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PHXI15:WAVES

354877 If the end correction of an open pipe is $0.8 c m$ , then the inner radius of that pipe will be

1

\frac{1}{3} c m

2

\frac{2}{3} c m

3

\frac{3}{2} c m

4

0.2 c m

Explanation:

For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by
$Δ l = 1.2 \times r$
[ $Δ l =$ end correction, $r =$ inner radius $]$
So, $r = \frac{Δ l}{1.2} = \frac{0.8}{1.2} [∵ Δ l = 0.8]$
$= \frac{2}{3} c m$

MHTCET - 2016

PHXI15:WAVES

354878 In a resonance pipe the first and second resonances are obtained at depths $22.7 c m$ and $70.2 c m$ respectively. The end correction will be

1

115.5 c m

2

1.05 c m

3

92.5 c m

4

113.5 c m

Explanation:

For end correction $x = 3$
$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (I_{2} + x)} x = \frac{l_{2} - 3 l_{1}}{2}$
$= \frac{70.2 - 3 \times 22.7}{2} = 1.05 c m$

PHXI15:WAVES

354879 A tube of diameter $d$ and of length $l$ unit is open at both ends. Its fundamental frequency of resonance is found to be $f_{1}$ . The velocity of sound in air is $330 m / \sec$ . One endof tube is now closed. The lowest frequency of resonance of tube is $f_{2}$ . Taking into consideration of end correction, $\frac{f_{2}}{f_{1}}$ is :

1

\frac{1}{2} \frac{(l + 0.3 d)}{(l + 0.6 d)}

2

\frac{(l + 0.6 d)}{(l + 0.3 d)}

3

\frac{1}{2} \frac{(l + 0.3 l)}{(l + 0.6 l)}

4

\frac{1}{2} \frac{(l + 0.6 d)}{(l + 0.3 d)}

Explanation:

supporting img

$\begin{aligned} f_{1} = \frac{v}{2 (l + 0.6 d)}, f_{2} = \frac{v}{4 (l + 0.3 d)} \\ \frac{f_{2}}{f_{1}} = \frac{l + 0.6 d}{2 (l + 0.3 d)} \end{aligned}$

PHXI15:WAVES

354880 In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is $0.1 m$ . When this length is changed to $0.35 m$ , the same tuning fork resonates with the first overnote. Calculate the end correction

1

0.025 m

2

0.012 m

3

0.024 m

4

0.05 m

Explanation:

Let $e$ be the end correction then according to question.
$\frac{v}{4 (l_{1} + e)} = \frac{3 v}{4 (l_{2} + e)} \Rightarrow e = 0.025 m$

PHXI15:WAVES

354881 In a pipe opened at both ends, $n_{1}$ and $n_{2}$ be the frequencies corresponding to vibrating lengths $l_{1}$ and $l_{2}$ respectively. The end correction is

1

\frac{n_{1} l_{1} - n_{2} l_{2}}{2 (n_{1} - n_{2})}

2

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{2} - n_{1})}

3

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{1} - n_{2})}

4

\frac{n_{2} l_{2} - n_{1} l_{1}}{(n_{1} - n_{2})}

Explanation:

Suppose pipe vibrates in fundamental mode,
For first pipe,
$l_{1} + d = \frac{λ_{1}}{2} (1)$
Where, $d$ is end correction,
Similarly for second pipe,
$l_{2} + d = \frac{λ_{2}}{2} (2)$
From eq.(1) and (2), we get
$\frac{l_{1} + d}{l_{2} + d} = \frac{λ_{1}}{λ_{2}} = \frac{1}{n_{1}} \times n^{2}$
$\Rightarrow n_{1} l_{1} + n_{1} d = n_{2} l_{2} + n_{2} d$
$\Rightarrow d [n_{1} - n_{2}] = n_{2} l_{2} - n_{1} l_{1}$
End correction, $d = \frac{n_{2} l_{2} - n_{1} l_{1}}{n_{1} - n_{2}}$

