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PHXI15:WAVES

354907 The fundamental frequency of a closed pipe is $400 H z .$ If $\frac{1}{3} r d$ pipe is filled with water, then the frequency of $2^{n d}$ harmonic of the pipe will be (neglect end correction)

1

1200 H z

2

1800 H z

3

600 H z

4

300 H z

Explanation:

Fundamental frequency of closed pipe
$n = \frac{v}{4 L} = 400$
$v = 400 \times 4 L$
If $\frac{1}{3} r d$ of pipe is filled with water, then remaining length of air column is $L - \frac{L}{3} = \frac{2 L}{3}$
Now, fundamental frequency
$= \frac{v}{4 (\frac{2 L}{3})} = \frac{3 v}{8 L}$
The second harmonic of the pipe $= 3 x$ fundamental frequency
$= 3 \times (\frac{3 v}{8 L}) = 3 \times \frac{3 \times 400 \times 4 L}{8 L} = 1800 H z$

MHTCET - 2020

PHXI15:WAVES

354908 If $ℓ_{1}$ and $ℓ_{2} (> ℓ_{1})$ be the lengths of an air column for the first and second resonance When a tuning fork of frequency $n$ is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

1

\frac{ℓ_{1} + ℓ_{2}}{2}

2

2 (ℓ_{2} + ℓ_{1})

3

\frac{ℓ_{2} - 3 ℓ_{1}}{2}

4

1 / 2 (2 ℓ_{1} + ℓ_{2})

Explanation:

Let $x$ be the correction $l_{1} + x = λ / 4$
supporting img

$\begin{aligned} l_{2} + x = \frac{3 λ}{4} \\ \Rightarrow x = \frac{l_{2} - 3 l_{1}}{2} \end{aligned}$

PHXI15:WAVES

354909 In the case of vibration of closed end organ pipe in fundamental mode of vibration, the pressure is maximum at

1 Open end

2 Closed end

3 At middle

4 Any where

PHXI15:WAVES

354910 A pipe closed at one end has length $0.8 m$ . At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 N$ and the speed of sound is $320 m / s$ , the mass of the string used is

1 4 gram

2 8 gram

3 10 gram

4 12 gram

Explanation:

Let, length of string, $l = 0.5 m$
Length of pipe, $l = 0.8 m$
As the frequency of string is vibrating in second harmonics
i.e., $v = (\frac{2}{2 l}) \sqrt{\frac{T}{μ}}$
( $μ =$ mass pen unit length of string)
Fundamental frequency of closed pipe is,
$f^{'} = \frac{v}{4 L}$ . Here ' $v$ ' is speed of sound.
At resonance, $f = f^{'}$
$\Rightarrow (\frac{2}{2 l}) \sqrt{\frac{T}{μ}} = \frac{v}{4 L}$
$\Rightarrow (\frac{1}{0.5}) \sqrt{\frac{50}{μ}} = (\frac{320}{4 \times 0.8}) \Rightarrow \frac{50}{μ} = 2500$
$\Rightarrow μ = \frac{1}{50} k g / m$
mass of string, $m = μ l = \frac{1}{50} \times 0.5$
$\Rightarrow m = (\frac{1}{100}) k g = 10 g m$

MHTCET - 2022

PHXI15:WAVES

354911 An open pipe is suddenly closed with the result that the second overtone of the closed pipe is found to be higher in frequency by $100 H z$ , than the first overtone of the original pipe. The fundamental frequency of open pipe will be

1

100 H z

2

300 H z

3

150 H z

4

200 H z

Explanation:

Second overtone of the closed pipe is $\frac{5 v}{4 L}$
First overtone of the open pipe is $\frac{2 v}{2 L}$
Given that $\frac{5 v}{4 L} - \frac{2 v}{2 L} = 100$
$\begin{aligned} \frac{v}{2 L} (\frac{5}{2} - 2) = 100 \\ \frac{v}{2 L} = 200 \end{aligned}$

