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PHXI15:WAVES

354920 A pipe closed at one end has length $83 c m$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 H z$ are ( take, velocity of sound in air $= 332 m / s$ )

1

3

2

4

3

5

4

6

Explanation:

Fundamental frequency of one end closed pipe,
$v = \frac{n v}{4 L}$
Here, $n = 1$
$or v_{1} = \frac{332}{4 \times 83 \times 10^{- 2}}$
or $v_{1} = 100$
As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are
$3 v_{1} = 300 H z, 5 v_{2} = 500 H z$
$7 v_{2} = 700 H z, 9 v_{2} = 900 H z$
Hence, the number of possible natural oscillations of air column is 5

MHTCET - 2018

PHXI15:WAVES

354921 A closed organ pipe has length ' $l$ '. The air in it is vibrating in $3^{rd}$ overtone with maximum displacement amplitude $a$ . The displacement amplitude at a distance of $\frac{l}{7}$ from closed end of the pipe is equal to

1

a

2

\frac{a}{2}

3

\frac{a \sqrt{3}}{2}

4 Zero

Explanation:

The figure shows variation of displacement of particles in a closed organ pipe for $3^{rd}$ overtone. For third overtone $l = \frac{7 λ}{4}$ or $λ = \frac{4 l}{7}$ or $\frac{λ}{4} = \frac{l}{7}$
supporting img

$l = \frac{7 λ}{4}$
Hence the amplitude at $P$ at a distance $\frac{l}{7}$ from closed end is $a$ because there is an antinode at that point.

PHXI15:WAVES

354922 The waves set up in a closed pipe are

1 longitudinal and progressive

2 longitudinal and stationary

3 transverse and stationary

4 transverse and progressive.

PHXI15:WAVES

354923 A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $P_{0}$ inside the tube. If the atmospheric pressure is $P_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be

1

\frac{(P_{A} + 2 P_{0})}{(P_{A} - 2 P_{0})}

2

\frac{(P_{A} + P_{0})}{(P_{A} - P_{0})}

3

\frac{(P_{A} + \frac{1}{2} P_{0})}{(P_{A} - \frac{1}{2} P_{0})}

4

\frac{P_{A}}{P_{0}}

Explanation:

Maximum pressure at closed end will be atmospheric pressure adding with acoustic wave pressure
So $P_{max} = P_{A} + P_{0}$ and $P_{min} = P_{A} - P_{0}$
Thus $\frac{P_{max}}{P_{min}} = \frac{P_{A} + P_{0}}{P_{A} - P_{0}}$

PHXI15:WAVES

354924 In a resonance tube experiment, a closed organ pipe of length $120 c m$ is used and tuned with a tuning fork of frequency $340 H z$ . If water is poured into the pipe, then incorrect statement among the following [velocity of sound in air is $340 m / s$ , neglect end correction]

1 Minimum length of water column to have the resonance is

45 c m

2 The distance between two successive nodes is

50 c m

3 The maximum length of water column to create the resonance is

95 c m

4 The distance between two successive nodes is

25 c m

Explanation:

From $v = f λ$
$\Rightarrow 340 = 340 \times λ = λ = 1 m$
For $1^{st}$ resonance, $l_{1} = \frac{λ}{4}$
$l_{1} = 25 c m$ , so length of water column is $95 c m$ .
For $2^{nd}$ resonance, $l_{2} = \frac{3 λ}{4} = 75 c m$ ,
Separation between consecutive nodes is, $l_{2} - l_{1} = 50 c m$

PHXI15:WAVES

354920 A pipe closed at one end has length $83 c m$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 H z$ are ( take, velocity of sound in air $= 332 m / s$ )

1

3

2

4

3

5

4

6

Explanation:

Fundamental frequency of one end closed pipe,
$v = \frac{n v}{4 L}$
Here, $n = 1$
$or v_{1} = \frac{332}{4 \times 83 \times 10^{- 2}}$
or $v_{1} = 100$
As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are
$3 v_{1} = 300 H z, 5 v_{2} = 500 H z$
$7 v_{2} = 700 H z, 9 v_{2} = 900 H z$
Hence, the number of possible natural oscillations of air column is 5

MHTCET - 2018

PHXI15:WAVES

354921 A closed organ pipe has length ' $l$ '. The air in it is vibrating in $3^{rd}$ overtone with maximum displacement amplitude $a$ . The displacement amplitude at a distance of $\frac{l}{7}$ from closed end of the pipe is equal to

1

a

2

\frac{a}{2}

3

\frac{a \sqrt{3}}{2}

4 Zero

Explanation:

The figure shows variation of displacement of particles in a closed organ pipe for $3^{rd}$ overtone. For third overtone $l = \frac{7 λ}{4}$ or $λ = \frac{4 l}{7}$ or $\frac{λ}{4} = \frac{l}{7}$
supporting img

$l = \frac{7 λ}{4}$
Hence the amplitude at $P$ at a distance $\frac{l}{7}$ from closed end is $a$ because there is an antinode at that point.

