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WAVES

172573 A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

1

3 : 4

2

2 : 3

3

1 : 2

4

4 : 5

Explanation:

A For frequency of first overtone of a closed $pipe (f) = \frac{3 v}{4 l}$
For frequency of first overtone of an open $pipe (f^{'}) = \frac{V}{l^{'}}$
According to the question their first overtone are identical-
$f = f^{'}$
$\frac{3 v}{4 l} = \frac{v}{l^{'}}$
$\frac{l}{l^{'}} = \frac{3}{4}$
Their lengths are in the ratio $l : l^{'} = 3 : 4$

MHT-CET 2020

WAVES

172574 A pipe open at one end has length $0.8 m$ . At the open end of the tube a string $0.5 m$ long is vibrating in its $1^{st}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 N$ , the mass of string is (speed of sound $= 320 m / s$ )

1 15 gram

2 20 gram

3 10 gram

4 25 gram

Explanation:

C Given that,
$v = 320 m / s$ , length of pipe $(L) = 0.8 m$ , length of string $(l) = 0.5 m$
The fundamental frequency of pipe open at one end is given by,
$f = \frac{v}{4 L} = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 Hz$
$∴$ The first overtone of the string $= 100 Hz$
$∴$ Fundamental frequency of the string $= \frac{100}{2} = 50 Hz$ For frequency of string $(f_{1}) = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$50 = \frac{1}{2 \times 0.5} \sqrt{\frac{50}{μ}}$
$50 = \sqrt{\frac{50}{μ}}$
Squaring both side, we get-
$(50)^{2} = \frac{50}{μ}$
$μ = \frac{1}{50} kg / m$
Hence, mass of string $(m) = μ L = \frac{1}{50} \times 0.5 = 10^{- 2} kg = 10 g$

MHT-CET 2020]**#IIT JEE-2010

WAVES

172575 On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 Hz$ more than fundamental frequency of vibration in open organ pipe. The fundamental frequency of open organ pipe is

1

250 Hz

2

100 Hz

3

50 Hz

4

200 Hz

Explanation:

B Fundamental frequency of closed pipe $(f_{1}) = \frac{V}{4 l}$
Frequency of open $pipe (f_{2}) = \frac{V}{2 l}$
Frequency of third harmonic $(f_{3}) = \frac{3 v}{4 l}$
$∴$ Difference between frequency of third harmonic and frequency of open $(f_{3} - f_{2}) = 50 Hz$
$\frac{3 v}{4 l} - \frac{v}{2 l} = 50$
$\frac{v}{4 l} = 50$
$\frac{v}{2 l} = 100 Hz$

MHT-CET 2020

WAVES

172576 The air column in pipe which is closed at one end will be in resonance with a vibrating turning fork at a frequency of $260 Hz$ , if the length of the air column is-

1

31.73 cm

2

62.5 cm

3

35.75 cm

4

12.5 cm

Explanation:

A One end closed air pipe acts as a closed organ pipe.
$∴ f = \frac{v}{4 l}$
$260 = \frac{330}{4 l} (given, f = 260 Hz}$
$l = \frac{33}{26 \times 4}$
$l = 0.3173 m$
$l = 31.73 cm$

BCECE-2013

WAVES

172577 If two air columns of lengths $100 cm$ and $101 cm$ sounding in their fundamental note gave 17 beats in 20 seconds, then the velocity of sound will be

1

277.8 m / s

2

300 m / s

3

250 m / s

4

343.4 m / s

Explanation:

D Given, $l_{1} = 100 cm = 1 m, l_{2} = 101 cm = 1.01 m$ We know that,
$f_{1} = \frac{v}{4 l_{1}}$
$f_{2} = \frac{v}{4 l_{2}}$
$f_{1} - f_{2} = \frac{17}{20}$
$\frac{v}{4 l_{1}} - \frac{v}{4 l_{2}} = \frac{17}{20}$
$v (\frac{1}{1} - \frac{1}{1.01}) = \frac{17}{5}$
$v = \frac{17}{5} \times \frac{1.01}{0.01} = 343.4 m / s$

