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PHXI15:WAVES

354937 A metre-long open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (A tunning fork of frequency $340 H z$ ) when the tube length is 25.5 $c m$ or $79.3 c m$ . Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

1

336 m / s

2

331 m / s

3

356 m / s

4

366 m / s

Explanation:

Length at which first resonance occurs $l_{1} = 25.5 c m = 25.5 \times 10^{- 2} c m$
Length at which second resonance occurs $l_{2} = 793 m$
$∴$ Wave length $λ = 2 (l_{2} - l_{1})$
$= 2 (793 - 25.5) = 2 \times 53.8 = 107.6 c m = 1.076 m$
Using $v = f λ, f = 340 H z$
$∴ Speed of sound in air = 365.84 m / s$

NCERT Exemplar

PHXI15:WAVES

354938 For a closed organ pipe, the fundamental frequency is $100 H z$ . The $5^{th}$ overtone is

1

1100 H z

2

900 H z

3

500 H z

4

600 H z

Explanation:

$f_{n} = (2 n + 1) f_{0}$
$= (2 \times 5 + 1) \times 100 = 1100 H z$

PHXI15:WAVES

354939 Two closed organ pipes $A$ and $B$ , have the same length. $A$ is wider than $B$ . The resonate in fundamental mode at frequencies $n_{A}$ and $n_{B}$ respectively then

1

n_{A} > n_{B}

2

n_{A} < n_{B}

3 Either (1) or (2) depending on the ratio of their diameters

4

n_{A} = n_{B}

Explanation:

In closed organ pipe. First resonance occurs at $λ / 4$ .
So, in fundamental mode of vibration of organ pipe $\frac{λ}{4} = (l + 0.3 d)$
Where $0.3 d$ is end correction.
Frequency of vibration $n = \frac{v}{λ} = \frac{v}{4 (l + 0.3 d)}$
As $l$ is same, wider pipe $A$ will resonate at a lower frequency, i.e., $n_{A} < n_{B}$ .

PHXI15:WAVES

354940 While measuring the speed of sound by performing a resonance column experiment a student gets the first resonance condition at a column length of $18 c m$ during winter. Repeating the same experiment during summer, she measures the coulmn length to be $x c m$ for the, second resonance. Then

1

18 > x

2

x > 54

3

54 > x > 36

4

36 > x > 18

Explanation:

In winter fundamental frequency
$f_{1} = \frac{v}{4 l_{1}}$
First overtone $f_{3} = \frac{3 v}{4 l_{3}}$
As $f_{1} = f_{3}$ or $\frac{v}{4 l_{1}} = \frac{3 v}{4 l_{3}}$
$l_{3} = 3 l_{1} = 54 c m$
In summer $f_{3}^{'} = \frac{3 v^{'}}{4 x}$
As $f_{3} = f_{3}^{'}$
$f_{3}^{'} = \frac{3 v}{4 (54 c m)} = \frac{3 v^{'}}{4 x}$
$x = \frac{v^{'}}{v} (54 c m)$ ]
In summer $v^{'} > v \Rightarrow x > 54 c m$

PHXI15:WAVES

354941 A tube with both ends closed has same set of natural frequency as

1 One end closed organ pipe

2 Both end open organ pipe

3 Vibratory string fixed at both ends

4 Vibratory string fixed at one end

PHXI15:WAVES

354937 A metre-long open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (A tunning fork of frequency $340 H z$ ) when the tube length is 25.5 $c m$ or $79.3 c m$ . Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

1

336 m / s

2

331 m / s

3

356 m / s

4

366 m / s

Explanation:

Length at which first resonance occurs $l_{1} = 25.5 c m = 25.5 \times 10^{- 2} c m$
Length at which second resonance occurs $l_{2} = 793 m$
$∴$ Wave length $λ = 2 (l_{2} - l_{1})$
$= 2 (793 - 25.5) = 2 \times 53.8 = 107.6 c m = 1.076 m$
Using $v = f λ, f = 340 H z$
$∴ Speed of sound in air = 365.84 m / s$