PHXI15:WAVES

354907 The fundamental frequency of a closed pipe is $400 H z .$ If $\frac{1}{3} r d$ pipe is filled with water, then the frequency of $2^{n d}$ harmonic of the pipe will be (neglect end correction)

1

1200 H z

2

1800 H z

3

600 H z

4

300 H z

Explanation:

Fundamental frequency of closed pipe
$n = \frac{v}{4 L} = 400$
$v = 400 \times 4 L$
If $\frac{1}{3} r d$ of pipe is filled with water, then remaining length of air column is $L - \frac{L}{3} = \frac{2 L}{3}$
Now, fundamental frequency
$= \frac{v}{4 (\frac{2 L}{3})} = \frac{3 v}{8 L}$
The second harmonic of the pipe $= 3 x$ fundamental frequency
$= 3 \times (\frac{3 v}{8 L}) = 3 \times \frac{3 \times 400 \times 4 L}{8 L} = 1800 H z$

MHTCET - 2020

PHXI15:WAVES

354908 If $ℓ_{1}$ and $ℓ_{2} (> ℓ_{1})$ be the lengths of an air column for the first and second resonance When a tuning fork of frequency $n$ is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

1

\frac{ℓ_{1} + ℓ_{2}}{2}

2

2 (ℓ_{2} + ℓ_{1})

3

\frac{ℓ_{2} - 3 ℓ_{1}}{2}

4

1 / 2 (2 ℓ_{1} + ℓ_{2})

Explanation:

Let $x$ be the correction $l_{1} + x = λ / 4$
supporting img

$\begin{aligned} l_{2} + x = \frac{3 λ}{4} \\ \Rightarrow x = \frac{l_{2} - 3 l_{1}}{2} \end{aligned}$

PHXI15:WAVES

354909 In the case of vibration of closed end organ pipe in fundamental mode of vibration, the pressure is maximum at

1 Open end

2 Closed end

3 At middle

4 Any where

PHXI15:WAVES

354910 A pipe closed at one end has length $0.8 m$ . At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 N$ and the speed of sound is $320 m / s$ , the mass of the string used is

1 4 gram

2 8 gram

3 10 gram

4 12 gram

Explanation:

Let, length of string, $l = 0.5 m$
Length of pipe, $l = 0.8 m$
As the frequency of string is vibrating in second harmonics
i.e., $v = (\frac{2}{2 l}) \sqrt{\frac{T}{μ}}$
( $μ =$ mass pen unit length of string)
Fundamental frequency of closed pipe is,
$f^{'} = \frac{v}{4 L}$ . Here ' $v$ ' is speed of sound.
At resonance, $f = f^{'}$
$\Rightarrow (\frac{2}{2 l}) \sqrt{\frac{T}{μ}} = \frac{v}{4 L}$
$\Rightarrow (\frac{1}{0.5}) \sqrt{\frac{50}{μ}} = (\frac{320}{4 \times 0.8}) \Rightarrow \frac{50}{μ} = 2500$
$\Rightarrow μ = \frac{1}{50} k g / m$
mass of string, $m = μ l = \frac{1}{50} \times 0.5$
$\Rightarrow m = (\frac{1}{100}) k g = 10 g m$

MHTCET - 2022

PHXI15:WAVES

354911 An open pipe is suddenly closed with the result that the second overtone of the closed pipe is found to be higher in frequency by $100 H z$ , than the first overtone of the original pipe. The fundamental frequency of open pipe will be

1

100 H z

2

300 H z

3

150 H z

4

200 H z

Explanation:

Second overtone of the closed pipe is $\frac{5 v}{4 L}$
First overtone of the open pipe is $\frac{2 v}{2 L}$
Given that $\frac{5 v}{4 L} - \frac{2 v}{2 L} = 100$
$\begin{aligned} \frac{v}{2 L} (\frac{5}{2} - 2) = 100 \\ \frac{v}{2 L} = 200 \end{aligned}$