JCECE-2013

WAVES

172573 A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

1

3 : 4

2

2 : 3

3

1 : 2

4

4 : 5

Explanation:

A For frequency of first overtone of a closed $pipe (f) = \frac{3 v}{4 l}$
For frequency of first overtone of an open $pipe (f^{'}) = \frac{V}{l^{'}}$
According to the question their first overtone are identical-
$f = f^{'}$
$\frac{3 v}{4 l} = \frac{v}{l^{'}}$
$\frac{l}{l^{'}} = \frac{3}{4}$
Their lengths are in the ratio $l : l^{'} = 3 : 4$

MHT-CET 2020

WAVES

172574 A pipe open at one end has length $0.8 m$ . At the open end of the tube a string $0.5 m$ long is vibrating in its $1^{st}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 N$ , the mass of string is (speed of sound $= 320 m / s$ )

1 15 gram

2 20 gram

3 10 gram

4 25 gram

Explanation:

C Given that,
$v = 320 m / s$ , length of pipe $(L) = 0.8 m$ , length of string $(l) = 0.5 m$
The fundamental frequency of pipe open at one end is given by,
$f = \frac{v}{4 L} = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 Hz$
$∴$ The first overtone of the string $= 100 Hz$
$∴$ Fundamental frequency of the string $= \frac{100}{2} = 50 Hz$ For frequency of string $(f_{1}) = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$50 = \frac{1}{2 \times 0.5} \sqrt{\frac{50}{μ}}$
$50 = \sqrt{\frac{50}{μ}}$
Squaring both side, we get-
$(50)^{2} = \frac{50}{μ}$
$μ = \frac{1}{50} kg / m$
Hence, mass of string $(m) = μ L = \frac{1}{50} \times 0.5 = 10^{- 2} kg = 10 g$

MHT-CET 2020]**#IIT JEE-2010

WAVES

172575 On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 Hz$ more than fundamental frequency of vibration in open organ pipe. The fundamental frequency of open organ pipe is

1

250 Hz

2

100 Hz

3

50 Hz

4

200 Hz

Explanation:

B Fundamental frequency of closed pipe $(f_{1}) = \frac{V}{4 l}$
Frequency of open $pipe (f_{2}) = \frac{V}{2 l}$
Frequency of third harmonic $(f_{3}) = \frac{3 v}{4 l}$
$∴$ Difference between frequency of third harmonic and frequency of open $(f_{3} - f_{2}) = 50 Hz$
$\frac{3 v}{4 l} - \frac{v}{2 l} = 50$
$\frac{v}{4 l} = 50$
$\frac{v}{2 l} = 100 Hz$

MHT-CET 2020

WAVES

172576 The air column in pipe which is closed at one end will be in resonance with a vibrating turning fork at a frequency of $260 Hz$ , if the length of the air column is-

1

31.73 cm

2

62.5 cm

3

35.75 cm

4

12.5 cm

Explanation:

A One end closed air pipe acts as a closed organ pipe.
$∴ f = \frac{v}{4 l}$
$260 = \frac{330}{4 l} (given, f = 260 Hz}$
$l = \frac{33}{26 \times 4}$
$l = 0.3173 m$
$l = 31.73 cm$

BCECE-2013

WAVES

172577 If two air columns of lengths $100 cm$ and $101 cm$ sounding in their fundamental note gave 17 beats in 20 seconds, then the velocity of sound will be

1

277.8 m / s

2

300 m / s

3

250 m / s

4

343.4 m / s

Explanation:

D Given, $l_{1} = 100 cm = 1 m, l_{2} = 101 cm = 1.01 m$ We know that,
$f_{1} = \frac{v}{4 l_{1}}$
$f_{2} = \frac{v}{4 l_{2}}$
$f_{1} - f_{2} = \frac{17}{20}$
$\frac{v}{4 l_{1}} - \frac{v}{4 l_{2}} = \frac{17}{20}$
$v (\frac{1}{1} - \frac{1}{1.01}) = \frac{17}{5}$
$v = \frac{17}{5} \times \frac{1.01}{0.01} = 343.4 m / s$