NCERT Exemplar

PHXI15:WAVES

354938 For a closed organ pipe, the fundamental frequency is $100 H z$ . The $5^{th}$ overtone is

1

1100 H z

2

900 H z

3

500 H z

4

600 H z

Explanation:

$f_{n} = (2 n + 1) f_{0}$
$= (2 \times 5 + 1) \times 100 = 1100 H z$

PHXI15:WAVES

354939 Two closed organ pipes $A$ and $B$ , have the same length. $A$ is wider than $B$ . The resonate in fundamental mode at frequencies $n_{A}$ and $n_{B}$ respectively then

1

n_{A} > n_{B}

2

n_{A} < n_{B}

3 Either (1) or (2) depending on the ratio of their diameters

4

n_{A} = n_{B}

Explanation:

In closed organ pipe. First resonance occurs at $λ / 4$ .
So, in fundamental mode of vibration of organ pipe $\frac{λ}{4} = (l + 0.3 d)$
Where $0.3 d$ is end correction.
Frequency of vibration $n = \frac{v}{λ} = \frac{v}{4 (l + 0.3 d)}$
As $l$ is same, wider pipe $A$ will resonate at a lower frequency, i.e., $n_{A} < n_{B}$ .

PHXI15:WAVES

354940 While measuring the speed of sound by performing a resonance column experiment a student gets the first resonance condition at a column length of $18 c m$ during winter. Repeating the same experiment during summer, she measures the coulmn length to be $x c m$ for the, second resonance. Then

1

18 > x

2

x > 54

3

54 > x > 36

4

36 > x > 18

Explanation:

In winter fundamental frequency
$f_{1} = \frac{v}{4 l_{1}}$
First overtone $f_{3} = \frac{3 v}{4 l_{3}}$
As $f_{1} = f_{3}$ or $\frac{v}{4 l_{1}} = \frac{3 v}{4 l_{3}}$
$l_{3} = 3 l_{1} = 54 c m$
In summer $f_{3}^{'} = \frac{3 v^{'}}{4 x}$
As $f_{3} = f_{3}^{'}$
$f_{3}^{'} = \frac{3 v}{4 (54 c m)} = \frac{3 v^{'}}{4 x}$
$x = \frac{v^{'}}{v} (54 c m)$ ]
In summer $v^{'} > v \Rightarrow x > 54 c m$

PHXI15:WAVES

354941 A tube with both ends closed has same set of natural frequency as

1 One end closed organ pipe

2 Both end open organ pipe

3 Vibratory string fixed at both ends

4 Vibratory string fixed at one end

PHXI15:WAVES

354937 A metre-long open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (A tunning fork of frequency $340 H z$ ) when the tube length is 25.5 $c m$ or $79.3 c m$ . Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

1

336 m / s

2

331 m / s

3

356 m / s

4

366 m / s

Explanation:

Length at which first resonance occurs $l_{1} = 25.5 c m = 25.5 \times 10^{- 2} c m$
Length at which second resonance occurs $l_{2} = 793 m$
$∴$ Wave length $λ = 2 (l_{2} - l_{1})$
$= 2 (793 - 25.5) = 2 \times 53.8 = 107.6 c m = 1.076 m$
Using $v = f λ, f = 340 H z$
$∴ Speed of sound in air = 365.84 m / s$

NCERT Exemplar

PHXI15:WAVES

354938 For a closed organ pipe, the fundamental frequency is $100 H z$ . The $5^{th}$ overtone is

1

1100 H z

2

900 H z

3

500 H z

4

600 H z

Explanation:

$f_{n} = (2 n + 1) f_{0}$
$= (2 \times 5 + 1) \times 100 = 1100 H z$

PHXI15:WAVES

354939 Two closed organ pipes $A$ and $B$ , have the same length. $A$ is wider than $B$ . The resonate in fundamental mode at frequencies $n_{A}$ and $n_{B}$ respectively then

1

n_{A} > n_{B}

2

n_{A} < n_{B}

3 Either (1) or (2) depending on the ratio of their diameters

4

n_{A} = n_{B}

Explanation:

In closed organ pipe. First resonance occurs at $λ / 4$ .
So, in fundamental mode of vibration of organ pipe $\frac{λ}{4} = (l + 0.3 d)$
Where $0.3 d$ is end correction.
Frequency of vibration $n = \frac{v}{λ} = \frac{v}{4 (l + 0.3 d)}$
As $l$ is same, wider pipe $A$ will resonate at a lower frequency, i.e., $n_{A} < n_{B}$ .