PHXI15:WAVES

354922 The waves set up in a closed pipe are

1 longitudinal and progressive

2 longitudinal and stationary

3 transverse and stationary

4 transverse and progressive.

PHXI15:WAVES

354923 A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $P_{0}$ inside the tube. If the atmospheric pressure is $P_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be

1

\frac{(P_{A} + 2 P_{0})}{(P_{A} - 2 P_{0})}

2

\frac{(P_{A} + P_{0})}{(P_{A} - P_{0})}

3

\frac{(P_{A} + \frac{1}{2} P_{0})}{(P_{A} - \frac{1}{2} P_{0})}

4

\frac{P_{A}}{P_{0}}

Explanation:

Maximum pressure at closed end will be atmospheric pressure adding with acoustic wave pressure
So $P_{max} = P_{A} + P_{0}$ and $P_{min} = P_{A} - P_{0}$
Thus $\frac{P_{max}}{P_{min}} = \frac{P_{A} + P_{0}}{P_{A} - P_{0}}$

PHXI15:WAVES

354924 In a resonance tube experiment, a closed organ pipe of length $120 c m$ is used and tuned with a tuning fork of frequency $340 H z$ . If water is poured into the pipe, then incorrect statement among the following [velocity of sound in air is $340 m / s$ , neglect end correction]

1 Minimum length of water column to have the resonance is

45 c m

2 The distance between two successive nodes is

50 c m

3 The maximum length of water column to create the resonance is

95 c m

4 The distance between two successive nodes is

25 c m

Explanation:

From $v = f λ$
$\Rightarrow 340 = 340 \times λ = λ = 1 m$
For $1^{st}$ resonance, $l_{1} = \frac{λ}{4}$
$l_{1} = 25 c m$ , so length of water column is $95 c m$ .
For $2^{nd}$ resonance, $l_{2} = \frac{3 λ}{4} = 75 c m$ ,
Separation between consecutive nodes is, $l_{2} - l_{1} = 50 c m$

PHXI15:WAVES

354920 A pipe closed at one end has length $83 c m$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 H z$ are ( take, velocity of sound in air $= 332 m / s$ )

1

3

2

4

3

5

4

6

Explanation:

Fundamental frequency of one end closed pipe,
$v = \frac{n v}{4 L}$
Here, $n = 1$
$or v_{1} = \frac{332}{4 \times 83 \times 10^{- 2}}$
or $v_{1} = 100$
As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are
$3 v_{1} = 300 H z, 5 v_{2} = 500 H z$
$7 v_{2} = 700 H z, 9 v_{2} = 900 H z$
Hence, the number of possible natural oscillations of air column is 5

MHTCET - 2018

PHXI15:WAVES

354921 A closed organ pipe has length ' $l$ '. The air in it is vibrating in $3^{rd}$ overtone with maximum displacement amplitude $a$ . The displacement amplitude at a distance of $\frac{l}{7}$ from closed end of the pipe is equal to

1

a

2

\frac{a}{2}

3

\frac{a \sqrt{3}}{2}

4 Zero

Explanation:

The figure shows variation of displacement of particles in a closed organ pipe for $3^{rd}$ overtone. For third overtone $l = \frac{7 λ}{4}$ or $λ = \frac{4 l}{7}$ or $\frac{λ}{4} = \frac{l}{7}$
supporting img

$l = \frac{7 λ}{4}$
Hence the amplitude at $P$ at a distance $\frac{l}{7}$ from closed end is $a$ because there is an antinode at that point.

PHXI15:WAVES

354922 The waves set up in a closed pipe are

1 longitudinal and progressive

2 longitudinal and stationary

3 transverse and stationary

4 transverse and progressive.