JCECE-2013

WAVES

172573 A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

1

3 : 4

2

2 : 3

3

1 : 2

4

4 : 5

Explanation:

A For frequency of first overtone of a closed $pipe (f) = \frac{3 v}{4 l}$
For frequency of first overtone of an open $pipe (f^{'}) = \frac{V}{l^{'}}$
According to the question their first overtone are identical-
$f = f^{'}$
$\frac{3 v}{4 l} = \frac{v}{l^{'}}$
$\frac{l}{l^{'}} = \frac{3}{4}$
Their lengths are in the ratio $l : l^{'} = 3 : 4$

MHT-CET 2020

WAVES

172574 A pipe open at one end has length $0.8 m$ . At the open end of the tube a string $0.5 m$ long is vibrating in its $1^{st}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 N$ , the mass of string is (speed of sound $= 320 m / s$ )

1 15 gram

2 20 gram

3 10 gram

4 25 gram

Explanation:

C Given that,
$v = 320 m / s$ , length of pipe $(L) = 0.8 m$ , length of string $(l) = 0.5 m$
The fundamental frequency of pipe open at one end is given by,
$f = \frac{v}{4 L} = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 Hz$
$∴$ The first overtone of the string $= 100 Hz$
$∴$ Fundamental frequency of the string $= \frac{100}{2} = 50 Hz$ For frequency of string $(f_{1}) = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$50 = \frac{1}{2 \times 0.5} \sqrt{\frac{50}{μ}}$
$50 = \sqrt{\frac{50}{μ}}$
Squaring both side, we get-
$(50)^{2} = \frac{50}{μ}$
$μ = \frac{1}{50} kg / m$
Hence, mass of string $(m) = μ L = \frac{1}{50} \times 0.5 = 10^{- 2} kg = 10 g$

MHT-CET 2020]**#IIT JEE-2010

WAVES

172575 On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 Hz$ more than fundamental frequency of vibration in open organ pipe. The fundamental frequency of open organ pipe is

1

250 Hz

2

100 Hz

3

50 Hz

4

200 Hz

Explanation:

B Fundamental frequency of closed pipe $(f_{1}) = \frac{V}{4 l}$
Frequency of open $pipe (f_{2}) = \frac{V}{2 l}$
Frequency of third harmonic $(f_{3}) = \frac{3 v}{4 l}$
$∴$ Difference between frequency of third harmonic and frequency of open $(f_{3} - f_{2}) = 50 Hz$
$\frac{3 v}{4 l} - \frac{v}{2 l} = 50$
$\frac{v}{4 l} = 50$
$\frac{v}{2 l} = 100 Hz$

MHT-CET 2020

WAVES

172576 The air column in pipe which is closed at one end will be in resonance with a vibrating turning fork at a frequency of $260 Hz$ , if the length of the air column is-

1

31.73 cm

2

62.5 cm

3

35.75 cm

4

12.5 cm

Explanation:

A One end closed air pipe acts as a closed organ pipe.
$∴ f = \frac{v}{4 l}$
$260 = \frac{330}{4 l} (given, f = 260 Hz}$
$l = \frac{33}{26 \times 4}$
$l = 0.3173 m$
$l = 31.73 cm$

BCECE-2013

WAVES

172577 If two air columns of lengths $100 cm$ and $101 cm$ sounding in their fundamental note gave 17 beats in 20 seconds, then the velocity of sound will be

1

277.8 m / s

2

300 m / s

3

250 m / s

4

343.4 m / s

Explanation:

D Given, $l_{1} = 100 cm = 1 m, l_{2} = 101 cm = 1.01 m$ We know that,
$f_{1} = \frac{v}{4 l_{1}}$
$f_{2} = \frac{v}{4 l_{2}}$
$f_{1} - f_{2} = \frac{17}{20}$
$\frac{v}{4 l_{1}} - \frac{v}{4 l_{2}} = \frac{17}{20}$
$v (\frac{1}{1} - \frac{1}{1.01}) = \frac{17}{5}$
$v = \frac{17}{5} \times \frac{1.01}{0.01} = 343.4 m / s$