PHXI15:WAVES

354940 While measuring the speed of sound by performing a resonance column experiment a student gets the first resonance condition at a column length of $18 c m$ during winter. Repeating the same experiment during summer, she measures the coulmn length to be $x c m$ for the, second resonance. Then

1

18 > x

2

x > 54

3

54 > x > 36

4

36 > x > 18

Explanation:

In winter fundamental frequency
$f_{1} = \frac{v}{4 l_{1}}$
First overtone $f_{3} = \frac{3 v}{4 l_{3}}$
As $f_{1} = f_{3}$ or $\frac{v}{4 l_{1}} = \frac{3 v}{4 l_{3}}$
$l_{3} = 3 l_{1} = 54 c m$
In summer $f_{3}^{'} = \frac{3 v^{'}}{4 x}$
As $f_{3} = f_{3}^{'}$
$f_{3}^{'} = \frac{3 v}{4 (54 c m)} = \frac{3 v^{'}}{4 x}$
$x = \frac{v^{'}}{v} (54 c m)$ ]
In summer $v^{'} > v \Rightarrow x > 54 c m$

PHXI15:WAVES

354941 A tube with both ends closed has same set of natural frequency as

1 One end closed organ pipe

2 Both end open organ pipe

3 Vibratory string fixed at both ends

4 Vibratory string fixed at one end

PHXI15:WAVES

354937 A metre-long open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (A tunning fork of frequency $340 H z$ ) when the tube length is 25.5 $c m$ or $79.3 c m$ . Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

1

336 m / s

2

331 m / s

3

356 m / s

4

366 m / s

Explanation:

Length at which first resonance occurs $l_{1} = 25.5 c m = 25.5 \times 10^{- 2} c m$
Length at which second resonance occurs $l_{2} = 793 m$
$∴$ Wave length $λ = 2 (l_{2} - l_{1})$
$= 2 (793 - 25.5) = 2 \times 53.8 = 107.6 c m = 1.076 m$
Using $v = f λ, f = 340 H z$
$∴ Speed of sound in air = 365.84 m / s$

NCERT Exemplar

PHXI15:WAVES

354938 For a closed organ pipe, the fundamental frequency is $100 H z$ . The $5^{th}$ overtone is

1

1100 H z

2

900 H z

3

500 H z

4

600 H z

Explanation:

$f_{n} = (2 n + 1) f_{0}$
$= (2 \times 5 + 1) \times 100 = 1100 H z$

PHXI15:WAVES

354939 Two closed organ pipes $A$ and $B$ , have the same length. $A$ is wider than $B$ . The resonate in fundamental mode at frequencies $n_{A}$ and $n_{B}$ respectively then

1

n_{A} > n_{B}

2

n_{A} < n_{B}

3 Either (1) or (2) depending on the ratio of their diameters

4

n_{A} = n_{B}

Explanation:

In closed organ pipe. First resonance occurs at $λ / 4$ .
So, in fundamental mode of vibration of organ pipe $\frac{λ}{4} = (l + 0.3 d)$
Where $0.3 d$ is end correction.
Frequency of vibration $n = \frac{v}{λ} = \frac{v}{4 (l + 0.3 d)}$
As $l$ is same, wider pipe $A$ will resonate at a lower frequency, i.e., $n_{A} < n_{B}$ .

