Answer
398.7k+ views
Hint:Here, we are given the frequency of the fork and the thing which we have to determine is the length of the pipe. Therefore, we have to find the frequency of the pipe first which can be obtained by the given frequency of the fork. After that, by using the equation of frequency we can get our answer.
Formulas used:
$f = \dfrac{{nv}}{{2l}}$, where, $f$ is the frequency of open pipe, $n$ is the number of normal mode of vibration, $v$ is the speed of sound of air and $l$ is the length of open pipe.
Complete step by step solution:
Our first step is to find the frequency of the open pipe from the frequency of vibrating fork which is given to us in the question.We know that,
Frequency of open pipe= Frequency of fork \[ \pm 1\]
$f = 256 \pm 1Hz$
$\Rightarrow f = 257Hz$ or $255Hz$
We have to calculate the length of the pipe for both these frequencies.
So let us first calculate for $f = 257Hz$
We know that $f = \dfrac{{nv}}{{2l}}$
Here, it is given that the fork gives one beat per second with the third normal mode of vibration of an open pipe. Therefore we will take $n = 3$ and Speed of sound of air is $340m{s^{ - 1}}$
$
\Rightarrow 255 = \dfrac{{3 \times 340}}{{2 \times l}} \\
\Rightarrow l = 2m = 200 cm$
Similarly for $f = 255Hz$, we get
$
\Rightarrow 255 = \dfrac{{3 \times 340}}{{2 \times l}} \\
\therefore l = 2m = 200 cm $
We can see that for both the cases, the value of the length is approximate $200cm$.
Hence, option D is the right answer.
Note: Here, we have used the formula $f = \dfrac{{nv}}{{2l}}$ for finding the length of the pipe. Similarly when the length of the pipe is given, we can find frequency of both open pipe and vibrating fork by using the same equation.
Formulas used:
$f = \dfrac{{nv}}{{2l}}$, where, $f$ is the frequency of open pipe, $n$ is the number of normal mode of vibration, $v$ is the speed of sound of air and $l$ is the length of open pipe.
Complete step by step solution:
Our first step is to find the frequency of the open pipe from the frequency of vibrating fork which is given to us in the question.We know that,
Frequency of open pipe= Frequency of fork \[ \pm 1\]
$f = 256 \pm 1Hz$
$\Rightarrow f = 257Hz$ or $255Hz$
We have to calculate the length of the pipe for both these frequencies.
So let us first calculate for $f = 257Hz$
We know that $f = \dfrac{{nv}}{{2l}}$
Here, it is given that the fork gives one beat per second with the third normal mode of vibration of an open pipe. Therefore we will take $n = 3$ and Speed of sound of air is $340m{s^{ - 1}}$
$
\Rightarrow 255 = \dfrac{{3 \times 340}}{{2 \times l}} \\
\Rightarrow l = 2m = 200 cm$
Similarly for $f = 255Hz$, we get
$
\Rightarrow 255 = \dfrac{{3 \times 340}}{{2 \times l}} \\
\therefore l = 2m = 200 cm $
We can see that for both the cases, the value of the length is approximate $200cm$.
Hence, option D is the right answer.
Note: Here, we have used the formula $f = \dfrac{{nv}}{{2l}}$ for finding the length of the pipe. Similarly when the length of the pipe is given, we can find frequency of both open pipe and vibrating fork by using the same equation.
Recently Updated Pages
In a flask the weight ratio of CH4g and SO2g at 298 class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a flask colourless N2O4 is in equilibrium with brown class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a first order reaction the concentration of the class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a first order reaction the concentration of the class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a fermentation tank molasses solution is mixed with class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In a face centred cubic unit cell what is the volume class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name 10 Living and Non living things class 9 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
List some examples of Rabi and Kharif crops class 8 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)