JCECE-2013

WAVES

172573 A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

1

3 : 4

2

2 : 3

3

1 : 2

4

4 : 5

Explanation:

A For frequency of first overtone of a closed $pipe (f) = \frac{3 v}{4 l}$
For frequency of first overtone of an open $pipe (f^{'}) = \frac{V}{l^{'}}$
According to the question their first overtone are identical-
$f = f^{'}$
$\frac{3 v}{4 l} = \frac{v}{l^{'}}$
$\frac{l}{l^{'}} = \frac{3}{4}$
Their lengths are in the ratio $l : l^{'} = 3 : 4$

MHT-CET 2020

WAVES

172574 A pipe open at one end has length $0.8 m$ . At the open end of the tube a string $0.5 m$ long is vibrating in its $1^{st}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 N$ , the mass of string is (speed of sound $= 320 m / s$ )

1 15 gram

2 20 gram

3 10 gram

4 25 gram

Explanation:

C Given that,
$v = 320 m / s$ , length of pipe $(L) = 0.8 m$ , length of string $(l) = 0.5 m$
The fundamental frequency of pipe open at one end is given by,
$f = \frac{v}{4 L} = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 Hz$
$∴$ The first overtone of the string $= 100 Hz$
$∴$ Fundamental frequency of the string $= \frac{100}{2} = 50 Hz$ For frequency of string $(f_{1}) = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$50 = \frac{1}{2 \times 0.5} \sqrt{\frac{50}{μ}}$
$50 = \sqrt{\frac{50}{μ}}$
Squaring both side, we get-
$(50)^{2} = \frac{50}{μ}$
$μ = \frac{1}{50} kg / m$
Hence, mass of string $(m) = μ L = \frac{1}{50} \times 0.5 = 10^{- 2} kg = 10 g$

MHT-CET 2020]**#IIT JEE-2010

WAVES

172575 On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 Hz$ more than fundamental frequency of vibration in open organ pipe. The fundamental frequency of open organ pipe is

1

250 Hz

2

100 Hz

3

50 Hz

4

200 Hz

Explanation:

B Fundamental frequency of closed pipe $(f_{1}) = \frac{V}{4 l}$
Frequency of open $pipe (f_{2}) = \frac{V}{2 l}$
Frequency of third harmonic $(f_{3}) = \frac{3 v}{4 l}$
$∴$ Difference between frequency of third harmonic and frequency of open $(f_{3} - f_{2}) = 50 Hz$
$\frac{3 v}{4 l} - \frac{v}{2 l} = 50$
$\frac{v}{4 l} = 50$
$\frac{v}{2 l} = 100 Hz$

MHT-CET 2020

WAVES

172576 The air column in pipe which is closed at one end will be in resonance with a vibrating turning fork at a frequency of $260 Hz$ , if the length of the air column is-

1

31.73 cm

2

62.5 cm

3

35.75 cm

4

12.5 cm

Explanation:

A One end closed air pipe acts as a closed organ pipe.
$∴ f = \frac{v}{4 l}$
$260 = \frac{330}{4 l} (given, f = 260 Hz}$
$l = \frac{33}{26 \times 4}$
$l = 0.3173 m$
$l = 31.73 cm$

BCECE-2013

WAVES

172577 If two air columns of lengths $100 cm$ and $101 cm$ sounding in their fundamental note gave 17 beats in 20 seconds, then the velocity of sound will be

1

277.8 m / s

2

300 m / s

3

250 m / s

4

343.4 m / s

Explanation:

D Given, $l_{1} = 100 cm = 1 m, l_{2} = 101 cm = 1.01 m$ We know that,
$f_{1} = \frac{v}{4 l_{1}}$
$f_{2} = \frac{v}{4 l_{2}}$
$f_{1} - f_{2} = \frac{17}{20}$
$\frac{v}{4 l_{1}} - \frac{v}{4 l_{2}} = \frac{17}{20}$
$v (\frac{1}{1} - \frac{1}{1.01}) = \frac{17}{5}$
$v = \frac{17}{5} \times \frac{1.01}{0.01} = 343.4 m / s$

