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PHXI15:WAVES

354925 An organ pipe $P_{1}$ , closed at one end vibrating in its fundamental harmonic and another pipe $P_{2}$ , open at both ends vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the lengths of $P_{1}$ and $P_{2}$ is:

1

\frac{1}{6}

2

\frac{8}{3}

3

\frac{1}{3}

4

\frac{1}{2}

Explanation:

$\frac{v}{4 ℓ_{1}} = \frac{3 v}{2 ℓ_{2}}, ∴ \frac{ℓ_{1}}{ℓ_{2}} = \frac{1}{6}$

PHXI15:WAVES

354926 The two nearest harmonics of a tube closed at one end and open at other end are $220 H z$ and $260 H z$ . What is the fundamental frequency of the system?

1

20 H z

2

30 H z

3

40 H z

4

10 H z

Explanation:

Difference between any two consecutive frequencies of
Closed organ pipe, $\frac{2 v}{4 l} = 260 - 220 = 40 H z$
$\Rightarrow \frac{v}{4 l} = 20 H z$
So fundamental frequency $= 20 H z$

PHXI15:WAVES

354927 If in an experiment for determination of velocity of sound by resonance tube method using a tuning fork of $500 H z,$ first resonance was observed at $30.52 c m$ and second was obtained at $64.13 c m,$ then maximum possible error in velocity of sound is
(Take actual speed of sound $= 332 m / s$

1

3.5 m / s

2

2.7 m / s

3

5.7 m / s

4

4.1 m / s

Explanation:

Using, $λ = 2 (l_{2} - l_{1})$
$\Rightarrow v^{'} = 2 f (l_{2} - l_{1})$
$= 2 \times 500 (64.13 - 30.52)$
$= 1000 [33.61] c m / s$
$= 336.1 m / s$
Error in velocity,
$Δ v = v^{'} - v$
$= (336.1 - 332) m / s = 4.1 m / s$

PHXI15:WAVES

354928 In the case of closed end organ pipe

1 The maximum possible wavelength is same as that of open end organ pipe

2 The maximum possible wavelength is less than that of open end organ pipe

3 The maximum possible wavelength may be less than that of open end organ pipe

4 The maximum possible wavelength is greater than that of open end organ pipe

PHXI15:WAVES

354925 An organ pipe $P_{1}$ , closed at one end vibrating in its fundamental harmonic and another pipe $P_{2}$ , open at both ends vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the lengths of $P_{1}$ and $P_{2}$ is:

1

\frac{1}{6}

2

\frac{8}{3}

3

\frac{1}{3}

4

\frac{1}{2}

Explanation:

$\frac{v}{4 ℓ_{1}} = \frac{3 v}{2 ℓ_{2}}, ∴ \frac{ℓ_{1}}{ℓ_{2}} = \frac{1}{6}$

PHXI15:WAVES

354926 The two nearest harmonics of a tube closed at one end and open at other end are $220 H z$ and $260 H z$ . What is the fundamental frequency of the system?

1

20 H z

2

30 H z

3

40 H z

4

10 H z

Explanation:

Difference between any two consecutive frequencies of
Closed organ pipe, $\frac{2 v}{4 l} = 260 - 220 = 40 H z$
$\Rightarrow \frac{v}{4 l} = 20 H z$
So fundamental frequency $= 20 H z$

PHXI15:WAVES

354927 If in an experiment for determination of velocity of sound by resonance tube method using a tuning fork of $500 H z,$ first resonance was observed at $30.52 c m$ and second was obtained at $64.13 c m,$ then maximum possible error in velocity of sound is
(Take actual speed of sound $= 332 m / s$

1

3.5 m / s

2

2.7 m / s

3

5.7 m / s

4

4.1 m / s

Explanation:

Using, $λ = 2 (l_{2} - l_{1})$
$\Rightarrow v^{'} = 2 f (l_{2} - l_{1})$
$= 2 \times 500 (64.13 - 30.52)$
$= 1000 [33.61] c m / s$
$= 336.1 m / s$
Error in velocity,
$Δ v = v^{'} - v$
$= (336.1 - 332) m / s = 4.1 m / s$

PHXI15:WAVES

354928 In the case of closed end organ pipe

1 The maximum possible wavelength is same as that of open end organ pipe

2 The maximum possible wavelength is less than that of open end organ pipe

3 The maximum possible wavelength may be less than that of open end organ pipe

4 The maximum possible wavelength is greater than that of open end organ pipe

PHXI15:WAVES

354925 An organ pipe $P_{1}$ , closed at one end vibrating in its fundamental harmonic and another pipe $P_{2}$ , open at both ends vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the lengths of $P_{1}$ and $P_{2}$ is:

1

\frac{1}{6}

2

\frac{8}{3}

3

\frac{1}{3}

4

\frac{1}{2}

Explanation:

$\frac{v}{4 ℓ_{1}} = \frac{3 v}{2 ℓ_{2}}, ∴ \frac{ℓ_{1}}{ℓ_{2}} = \frac{1}{6}$

PHXI15:WAVES

354926 The two nearest harmonics of a tube closed at one end and open at other end are $220 H z$ and $260 H z$ . What is the fundamental frequency of the system?

1

20 H z

2

30 H z

3

40 H z

4

10 H z

Explanation:

Difference between any two consecutive frequencies of
Closed organ pipe, $\frac{2 v}{4 l} = 260 - 220 = 40 H z$
$\Rightarrow \frac{v}{4 l} = 20 H z$
So fundamental frequency $= 20 H z$

PHXI15:WAVES

354927 If in an experiment for determination of velocity of sound by resonance tube method using a tuning fork of $500 H z,$ first resonance was observed at $30.52 c m$ and second was obtained at $64.13 c m,$ then maximum possible error in velocity of sound is
(Take actual speed of sound $= 332 m / s$

1

3.5 m / s

2

2.7 m / s

3

5.7 m / s

4

4.1 m / s

Explanation:

Using, $λ = 2 (l_{2} - l_{1})$
$\Rightarrow v^{'} = 2 f (l_{2} - l_{1})$
$= 2 \times 500 (64.13 - 30.52)$
$= 1000 [33.61] c m / s$
$= 336.1 m / s$
Error in velocity,
$Δ v = v^{'} - v$
$= (336.1 - 332) m / s = 4.1 m / s$

PHXI15:WAVES

354928 In the case of closed end organ pipe

1 The maximum possible wavelength is same as that of open end organ pipe

2 The maximum possible wavelength is less than that of open end organ pipe

3 The maximum possible wavelength may be less than that of open end organ pipe

4 The maximum possible wavelength is greater than that of open end organ pipe

PHXI15:WAVES

354925 An organ pipe $P_{1}$ , closed at one end vibrating in its fundamental harmonic and another pipe $P_{2}$ , open at both ends vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the lengths of $P_{1}$ and $P_{2}$ is:

1

\frac{1}{6}

2

\frac{8}{3}

3

\frac{1}{3}

4

\frac{1}{2}

Explanation:

$\frac{v}{4 ℓ_{1}} = \frac{3 v}{2 ℓ_{2}}, ∴ \frac{ℓ_{1}}{ℓ_{2}} = \frac{1}{6}$

PHXI15:WAVES

354926 The two nearest harmonics of a tube closed at one end and open at other end are $220 H z$ and $260 H z$ . What is the fundamental frequency of the system?

1

20 H z

2

30 H z

3

40 H z

4

10 H z

Explanation:

Difference between any two consecutive frequencies of
Closed organ pipe, $\frac{2 v}{4 l} = 260 - 220 = 40 H z$
$\Rightarrow \frac{v}{4 l} = 20 H z$
So fundamental frequency $= 20 H z$

PHXI15:WAVES

354927 If in an experiment for determination of velocity of sound by resonance tube method using a tuning fork of $500 H z,$ first resonance was observed at $30.52 c m$ and second was obtained at $64.13 c m,$ then maximum possible error in velocity of sound is
(Take actual speed of sound $= 332 m / s$

1

3.5 m / s

2

2.7 m / s

3

5.7 m / s

4

4.1 m / s

Explanation:

Using, $λ = 2 (l_{2} - l_{1})$
$\Rightarrow v^{'} = 2 f (l_{2} - l_{1})$
$= 2 \times 500 (64.13 - 30.52)$
$= 1000 [33.61] c m / s$
$= 336.1 m / s$
Error in velocity,
$Δ v = v^{'} - v$
$= (336.1 - 332) m / s = 4.1 m / s$

PHXI15:WAVES

354928 In the case of closed end organ pipe

1 The maximum possible wavelength is same as that of open end organ pipe

2 The maximum possible wavelength is less than that of open end organ pipe

3 The maximum possible wavelength may be less than that of open end organ pipe

4 The maximum possible wavelength is greater than that of open end organ pipe

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