MHTCET - 2015

PHXI15:WAVES

354877 If the end correction of an open pipe is $0.8 c m$ , then the inner radius of that pipe will be

1

\frac{1}{3} c m

2

\frac{2}{3} c m

3

\frac{3}{2} c m

4

0.2 c m

Explanation:

For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by
$Δ l = 1.2 \times r$
[ $Δ l =$ end correction, $r =$ inner radius $]$
So, $r = \frac{Δ l}{1.2} = \frac{0.8}{1.2} [∵ Δ l = 0.8]$
$= \frac{2}{3} c m$

MHTCET - 2016

PHXI15:WAVES

354878 In a resonance pipe the first and second resonances are obtained at depths $22.7 c m$ and $70.2 c m$ respectively. The end correction will be

1

115.5 c m

2

1.05 c m

3

92.5 c m

4

113.5 c m

Explanation:

For end correction $x = 3$
$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (I_{2} + x)} x = \frac{l_{2} - 3 l_{1}}{2}$
$= \frac{70.2 - 3 \times 22.7}{2} = 1.05 c m$

PHXI15:WAVES

354879 A tube of diameter $d$ and of length $l$ unit is open at both ends. Its fundamental frequency of resonance is found to be $f_{1}$ . The velocity of sound in air is $330 m / \sec$ . One endof tube is now closed. The lowest frequency of resonance of tube is $f_{2}$ . Taking into consideration of end correction, $\frac{f_{2}}{f_{1}}$ is :

1

\frac{1}{2} \frac{(l + 0.3 d)}{(l + 0.6 d)}

2

\frac{(l + 0.6 d)}{(l + 0.3 d)}

3

\frac{1}{2} \frac{(l + 0.3 l)}{(l + 0.6 l)}

4

\frac{1}{2} \frac{(l + 0.6 d)}{(l + 0.3 d)}

Explanation:

supporting img

$\begin{aligned} f_{1} = \frac{v}{2 (l + 0.6 d)}, f_{2} = \frac{v}{4 (l + 0.3 d)} \\ \frac{f_{2}}{f_{1}} = \frac{l + 0.6 d}{2 (l + 0.3 d)} \end{aligned}$

PHXI15:WAVES

354880 In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is $0.1 m$ . When this length is changed to $0.35 m$ , the same tuning fork resonates with the first overnote. Calculate the end correction

1

0.025 m

2

0.012 m

3

0.024 m

4

0.05 m

Explanation:

Let $e$ be the end correction then according to question.
$\frac{v}{4 (l_{1} + e)} = \frac{3 v}{4 (l_{2} + e)} \Rightarrow e = 0.025 m$

PHXI15:WAVES

354881 In a pipe opened at both ends, $n_{1}$ and $n_{2}$ be the frequencies corresponding to vibrating lengths $l_{1}$ and $l_{2}$ respectively. The end correction is

1

\frac{n_{1} l_{1} - n_{2} l_{2}}{2 (n_{1} - n_{2})}

2

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{2} - n_{1})}

3

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{1} - n_{2})}

4

\frac{n_{2} l_{2} - n_{1} l_{1}}{(n_{1} - n_{2})}

Explanation:

Suppose pipe vibrates in fundamental mode,
For first pipe,
$l_{1} + d = \frac{λ_{1}}{2} (1)$
Where, $d$ is end correction,
Similarly for second pipe,
$l_{2} + d = \frac{λ_{2}}{2} (2)$
From eq.(1) and (2), we get
$\frac{l_{1} + d}{l_{2} + d} = \frac{λ_{1}}{λ_{2}} = \frac{1}{n_{1}} \times n^{2}$
$\Rightarrow n_{1} l_{1} + n_{1} d = n_{2} l_{2} + n_{2} d$
$\Rightarrow d [n_{1} - n_{2}] = n_{2} l_{2} - n_{1} l_{1}$
End correction, $d = \frac{n_{2} l_{2} - n_{1} l_{1}}{n_{1} - n_{2}}$

