0.4 \mathrm{~m}$
4 $0.33 \mathrm{~cm}$
Explanation:
A Given, $\begin{aligned} l_{1}=0.2 \mathrm{~m} \text { and } l_{2}=0.62 \mathrm{~m} \\ \text { So, } \quad \text { end correction }(\mathrm{e}) =\frac{l_{2}-3 l_{1}}{2} \\ = \frac{0.62-0.60}{2}=0.01 \mathrm{~m}=1 \mathrm{~cm}\end{aligned}$ $\because \quad \mathrm{e} =0.3 \mathrm{D}$ $\mathrm{D} =\frac{\mathrm{e}}{0.3}=\frac{1}{0.3}=\frac{10}{3}$ The inner diameter of the tube, $\mathrm{D}=3.33 \mathrm{~cm}$
MHT-CET 2020
WAVES
173137
In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction ?
1 $1.05 \mathrm{~cm}$
2 $115.5 \mathrm{~cm}$
3 $92.5 \mathrm{~cm}$
4 $113.5 \mathrm{~cm}$
Explanation:
A Let, $l_{1}$ be the depth of first resonance of the pipe $\left(l_{1}\right)=22.7 \mathrm{~cm}$ and $l_{2}$ be the depth of second resonance of the pipe $\left(l_{2}\right)=70.2 \mathrm{~cm}$ Now for end correction $\mathrm{x}$, $\frac{l_{2}+\mathrm{x}}{l_{1}+\mathrm{x}}=\frac{3 \lambda / 4}{\lambda / 4}=3$ $\mathrm{x}=\frac{l_{2}-3 l_{1}}{2}=\frac{70.2-3 \times 22.7}{2}$ $\mathrm{x}=1.05 \mathrm{~cm}$
MHT-CET 2009
WAVES
173141
A car moving with a velocity of $36 \mathrm{~km} / \mathrm{hr}$ crosses a siren of frequency $500 \mathrm{~Hz}$. The apparent frequency of siren after passing it will be
1 $520 \mathrm{~Hz}$
2 $485 \mathrm{~Hz}$
3 $540 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
B Given, velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=36 \mathrm{~km} / \mathrm{hr}$ $=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{sec}, \mathrm{f}=500 \mathrm{~Hz}$ Velocity of sound frequency $\left(\mathrm{v}_{\mathrm{s}}\right)=330 \mathrm{~m} / \mathrm{sec}$ $\therefore$ Apparent frequency $\mathrm{v}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{s}}} \times f$ $=\frac{330-10}{330} \times 500=484.84=485 \mathrm{~Hz}$
173122
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
1 decrease
2 become zero
3 increase
4 not changes
Explanation:
C We know $\text { Frequency }(n)=\frac{1}{2 \operatorname{Lr}} \sqrt{\frac{T}{\pi \rho}}$ $\therefore$ If $\mathrm{L}$ and $\mathrm{r}$ decrease, the frequency will increase. Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.
MHT-CET 2020
WAVES
173127
The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency
1 will not change.
2 may increase or decrease.
3 will decrease.
4 will increase.
Explanation:
A Frequency depends upon the speed of the sound and length of the column. The frequency is independent of the material in the resonance. Hence option (a) is correct - resonating frequency will be not change.
MHT-CET 2020
WAVES
173128
In resonance tube experiment, the first and second resonating length of air column is $0.2 \mathrm{~m}$ and $0.62 \mathrm{~m}$ respectively. The inner diameter of the tube is
1 $3.33 \mathrm{~cm}$
2 $0.2 \mathrm{~m}$
3 $0.4 \mathrm{~m}$
4 $0.33 \mathrm{~cm}$
Explanation:
A Given, $\begin{aligned} l_{1}=0.2 \mathrm{~m} \text { and } l_{2}=0.62 \mathrm{~m} \\ \text { So, } \quad \text { end correction }(\mathrm{e}) =\frac{l_{2}-3 l_{1}}{2} \\ = \frac{0.62-0.60}{2}=0.01 \mathrm{~m}=1 \mathrm{~cm}\end{aligned}$ $\because \quad \mathrm{e} =0.3 \mathrm{D}$ $\mathrm{D} =\frac{\mathrm{e}}{0.3}=\frac{1}{0.3}=\frac{10}{3}$ The inner diameter of the tube, $\mathrm{D}=3.33 \mathrm{~cm}$
MHT-CET 2020
WAVES
173137
In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction ?