JCECE-2013

WAVES

172573 A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

1

3 : 4

2

2 : 3

3

1 : 2

4

4 : 5

Explanation:

A For frequency of first overtone of a closed $pipe (f) = \frac{3 v}{4 l}$
For frequency of first overtone of an open $pipe (f^{'}) = \frac{V}{l^{'}}$
According to the question their first overtone are identical-
$f = f^{'}$
$\frac{3 v}{4 l} = \frac{v}{l^{'}}$
$\frac{l}{l^{'}} = \frac{3}{4}$
Their lengths are in the ratio $l : l^{'} = 3 : 4$

MHT-CET 2020

WAVES

172574 A pipe open at one end has length $0.8 m$ . At the open end of the tube a string $0.5 m$ long is vibrating in its $1^{st}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 N$ , the mass of string is (speed of sound $= 320 m / s$ )

1 15 gram

2 20 gram

3 10 gram

4 25 gram

Explanation:

C Given that,
$v = 320 m / s$ , length of pipe $(L) = 0.8 m$ , length of string $(l) = 0.5 m$
The fundamental frequency of pipe open at one end is given by,
$f = \frac{v}{4 L} = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 Hz$
$∴$ The first overtone of the string $= 100 Hz$
$∴$ Fundamental frequency of the string $= \frac{100}{2} = 50 Hz$ For frequency of string $(f_{1}) = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$50 = \frac{1}{2 \times 0.5} \sqrt{\frac{50}{μ}}$
$50 = \sqrt{\frac{50}{μ}}$
Squaring both side, we get-
$(50)^{2} = \frac{50}{μ}$
$μ = \frac{1}{50} kg / m$
Hence, mass of string $(m) = μ L = \frac{1}{50} \times 0.5 = 10^{- 2} kg = 10 g$

MHT-CET 2020]**#IIT JEE-2010

WAVES

172575 On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 Hz$ more than fundamental frequency of vibration in open organ pipe. The fundamental frequency of open organ pipe is

1

250 Hz

2

100 Hz

3

50 Hz

4

200 Hz

Explanation:

B Fundamental frequency of closed pipe $(f_{1}) = \frac{V}{4 l}$
Frequency of open $pipe (f_{2}) = \frac{V}{2 l}$
Frequency of third harmonic $(f_{3}) = \frac{3 v}{4 l}$
$∴$ Difference between frequency of third harmonic and frequency of open $(f_{3} - f_{2}) = 50 Hz$
$\frac{3 v}{4 l} - \frac{v}{2 l} = 50$
$\frac{v}{4 l} = 50$
$\frac{v}{2 l} = 100 Hz$

MHT-CET 2020

WAVES

172576 The air column in pipe which is closed at one end will be in resonance with a vibrating turning fork at a frequency of $260 Hz$ , if the length of the air column is-

1

31.73 cm

2

62.5 cm

3

35.75 cm

4

12.5 cm

Explanation:

A One end closed air pipe acts as a closed organ pipe.
$∴ f = \frac{v}{4 l}$
$260 = \frac{330}{4 l} (given, f = 260 Hz}$
$l = \frac{33}{26 \times 4}$
$l = 0.3173 m$
$l = 31.73 cm$

BCECE-2013

WAVES

172577 If two air columns of lengths $100 cm$ and $101 cm$ sounding in their fundamental note gave 17 beats in 20 seconds, then the velocity of sound will be

1

277.8 m / s

2

300 m / s

3

250 m / s

4

343.4 m / s

Explanation:

D Given, $l_{1} = 100 cm = 1 m, l_{2} = 101 cm = 1.01 m$ We know that,
$f_{1} = \frac{v}{4 l_{1}}$
$f_{2} = \frac{v}{4 l_{2}}$
$f_{1} - f_{2} = \frac{17}{20}$
$\frac{v}{4 l_{1}} - \frac{v}{4 l_{2}} = \frac{17}{20}$
$v (\frac{1}{1} - \frac{1}{1.01}) = \frac{17}{5}$
$v = \frac{17}{5} \times \frac{1.01}{0.01} = 343.4 m / s$

JCECE-2013

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