MHTCET - 2015

PHXI15:WAVES

354877 If the end correction of an open pipe is $0.8 c m$ , then the inner radius of that pipe will be

1

\frac{1}{3} c m

2

\frac{2}{3} c m

3

\frac{3}{2} c m

4

0.2 c m

Explanation:

For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by
$Δ l = 1.2 \times r$
[ $Δ l =$ end correction, $r =$ inner radius $]$
So, $r = \frac{Δ l}{1.2} = \frac{0.8}{1.2} [∵ Δ l = 0.8]$
$= \frac{2}{3} c m$

MHTCET - 2016

PHXI15:WAVES

354878 In a resonance pipe the first and second resonances are obtained at depths $22.7 c m$ and $70.2 c m$ respectively. The end correction will be

1

115.5 c m

2

1.05 c m

3

92.5 c m

4

113.5 c m

Explanation:

For end correction $x = 3$
$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (I_{2} + x)} x = \frac{l_{2} - 3 l_{1}}{2}$
$= \frac{70.2 - 3 \times 22.7}{2} = 1.05 c m$

PHXI15:WAVES

354879 A tube of diameter $d$ and of length $l$ unit is open at both ends. Its fundamental frequency of resonance is found to be $f_{1}$ . The velocity of sound in air is $330 m / \sec$ . One endof tube is now closed. The lowest frequency of resonance of tube is $f_{2}$ . Taking into consideration of end correction, $\frac{f_{2}}{f_{1}}$ is :

1

\frac{1}{2} \frac{(l + 0.3 d)}{(l + 0.6 d)}

2

\frac{(l + 0.6 d)}{(l + 0.3 d)}

3

\frac{1}{2} \frac{(l + 0.3 l)}{(l + 0.6 l)}

4

\frac{1}{2} \frac{(l + 0.6 d)}{(l + 0.3 d)}

Explanation:

supporting img

$\begin{aligned} f_{1} = \frac{v}{2 (l + 0.6 d)}, f_{2} = \frac{v}{4 (l + 0.3 d)} \\ \frac{f_{2}}{f_{1}} = \frac{l + 0.6 d}{2 (l + 0.3 d)} \end{aligned}$

PHXI15:WAVES

354880 In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is $0.1 m$ . When this length is changed to $0.35 m$ , the same tuning fork resonates with the first overnote. Calculate the end correction

1

0.025 m

2

0.012 m

3

0.024 m

4

0.05 m

Explanation:

Let $e$ be the end correction then according to question.
$\frac{v}{4 (l_{1} + e)} = \frac{3 v}{4 (l_{2} + e)} \Rightarrow e = 0.025 m$

PHXI15:WAVES

354881 In a pipe opened at both ends, $n_{1}$ and $n_{2}$ be the frequencies corresponding to vibrating lengths $l_{1}$ and $l_{2}$ respectively. The end correction is

1

\frac{n_{1} l_{1} - n_{2} l_{2}}{2 (n_{1} - n_{2})}

2

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{2} - n_{1})}

3

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{1} - n_{2})}

4

\frac{n_{2} l_{2} - n_{1} l_{1}}{(n_{1} - n_{2})}

Explanation:

Suppose pipe vibrates in fundamental mode,
For first pipe,
$l_{1} + d = \frac{λ_{1}}{2} (1)$
Where, $d$ is end correction,
Similarly for second pipe,
$l_{2} + d = \frac{λ_{2}}{2} (2)$
From eq.(1) and (2), we get
$\frac{l_{1} + d}{l_{2} + d} = \frac{λ_{1}}{λ_{2}} = \frac{1}{n_{1}} \times n^{2}$
$\Rightarrow n_{1} l_{1} + n_{1} d = n_{2} l_{2} + n_{2} d$
$\Rightarrow d [n_{1} - n_{2}] = n_{2} l_{2} - n_{1} l_{1}$
End correction, $d = \frac{n_{2} l_{2} - n_{1} l_{1}}{n_{1} - n_{2}}$