1 $1.05 \mathrm{~cm}$
2 $115.5 \mathrm{~cm}$
3 $92.5 \mathrm{~cm}$
4 $113.5 \mathrm{~cm}$
Explanation:
A Let, $l_{1}$ be the depth of first resonance of the pipe $\left(l_{1}\right)=22.7 \mathrm{~cm}$ and $l_{2}$ be the depth of second resonance of the pipe $\left(l_{2}\right)=70.2 \mathrm{~cm}$ Now for end correction $\mathrm{x}$, $\frac{l_{2}+\mathrm{x}}{l_{1}+\mathrm{x}}=\frac{3 \lambda / 4}{\lambda / 4}=3$ $\mathrm{x}=\frac{l_{2}-3 l_{1}}{2}=\frac{70.2-3 \times 22.7}{2}$ $\mathrm{x}=1.05 \mathrm{~cm}$
MHT-CET 2009
WAVES
173141
A car moving with a velocity of $36 \mathrm{~km} / \mathrm{hr}$ crosses a siren of frequency $500 \mathrm{~Hz}$. The apparent frequency of siren after passing it will be
1 $520 \mathrm{~Hz}$
2 $485 \mathrm{~Hz}$
3 $540 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
B Given, velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=36 \mathrm{~km} / \mathrm{hr}$ $=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{sec}, \mathrm{f}=500 \mathrm{~Hz}$ Velocity of sound frequency $\left(\mathrm{v}_{\mathrm{s}}\right)=330 \mathrm{~m} / \mathrm{sec}$ $\therefore$ Apparent frequency $\mathrm{v}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{s}}} \times f$ $=\frac{330-10}{330} \times 500=484.84=485 \mathrm{~Hz}$
173122
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
1 decrease
2 become zero
3 increase
4 not changes
Explanation:
C We know $\text { Frequency }(n)=\frac{1}{2 \operatorname{Lr}} \sqrt{\frac{T}{\pi \rho}}$ $\therefore$ If $\mathrm{L}$ and $\mathrm{r}$ decrease, the frequency will increase. Thus, if the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will increase.
MHT-CET 2020
WAVES
173127
The resonance tube is filled with a liquid of density higher than that of water, then resonating frequency
1 will not change.
2 may increase or decrease.
3 will decrease.
4 will increase.
Explanation:
A Frequency depends upon the speed of the sound and length of the column. The frequency is independent of the material in the resonance. Hence option (a) is correct - resonating frequency will be not change.
MHT-CET 2020
WAVES
173128
In resonance tube experiment, the first and second resonating length of air column is $0.2 \mathrm{~m}$ and $0.62 \mathrm{~m}$ respectively. The inner diameter of the tube is
1 $3.33 \mathrm{~cm}$
2 $0.2 \mathrm{~m}$
3 $0.4 \mathrm{~m}$
4 $0.33 \mathrm{~cm}$
Explanation:
A Given, $\begin{aligned} l_{1}=0.2 \mathrm{~m} \text { and } l_{2}=0.62 \mathrm{~m} \\ \text { So, } \quad \text { end correction }(\mathrm{e}) =\frac{l_{2}-3 l_{1}}{2} \\ = \frac{0.62-0.60}{2}=0.01 \mathrm{~m}=1 \mathrm{~cm}\end{aligned}$ $\because \quad \mathrm{e} =0.3 \mathrm{D}$ $\mathrm{D} =\frac{\mathrm{e}}{0.3}=\frac{1}{0.3}=\frac{10}{3}$ The inner diameter of the tube, $\mathrm{D}=3.33 \mathrm{~cm}$
MHT-CET 2020
WAVES
173137
In a resonance pipe the first and second resonances are obtained at depths $22.7 \mathrm{~cm}$ and $70.2 \mathrm{~cm}$ respectively. What will be the end correction ?
1 $1.05 \mathrm{~cm}$
2 $115.5 \mathrm{~cm}$
3 $92.5 \mathrm{~cm}$
4 $113.5 \mathrm{~cm}$
Explanation:
A Let, $l_{1}$ be the depth of first resonance of the pipe $\left(l_{1}\right)=22.7 \mathrm{~cm}$ and $l_{2}$ be the depth of second resonance of the pipe $\left(l_{2}\right)=70.2 \mathrm{~cm}$ Now for end correction $\mathrm{x}$, $\frac{l_{2}+\mathrm{x}}{l_{1}+\mathrm{x}}=\frac{3 \lambda / 4}{\lambda / 4}=3$ $\mathrm{x}=\frac{l_{2}-3 l_{1}}{2}=\frac{70.2-3 \times 22.7}{2}$ $\mathrm{x}=1.05 \mathrm{~cm}$
MHT-CET 2009
WAVES
173141
A car moving with a velocity of $36 \mathrm{~km} / \mathrm{hr}$ crosses a siren of frequency $500 \mathrm{~Hz}$. The apparent frequency of siren after passing it will be
1 $520 \mathrm{~Hz}$
2 $485 \mathrm{~Hz}$
3 $540 \mathrm{~Hz}$
4 $460 \mathrm{~Hz}$
Explanation:
B Given, velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=36 \mathrm{~km} / \mathrm{hr}$ $=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{sec}, \mathrm{f}=500 \mathrm{~Hz}$ Velocity of sound frequency $\left(\mathrm{v}_{\mathrm{s}}\right)=330 \mathrm{~m} / \mathrm{sec}$ $\therefore$ Apparent frequency $\mathrm{v}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{s}}} \times f$ $=\frac{330-10}{330} \times 500=484.84=485 \mathrm{~Hz}$