PHXI15:WAVES

354907 The fundamental frequency of a closed pipe is $400 H z .$ If $\frac{1}{3} r d$ pipe is filled with water, then the frequency of $2^{n d}$ harmonic of the pipe will be (neglect end correction)

1

1200 H z

2

1800 H z

3

600 H z

4

300 H z

Explanation:

Fundamental frequency of closed pipe
$n = \frac{v}{4 L} = 400$
$v = 400 \times 4 L$
If $\frac{1}{3} r d$ of pipe is filled with water, then remaining length of air column is $L - \frac{L}{3} = \frac{2 L}{3}$
Now, fundamental frequency
$= \frac{v}{4 (\frac{2 L}{3})} = \frac{3 v}{8 L}$
The second harmonic of the pipe $= 3 x$ fundamental frequency
$= 3 \times (\frac{3 v}{8 L}) = 3 \times \frac{3 \times 400 \times 4 L}{8 L} = 1800 H z$

MHTCET - 2020

PHXI15:WAVES

354908 If $ℓ_{1}$ and $ℓ_{2} (> ℓ_{1})$ be the lengths of an air column for the first and second resonance When a tuning fork of frequency $n$ is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

1

\frac{ℓ_{1} + ℓ_{2}}{2}

2

2 (ℓ_{2} + ℓ_{1})

3

\frac{ℓ_{2} - 3 ℓ_{1}}{2}

4

1 / 2 (2 ℓ_{1} + ℓ_{2})

Explanation:

Let $x$ be the correction $l_{1} + x = λ / 4$
supporting img

$\begin{aligned} l_{2} + x = \frac{3 λ}{4} \\ \Rightarrow x = \frac{l_{2} - 3 l_{1}}{2} \end{aligned}$

PHXI15:WAVES

354909 In the case of vibration of closed end organ pipe in fundamental mode of vibration, the pressure is maximum at

1 Open end

2 Closed end

3 At middle

4 Any where

PHXI15:WAVES

354910 A pipe closed at one end has length $0.8 m$ . At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 N$ and the speed of sound is $320 m / s$ , the mass of the string used is

1 4 gram

2 8 gram

3 10 gram

4 12 gram

Explanation:

Let, length of string, $l = 0.5 m$
Length of pipe, $l = 0.8 m$
As the frequency of string is vibrating in second harmonics
i.e., $v = (\frac{2}{2 l}) \sqrt{\frac{T}{μ}}$
( $μ =$ mass pen unit length of string)
Fundamental frequency of closed pipe is,
$f^{'} = \frac{v}{4 L}$ . Here ' $v$ ' is speed of sound.
At resonance, $f = f^{'}$
$\Rightarrow (\frac{2}{2 l}) \sqrt{\frac{T}{μ}} = \frac{v}{4 L}$
$\Rightarrow (\frac{1}{0.5}) \sqrt{\frac{50}{μ}} = (\frac{320}{4 \times 0.8}) \Rightarrow \frac{50}{μ} = 2500$
$\Rightarrow μ = \frac{1}{50} k g / m$
mass of string, $m = μ l = \frac{1}{50} \times 0.5$
$\Rightarrow m = (\frac{1}{100}) k g = 10 g m$

MHTCET - 2022

PHXI15:WAVES

354911 An open pipe is suddenly closed with the result that the second overtone of the closed pipe is found to be higher in frequency by $100 H z$ , than the first overtone of the original pipe. The fundamental frequency of open pipe will be

1

100 H z

2

300 H z

3

150 H z

4

200 H z

Explanation:

Second overtone of the closed pipe is $\frac{5 v}{4 L}$
First overtone of the open pipe is $\frac{2 v}{2 L}$
Given that $\frac{5 v}{4 L} - \frac{2 v}{2 L} = 100$
$\begin{aligned} \frac{v}{2 L} (\frac{5}{2} - 2) = 100 \\ \frac{v}{2 L} = 200 \end{aligned}$