PHXI15:WAVES

354923 A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $P_{0}$ inside the tube. If the atmospheric pressure is $P_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be

1

\frac{(P_{A} + 2 P_{0})}{(P_{A} - 2 P_{0})}

2

\frac{(P_{A} + P_{0})}{(P_{A} - P_{0})}

3

\frac{(P_{A} + \frac{1}{2} P_{0})}{(P_{A} - \frac{1}{2} P_{0})}

4

\frac{P_{A}}{P_{0}}

Explanation:

Maximum pressure at closed end will be atmospheric pressure adding with acoustic wave pressure
So $P_{max} = P_{A} + P_{0}$ and $P_{min} = P_{A} - P_{0}$
Thus $\frac{P_{max}}{P_{min}} = \frac{P_{A} + P_{0}}{P_{A} - P_{0}}$

PHXI15:WAVES

354924 In a resonance tube experiment, a closed organ pipe of length $120 c m$ is used and tuned with a tuning fork of frequency $340 H z$ . If water is poured into the pipe, then incorrect statement among the following [velocity of sound in air is $340 m / s$ , neglect end correction]

1 Minimum length of water column to have the resonance is

45 c m

2 The distance between two successive nodes is

50 c m

3 The maximum length of water column to create the resonance is

95 c m

4 The distance between two successive nodes is

25 c m

Explanation:

From $v = f λ$
$\Rightarrow 340 = 340 \times λ = λ = 1 m$
For $1^{st}$ resonance, $l_{1} = \frac{λ}{4}$
$l_{1} = 25 c m$ , so length of water column is $95 c m$ .
For $2^{nd}$ resonance, $l_{2} = \frac{3 λ}{4} = 75 c m$ ,
Separation between consecutive nodes is, $l_{2} - l_{1} = 50 c m$

PHXI15:WAVES

354920 A pipe closed at one end has length $83 c m$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 H z$ are ( take, velocity of sound in air $= 332 m / s$ )

1

3

2

4

3

5

4

6

Explanation:

Fundamental frequency of one end closed pipe,
$v = \frac{n v}{4 L}$
Here, $n = 1$
$or v_{1} = \frac{332}{4 \times 83 \times 10^{- 2}}$
or $v_{1} = 100$
As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are
$3 v_{1} = 300 H z, 5 v_{2} = 500 H z$
$7 v_{2} = 700 H z, 9 v_{2} = 900 H z$
Hence, the number of possible natural oscillations of air column is 5

MHTCET - 2018

PHXI15:WAVES

354921 A closed organ pipe has length ' $l$ '. The air in it is vibrating in $3^{rd}$ overtone with maximum displacement amplitude $a$ . The displacement amplitude at a distance of $\frac{l}{7}$ from closed end of the pipe is equal to

1

a

2

\frac{a}{2}

3

\frac{a \sqrt{3}}{2}

4 Zero

Explanation:

The figure shows variation of displacement of particles in a closed organ pipe for $3^{rd}$ overtone. For third overtone $l = \frac{7 λ}{4}$ or $λ = \frac{4 l}{7}$ or $\frac{λ}{4} = \frac{l}{7}$
supporting img

$l = \frac{7 λ}{4}$
Hence the amplitude at $P$ at a distance $\frac{l}{7}$ from closed end is $a$ because there is an antinode at that point.

PHXI15:WAVES

354922 The waves set up in a closed pipe are

1 longitudinal and progressive

2 longitudinal and stationary

3 transverse and stationary

4 transverse and progressive.

PHXI15:WAVES

354923 A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $P_{0}$ inside the tube. If the atmospheric pressure is $P_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be

1

\frac{(P_{A} + 2 P_{0})}{(P_{A} - 2 P_{0})}

2

\frac{(P_{A} + P_{0})}{(P_{A} - P_{0})}

3

\frac{(P_{A} + \frac{1}{2} P_{0})}{(P_{A} - \frac{1}{2} P_{0})}

4

\frac{P_{A}}{P_{0}}

Explanation:

Maximum pressure at closed end will be atmospheric pressure adding with acoustic wave pressure
So $P_{max} = P_{A} + P_{0}$ and $P_{min} = P_{A} - P_{0}$
Thus $\frac{P_{max}}{P_{min}} = \frac{P_{A} + P_{0}}{P_{A} - P_{0}}$