PHXI15:WAVES

354940 While measuring the speed of sound by performing a resonance column experiment a student gets the first resonance condition at a column length of $18 c m$ during winter. Repeating the same experiment during summer, she measures the coulmn length to be $x c m$ for the, second resonance. Then

1

18 > x

2

x > 54

3

54 > x > 36

4

36 > x > 18

Explanation:

In winter fundamental frequency
$f_{1} = \frac{v}{4 l_{1}}$
First overtone $f_{3} = \frac{3 v}{4 l_{3}}$
As $f_{1} = f_{3}$ or $\frac{v}{4 l_{1}} = \frac{3 v}{4 l_{3}}$
$l_{3} = 3 l_{1} = 54 c m$
In summer $f_{3}^{'} = \frac{3 v^{'}}{4 x}$
As $f_{3} = f_{3}^{'}$
$f_{3}^{'} = \frac{3 v}{4 (54 c m)} = \frac{3 v^{'}}{4 x}$
$x = \frac{v^{'}}{v} (54 c m)$ ]
In summer $v^{'} > v \Rightarrow x > 54 c m$

PHXI15:WAVES

354941 A tube with both ends closed has same set of natural frequency as

1 One end closed organ pipe

2 Both end open organ pipe

3 Vibratory string fixed at both ends

4 Vibratory string fixed at one end

PHXI15:WAVES

354937 A metre-long open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (A tunning fork of frequency $340 H z$ ) when the tube length is 25.5 $c m$ or $79.3 c m$ . Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

1

336 m / s

2

331 m / s

3

356 m / s

4

366 m / s

Explanation:

Length at which first resonance occurs $l_{1} = 25.5 c m = 25.5 \times 10^{- 2} c m$
Length at which second resonance occurs $l_{2} = 793 m$
$∴$ Wave length $λ = 2 (l_{2} - l_{1})$
$= 2 (793 - 25.5) = 2 \times 53.8 = 107.6 c m = 1.076 m$
Using $v = f λ, f = 340 H z$
$∴ Speed of sound in air = 365.84 m / s$

NCERT Exemplar

PHXI15:WAVES

354938 For a closed organ pipe, the fundamental frequency is $100 H z$ . The $5^{th}$ overtone is

1

1100 H z

2

900 H z

3

500 H z

4

600 H z

Explanation:

$f_{n} = (2 n + 1) f_{0}$
$= (2 \times 5 + 1) \times 100 = 1100 H z$

PHXI15:WAVES

354939 Two closed organ pipes $A$ and $B$ , have the same length. $A$ is wider than $B$ . The resonate in fundamental mode at frequencies $n_{A}$ and $n_{B}$ respectively then

1

n_{A} > n_{B}

2

n_{A} < n_{B}

3 Either (1) or (2) depending on the ratio of their diameters

4

n_{A} = n_{B}

Explanation:

In closed organ pipe. First resonance occurs at $λ / 4$ .
So, in fundamental mode of vibration of organ pipe $\frac{λ}{4} = (l + 0.3 d)$
Where $0.3 d$ is end correction.
Frequency of vibration $n = \frac{v}{λ} = \frac{v}{4 (l + 0.3 d)}$
As $l$ is same, wider pipe $A$ will resonate at a lower frequency, i.e., $n_{A} < n_{B}$ .

PHXI15:WAVES

354940 While measuring the speed of sound by performing a resonance column experiment a student gets the first resonance condition at a column length of $18 c m$ during winter. Repeating the same experiment during summer, she measures the coulmn length to be $x c m$ for the, second resonance. Then

1

18 > x

2

x > 54

3

54 > x > 36

4

36 > x > 18

Explanation:

In winter fundamental frequency
$f_{1} = \frac{v}{4 l_{1}}$
First overtone $f_{3} = \frac{3 v}{4 l_{3}}$
As $f_{1} = f_{3}$ or $\frac{v}{4 l_{1}} = \frac{3 v}{4 l_{3}}$
$l_{3} = 3 l_{1} = 54 c m$
In summer $f_{3}^{'} = \frac{3 v^{'}}{4 x}$
As $f_{3} = f_{3}^{'}$
$f_{3}^{'} = \frac{3 v}{4 (54 c m)} = \frac{3 v^{'}}{4 x}$
$x = \frac{v^{'}}{v} (54 c m)$ ]
In summer $v^{'} > v \Rightarrow x > 54 c m$

PHXI15:WAVES

354941 A tube with both ends closed has same set of natural frequency as

1 One end closed organ pipe

2 Both end open organ pipe

3 Vibratory string fixed at both ends

4 Vibratory string fixed at one end

Explanation:

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