PHXI15:WAVES

354907 The fundamental frequency of a closed pipe is $400 H z .$ If $\frac{1}{3} r d$ pipe is filled with water, then the frequency of $2^{n d}$ harmonic of the pipe will be (neglect end correction)

1

1200 H z

2

1800 H z

3

600 H z

4

300 H z

Explanation:

Fundamental frequency of closed pipe
$n = \frac{v}{4 L} = 400$
$v = 400 \times 4 L$
If $\frac{1}{3} r d$ of pipe is filled with water, then remaining length of air column is $L - \frac{L}{3} = \frac{2 L}{3}$
Now, fundamental frequency
$= \frac{v}{4 (\frac{2 L}{3})} = \frac{3 v}{8 L}$
The second harmonic of the pipe $= 3 x$ fundamental frequency
$= 3 \times (\frac{3 v}{8 L}) = 3 \times \frac{3 \times 400 \times 4 L}{8 L} = 1800 H z$

MHTCET - 2020

PHXI15:WAVES

354908 If $ℓ_{1}$ and $ℓ_{2} (> ℓ_{1})$ be the lengths of an air column for the first and second resonance When a tuning fork of frequency $n$ is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

1

\frac{ℓ_{1} + ℓ_{2}}{2}

2

2 (ℓ_{2} + ℓ_{1})

3

\frac{ℓ_{2} - 3 ℓ_{1}}{2}

4

1 / 2 (2 ℓ_{1} + ℓ_{2})

Explanation:

Let $x$ be the correction $l_{1} + x = λ / 4$
supporting img

$\begin{aligned} l_{2} + x = \frac{3 λ}{4} \\ \Rightarrow x = \frac{l_{2} - 3 l_{1}}{2} \end{aligned}$

PHXI15:WAVES

354909 In the case of vibration of closed end organ pipe in fundamental mode of vibration, the pressure is maximum at

1 Open end

2 Closed end

3 At middle

4 Any where

PHXI15:WAVES

354910 A pipe closed at one end has length $0.8 m$ . At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 N$ and the speed of sound is $320 m / s$ , the mass of the string used is

1 4 gram

2 8 gram

3 10 gram

4 12 gram

Explanation:

Let, length of string, $l = 0.5 m$
Length of pipe, $l = 0.8 m$
As the frequency of string is vibrating in second harmonics
i.e., $v = (\frac{2}{2 l}) \sqrt{\frac{T}{μ}}$
( $μ =$ mass pen unit length of string)
Fundamental frequency of closed pipe is,
$f^{'} = \frac{v}{4 L}$ . Here ' $v$ ' is speed of sound.
At resonance, $f = f^{'}$
$\Rightarrow (\frac{2}{2 l}) \sqrt{\frac{T}{μ}} = \frac{v}{4 L}$
$\Rightarrow (\frac{1}{0.5}) \sqrt{\frac{50}{μ}} = (\frac{320}{4 \times 0.8}) \Rightarrow \frac{50}{μ} = 2500$
$\Rightarrow μ = \frac{1}{50} k g / m$
mass of string, $m = μ l = \frac{1}{50} \times 0.5$
$\Rightarrow m = (\frac{1}{100}) k g = 10 g m$

MHTCET - 2022

PHXI15:WAVES

354911 An open pipe is suddenly closed with the result that the second overtone of the closed pipe is found to be higher in frequency by $100 H z$ , than the first overtone of the original pipe. The fundamental frequency of open pipe will be

1

100 H z

2

300 H z

3

150 H z

4

200 H z

Explanation:

Second overtone of the closed pipe is $\frac{5 v}{4 L}$
First overtone of the open pipe is $\frac{2 v}{2 L}$
Given that $\frac{5 v}{4 L} - \frac{2 v}{2 L} = 100$
$\begin{aligned} \frac{v}{2 L} (\frac{5}{2} - 2) = 100 \\ \frac{v}{2 L} = 200 \end{aligned}$