PHXI15:WAVES

354924 In a resonance tube experiment, a closed organ pipe of length $120 c m$ is used and tuned with a tuning fork of frequency $340 H z$ . If water is poured into the pipe, then incorrect statement among the following [velocity of sound in air is $340 m / s$ , neglect end correction]

1 Minimum length of water column to have the resonance is

45 c m

2 The distance between two successive nodes is

50 c m

3 The maximum length of water column to create the resonance is

95 c m

4 The distance between two successive nodes is

25 c m

Explanation:

From $v = f λ$
$\Rightarrow 340 = 340 \times λ = λ = 1 m$
For $1^{st}$ resonance, $l_{1} = \frac{λ}{4}$
$l_{1} = 25 c m$ , so length of water column is $95 c m$ .
For $2^{nd}$ resonance, $l_{2} = \frac{3 λ}{4} = 75 c m$ ,
Separation between consecutive nodes is, $l_{2} - l_{1} = 50 c m$

PHXI15:WAVES

354920 A pipe closed at one end has length $83 c m$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 H z$ are ( take, velocity of sound in air $= 332 m / s$ )

1

3

2

4

3

5

4

6

Explanation:

Fundamental frequency of one end closed pipe,
$v = \frac{n v}{4 L}$
Here, $n = 1$
$or v_{1} = \frac{332}{4 \times 83 \times 10^{- 2}}$
or $v_{1} = 100$
As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are
$3 v_{1} = 300 H z, 5 v_{2} = 500 H z$
$7 v_{2} = 700 H z, 9 v_{2} = 900 H z$
Hence, the number of possible natural oscillations of air column is 5

MHTCET - 2018

PHXI15:WAVES

354921 A closed organ pipe has length ' $l$ '. The air in it is vibrating in $3^{rd}$ overtone with maximum displacement amplitude $a$ . The displacement amplitude at a distance of $\frac{l}{7}$ from closed end of the pipe is equal to

1

a

2

\frac{a}{2}

3

\frac{a \sqrt{3}}{2}

4 Zero

Explanation:

The figure shows variation of displacement of particles in a closed organ pipe for $3^{rd}$ overtone. For third overtone $l = \frac{7 λ}{4}$ or $λ = \frac{4 l}{7}$ or $\frac{λ}{4} = \frac{l}{7}$
supporting img

$l = \frac{7 λ}{4}$
Hence the amplitude at $P$ at a distance $\frac{l}{7}$ from closed end is $a$ because there is an antinode at that point.

PHXI15:WAVES

354922 The waves set up in a closed pipe are

1 longitudinal and progressive

2 longitudinal and stationary

3 transverse and stationary

4 transverse and progressive.

PHXI15:WAVES

354923 A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $P_{0}$ inside the tube. If the atmospheric pressure is $P_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be

1

\frac{(P_{A} + 2 P_{0})}{(P_{A} - 2 P_{0})}

2

\frac{(P_{A} + P_{0})}{(P_{A} - P_{0})}

3

\frac{(P_{A} + \frac{1}{2} P_{0})}{(P_{A} - \frac{1}{2} P_{0})}

4

\frac{P_{A}}{P_{0}}

Explanation:

Maximum pressure at closed end will be atmospheric pressure adding with acoustic wave pressure
So $P_{max} = P_{A} + P_{0}$ and $P_{min} = P_{A} - P_{0}$
Thus $\frac{P_{max}}{P_{min}} = \frac{P_{A} + P_{0}}{P_{A} - P_{0}}$

PHXI15:WAVES

354924 In a resonance tube experiment, a closed organ pipe of length $120 c m$ is used and tuned with a tuning fork of frequency $340 H z$ . If water is poured into the pipe, then incorrect statement among the following [velocity of sound in air is $340 m / s$ , neglect end correction]

1 Minimum length of water column to have the resonance is

45 c m

2 The distance between two successive nodes is

50 c m

3 The maximum length of water column to create the resonance is

95 c m

4 The distance between two successive nodes is

25 c m

Explanation:

From $v = f λ$
$\Rightarrow 340 = 340 \times λ = λ = 1 m$
For $1^{st}$ resonance, $l_{1} = \frac{λ}{4}$
$l_{1} = 25 c m$ , so length of water column is $95 c m$ .
For $2^{nd}$ resonance, $l_{2} = \frac{3 λ}{4} = 75 c m$ ,
Separation between consecutive nodes is, $l_{2} - l_{1} = 50 c m$

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