MHTCET - 2015

PHXI15:WAVES

354877 If the end correction of an open pipe is $0.8 c m$ , then the inner radius of that pipe will be

1

\frac{1}{3} c m

2

\frac{2}{3} c m

3

\frac{3}{2} c m

4

0.2 c m

Explanation:

For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by
$Δ l = 1.2 \times r$
[ $Δ l =$ end correction, $r =$ inner radius $]$
So, $r = \frac{Δ l}{1.2} = \frac{0.8}{1.2} [∵ Δ l = 0.8]$
$= \frac{2}{3} c m$

MHTCET - 2016

PHXI15:WAVES

354878 In a resonance pipe the first and second resonances are obtained at depths $22.7 c m$ and $70.2 c m$ respectively. The end correction will be

1

115.5 c m

2

1.05 c m

3

92.5 c m

4

113.5 c m

Explanation:

For end correction $x = 3$
$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (I_{2} + x)} x = \frac{l_{2} - 3 l_{1}}{2}$
$= \frac{70.2 - 3 \times 22.7}{2} = 1.05 c m$

PHXI15:WAVES

354879 A tube of diameter $d$ and of length $l$ unit is open at both ends. Its fundamental frequency of resonance is found to be $f_{1}$ . The velocity of sound in air is $330 m / \sec$ . One endof tube is now closed. The lowest frequency of resonance of tube is $f_{2}$ . Taking into consideration of end correction, $\frac{f_{2}}{f_{1}}$ is :

1

\frac{1}{2} \frac{(l + 0.3 d)}{(l + 0.6 d)}

2

\frac{(l + 0.6 d)}{(l + 0.3 d)}

3

\frac{1}{2} \frac{(l + 0.3 l)}{(l + 0.6 l)}

4

\frac{1}{2} \frac{(l + 0.6 d)}{(l + 0.3 d)}

Explanation:

supporting img

$\begin{aligned} f_{1} = \frac{v}{2 (l + 0.6 d)}, f_{2} = \frac{v}{4 (l + 0.3 d)} \\ \frac{f_{2}}{f_{1}} = \frac{l + 0.6 d}{2 (l + 0.3 d)} \end{aligned}$

PHXI15:WAVES

354880 In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is $0.1 m$ . When this length is changed to $0.35 m$ , the same tuning fork resonates with the first overnote. Calculate the end correction

1

0.025 m

2

0.012 m

3

0.024 m

4

0.05 m

Explanation:

Let $e$ be the end correction then according to question.
$\frac{v}{4 (l_{1} + e)} = \frac{3 v}{4 (l_{2} + e)} \Rightarrow e = 0.025 m$

PHXI15:WAVES

354881 In a pipe opened at both ends, $n_{1}$ and $n_{2}$ be the frequencies corresponding to vibrating lengths $l_{1}$ and $l_{2}$ respectively. The end correction is

1

\frac{n_{1} l_{1} - n_{2} l_{2}}{2 (n_{1} - n_{2})}

2

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{2} - n_{1})}

3

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{1} - n_{2})}

4

\frac{n_{2} l_{2} - n_{1} l_{1}}{(n_{1} - n_{2})}

Explanation:

Suppose pipe vibrates in fundamental mode,
For first pipe,
$l_{1} + d = \frac{λ_{1}}{2} (1)$
Where, $d$ is end correction,
Similarly for second pipe,
$l_{2} + d = \frac{λ_{2}}{2} (2)$
From eq.(1) and (2), we get
$\frac{l_{1} + d}{l_{2} + d} = \frac{λ_{1}}{λ_{2}} = \frac{1}{n_{1}} \times n^{2}$
$\Rightarrow n_{1} l_{1} + n_{1} d = n_{2} l_{2} + n_{2} d$
$\Rightarrow d [n_{1} - n_{2}] = n_{2} l_{2} - n_{1} l_{1}$
End correction, $d = \frac{n_{2} l_{2} - n_{1} l_{1}}{n_{1} - n_{2}}$