PHXI15:WAVES

354907 The fundamental frequency of a closed pipe is $400 H z .$ If $\frac{1}{3} r d$ pipe is filled with water, then the frequency of $2^{n d}$ harmonic of the pipe will be (neglect end correction)

1

1200 H z

2

1800 H z

3

600 H z

4

300 H z

Explanation:

Fundamental frequency of closed pipe
$n = \frac{v}{4 L} = 400$
$v = 400 \times 4 L$
If $\frac{1}{3} r d$ of pipe is filled with water, then remaining length of air column is $L - \frac{L}{3} = \frac{2 L}{3}$
Now, fundamental frequency
$= \frac{v}{4 (\frac{2 L}{3})} = \frac{3 v}{8 L}$
The second harmonic of the pipe $= 3 x$ fundamental frequency
$= 3 \times (\frac{3 v}{8 L}) = 3 \times \frac{3 \times 400 \times 4 L}{8 L} = 1800 H z$

MHTCET - 2020

PHXI15:WAVES

354908 If $ℓ_{1}$ and $ℓ_{2} (> ℓ_{1})$ be the lengths of an air column for the first and second resonance When a tuning fork of frequency $n$ is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

1

\frac{ℓ_{1} + ℓ_{2}}{2}

2

2 (ℓ_{2} + ℓ_{1})

3

\frac{ℓ_{2} - 3 ℓ_{1}}{2}

4

1 / 2 (2 ℓ_{1} + ℓ_{2})

Explanation:

Let $x$ be the correction $l_{1} + x = λ / 4$
supporting img

$\begin{aligned} l_{2} + x = \frac{3 λ}{4} \\ \Rightarrow x = \frac{l_{2} - 3 l_{1}}{2} \end{aligned}$

PHXI15:WAVES

354909 In the case of vibration of closed end organ pipe in fundamental mode of vibration, the pressure is maximum at

1 Open end

2 Closed end

3 At middle

4 Any where

PHXI15:WAVES

354910 A pipe closed at one end has length $0.8 m$ . At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 N$ and the speed of sound is $320 m / s$ , the mass of the string used is

1 4 gram

2 8 gram

3 10 gram

4 12 gram

Explanation:

Let, length of string, $l = 0.5 m$
Length of pipe, $l = 0.8 m$
As the frequency of string is vibrating in second harmonics
i.e., $v = (\frac{2}{2 l}) \sqrt{\frac{T}{μ}}$
( $μ =$ mass pen unit length of string)
Fundamental frequency of closed pipe is,
$f^{'} = \frac{v}{4 L}$ . Here ' $v$ ' is speed of sound.
At resonance, $f = f^{'}$
$\Rightarrow (\frac{2}{2 l}) \sqrt{\frac{T}{μ}} = \frac{v}{4 L}$
$\Rightarrow (\frac{1}{0.5}) \sqrt{\frac{50}{μ}} = (\frac{320}{4 \times 0.8}) \Rightarrow \frac{50}{μ} = 2500$
$\Rightarrow μ = \frac{1}{50} k g / m$
mass of string, $m = μ l = \frac{1}{50} \times 0.5$
$\Rightarrow m = (\frac{1}{100}) k g = 10 g m$

MHTCET - 2022

PHXI15:WAVES

354911 An open pipe is suddenly closed with the result that the second overtone of the closed pipe is found to be higher in frequency by $100 H z$ , than the first overtone of the original pipe. The fundamental frequency of open pipe will be

1

100 H z

2

300 H z

3

150 H z

4

200 H z

Explanation:

Second overtone of the closed pipe is $\frac{5 v}{4 L}$
First overtone of the open pipe is $\frac{2 v}{2 L}$
Given that $\frac{5 v}{4 L} - \frac{2 v}{2 L} = 100$
$\begin{aligned} \frac{v}{2 L} (\frac{5}{2} - 2) = 100 \\ \frac{v}{2 L} = 200 \end{aligned}$