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PHXI15:WAVES

355033 Two tuning forks of frequency $250 H z$ and $256 H z$ produce beats. If maximum of intensity is observed just now, after how much time the minimum is observed at the same place

1

\frac{1}{18} \sec

2

\frac{1}{4} \sec

3

\frac{1}{8} \sec

4

\frac{1}{12} \sec

Explanation:

The time interval between two successive maximas is equal to $\frac{1}{6} s$
For every $\frac{1}{12}$ th sec
The minimum will be observed.

PHXI15:WAVES

355034 A tuning fork vibrates with frequency $256 H z$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is $340 m s^{- 1}$ )

1

220 c m

2

200 c m

3

190 c m

4

180 c m

Explanation:

Organ pipe will have frequency either 255 or $257 H z$
Using $255 H z$
$255 = \frac{3 v}{2 l} l = \frac{3 \times 340}{2 \times 255} m$
$l = 200 c m$

JEE - 2018

PHXI15:WAVES

355035 Beats are produced by two waves given $y_{1} = a \sin 2000 π t$ and $y_{2} = a \sin 2008 π t$ .
The number of beats heard per second is

1 One

2 Zero

3 Eight

4 Four

Explanation:

Number of beats per second $= n_{1} - n_{2}$
$ω_{1} = 2000 π = 2 π n_{1} \Rightarrow n_{1} = 1000$
and $ω_{2} = 2008 π = 2 π n_{2} \Rightarrow n_{2} = 1004$
Number of beats heard per sec $= 1004 - 1000 = 4$

PHXI15:WAVES

355036 Assertion :
To hear distinct beats, difference in frequencies of two sources should be less than 10.
Reason :
More the number of beats per sec more difficult to hear them.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI15:WAVES

355033 Two tuning forks of frequency $250 H z$ and $256 H z$ produce beats. If maximum of intensity is observed just now, after how much time the minimum is observed at the same place

1

\frac{1}{18} \sec

2

\frac{1}{4} \sec

3

\frac{1}{8} \sec

4

\frac{1}{12} \sec

Explanation:

The time interval between two successive maximas is equal to $\frac{1}{6} s$
For every $\frac{1}{12}$ th sec
The minimum will be observed.

PHXI15:WAVES

355034 A tuning fork vibrates with frequency $256 H z$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is $340 m s^{- 1}$ )

1

220 c m

2

200 c m

3

190 c m

4

180 c m

Explanation:

Organ pipe will have frequency either 255 or $257 H z$
Using $255 H z$
$255 = \frac{3 v}{2 l} l = \frac{3 \times 340}{2 \times 255} m$
$l = 200 c m$

JEE - 2018

PHXI15:WAVES

355035 Beats are produced by two waves given $y_{1} = a \sin 2000 π t$ and $y_{2} = a \sin 2008 π t$ .
The number of beats heard per second is

1 One

2 Zero

3 Eight

4 Four

Explanation:

Number of beats per second $= n_{1} - n_{2}$
$ω_{1} = 2000 π = 2 π n_{1} \Rightarrow n_{1} = 1000$
and $ω_{2} = 2008 π = 2 π n_{2} \Rightarrow n_{2} = 1004$
Number of beats heard per sec $= 1004 - 1000 = 4$

PHXI15:WAVES

355036 Assertion :
To hear distinct beats, difference in frequencies of two sources should be less than 10.
Reason :
More the number of beats per sec more difficult to hear them.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI15:WAVES

355033 Two tuning forks of frequency $250 H z$ and $256 H z$ produce beats. If maximum of intensity is observed just now, after how much time the minimum is observed at the same place

1

\frac{1}{18} \sec

2

\frac{1}{4} \sec

3

\frac{1}{8} \sec

4

\frac{1}{12} \sec

Explanation:

The time interval between two successive maximas is equal to $\frac{1}{6} s$
For every $\frac{1}{12}$ th sec
The minimum will be observed.

PHXI15:WAVES

355034 A tuning fork vibrates with frequency $256 H z$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is $340 m s^{- 1}$ )

1

220 c m

2

200 c m

3

190 c m

4

180 c m

Explanation:

Organ pipe will have frequency either 255 or $257 H z$
Using $255 H z$
$255 = \frac{3 v}{2 l} l = \frac{3 \times 340}{2 \times 255} m$
$l = 200 c m$

JEE - 2018

PHXI15:WAVES

355035 Beats are produced by two waves given $y_{1} = a \sin 2000 π t$ and $y_{2} = a \sin 2008 π t$ .
The number of beats heard per second is

1 One

2 Zero

3 Eight

4 Four

Explanation:

Number of beats per second $= n_{1} - n_{2}$
$ω_{1} = 2000 π = 2 π n_{1} \Rightarrow n_{1} = 1000$
and $ω_{2} = 2008 π = 2 π n_{2} \Rightarrow n_{2} = 1004$
Number of beats heard per sec $= 1004 - 1000 = 4$

PHXI15:WAVES

355036 Assertion :
To hear distinct beats, difference in frequencies of two sources should be less than 10.
Reason :
More the number of beats per sec more difficult to hear them.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI15:WAVES

355033 Two tuning forks of frequency $250 H z$ and $256 H z$ produce beats. If maximum of intensity is observed just now, after how much time the minimum is observed at the same place

1

\frac{1}{18} \sec

2

\frac{1}{4} \sec

3

\frac{1}{8} \sec

4

\frac{1}{12} \sec

Explanation:

The time interval between two successive maximas is equal to $\frac{1}{6} s$
For every $\frac{1}{12}$ th sec
The minimum will be observed.

PHXI15:WAVES

355034 A tuning fork vibrates with frequency $256 H z$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is $340 m s^{- 1}$ )

1

220 c m

2

200 c m

3

190 c m

4

180 c m

Explanation:

Organ pipe will have frequency either 255 or $257 H z$
Using $255 H z$
$255 = \frac{3 v}{2 l} l = \frac{3 \times 340}{2 \times 255} m$
$l = 200 c m$

JEE - 2018

PHXI15:WAVES

355035 Beats are produced by two waves given $y_{1} = a \sin 2000 π t$ and $y_{2} = a \sin 2008 π t$ .
The number of beats heard per second is

1 One

2 Zero

3 Eight

4 Four

Explanation:

Number of beats per second $= n_{1} - n_{2}$
$ω_{1} = 2000 π = 2 π n_{1} \Rightarrow n_{1} = 1000$
and $ω_{2} = 2008 π = 2 π n_{2} \Rightarrow n_{2} = 1004$
Number of beats heard per sec $= 1004 - 1000 = 4$

PHXI15:WAVES

355036 Assertion :
To hear distinct beats, difference in frequencies of two sources should be less than 10.
Reason :
More the number of beats per sec more difficult to hear them.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

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