MHTCET - 2015

PHXI15:WAVES

354877 If the end correction of an open pipe is $0.8 c m$ , then the inner radius of that pipe will be

1

\frac{1}{3} c m

2

\frac{2}{3} c m

3

\frac{3}{2} c m

4

0.2 c m

Explanation:

For an open organ pipe, the relation between end correction and inner radius of the organ pipe is given by
$Δ l = 1.2 \times r$
[ $Δ l =$ end correction, $r =$ inner radius $]$
So, $r = \frac{Δ l}{1.2} = \frac{0.8}{1.2} [∵ Δ l = 0.8]$
$= \frac{2}{3} c m$

MHTCET - 2016

PHXI15:WAVES

354878 In a resonance pipe the first and second resonances are obtained at depths $22.7 c m$ and $70.2 c m$ respectively. The end correction will be

1

115.5 c m

2

1.05 c m

3

92.5 c m

4

113.5 c m

Explanation:

For end correction $x = 3$
$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (I_{2} + x)} x = \frac{l_{2} - 3 l_{1}}{2}$
$= \frac{70.2 - 3 \times 22.7}{2} = 1.05 c m$

PHXI15:WAVES

354879 A tube of diameter $d$ and of length $l$ unit is open at both ends. Its fundamental frequency of resonance is found to be $f_{1}$ . The velocity of sound in air is $330 m / \sec$ . One endof tube is now closed. The lowest frequency of resonance of tube is $f_{2}$ . Taking into consideration of end correction, $\frac{f_{2}}{f_{1}}$ is :

1

\frac{1}{2} \frac{(l + 0.3 d)}{(l + 0.6 d)}

2

\frac{(l + 0.6 d)}{(l + 0.3 d)}

3

\frac{1}{2} \frac{(l + 0.3 l)}{(l + 0.6 l)}

4

\frac{1}{2} \frac{(l + 0.6 d)}{(l + 0.3 d)}

Explanation:

supporting img

$\begin{aligned} f_{1} = \frac{v}{2 (l + 0.6 d)}, f_{2} = \frac{v}{4 (l + 0.3 d)} \\ \frac{f_{2}}{f_{1}} = \frac{l + 0.6 d}{2 (l + 0.3 d)} \end{aligned}$

PHXI15:WAVES

354880 In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is $0.1 m$ . When this length is changed to $0.35 m$ , the same tuning fork resonates with the first overnote. Calculate the end correction

1

0.025 m

2

0.012 m

3

0.024 m

4

0.05 m

Explanation:

Let $e$ be the end correction then according to question.
$\frac{v}{4 (l_{1} + e)} = \frac{3 v}{4 (l_{2} + e)} \Rightarrow e = 0.025 m$

PHXI15:WAVES

354881 In a pipe opened at both ends, $n_{1}$ and $n_{2}$ be the frequencies corresponding to vibrating lengths $l_{1}$ and $l_{2}$ respectively. The end correction is

1

\frac{n_{1} l_{1} - n_{2} l_{2}}{2 (n_{1} - n_{2})}

2

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{2} - n_{1})}

3

\frac{n_{2} l_{2} - n_{1} l_{1}}{2 (n_{1} - n_{2})}

4

\frac{n_{2} l_{2} - n_{1} l_{1}}{(n_{1} - n_{2})}

Explanation:

Suppose pipe vibrates in fundamental mode,
For first pipe,
$l_{1} + d = \frac{λ_{1}}{2} (1)$
Where, $d$ is end correction,
Similarly for second pipe,
$l_{2} + d = \frac{λ_{2}}{2} (2)$
From eq.(1) and (2), we get
$\frac{l_{1} + d}{l_{2} + d} = \frac{λ_{1}}{λ_{2}} = \frac{1}{n_{1}} \times n^{2}$
$\Rightarrow n_{1} l_{1} + n_{1} d = n_{2} l_{2} + n_{2} d$
$\Rightarrow d [n_{1} - n_{2}] = n_{2} l_{2} - n_{1} l_{1}$
End correction, $d = \frac{n_{2} l_{2} - n_{1} l_{1}}{n_{1} - n_{2}}$

MHTCET - 2015