Answer
414.6k+ views
Hint:Refer to the ideal gas law to determine the pressure at arm A. Use the formula to calculate the pressure below the height h of the liquid column of density \[\rho \].
Complete step by step answer:
According to Ideal gas law, the product of pressure and volume is constant.
Therefore, we can write,
\[\Rightarrow{P_1}{V_1} = {P_2}{V_2}\]
Here, \[{V_1}\] is the volume of the arm A at atmospheric pressure \[{P_1}\] and \[{V_2}\] is the volume of the arm A at pressure \[{P_2}\].
The initial volume is V and the final volume is \[\dfrac{V}{2}\]. The pressure \[{P_1}\] is the atmospheric pressure P.
Therefore, the above equation becomes,
\[\Rightarrow PV = {P_2}\dfrac{V}{2}\]
\[ \Rightarrow {p_2} = 2P\]
We know that the pressure below the height H is,
\[ \Rightarrow P = H\rho g\]
Therefore,
\[\Rightarrow {p_2} = 2H\rho g\]
Here, \[\rho \] is the density of the liquid and g is the acceleration due to gravity.
Let the height of the mercury column is x. The pressure below the height x is the sum of atmospheric pressure and the pressure due to the mercury column above it. We have determined the pressure at the arm A which is 2P.
Therefore,
\[\Rightarrow 2P = {P_0} + x\rho g\]
We have given, the atmospheric pressure is H cm of mercury. Therefore, the atmospheric pressure is,
\[\Rightarrow{P_0} = H\rho g\]
Therefore, the pressure at arm A is,
\[ \Rightarrow 2H\rho g = H\rho g + x\rho g\]
\[ \Rightarrow 2H\rho g = \rho g\left( {H + x} \right)\]
\[ \Rightarrow 2H = \left( {H + x} \right)\]
\[ \Rightarrow\therefore x = h\]
So, the correct answer is option (A).
Note:The pressure inside an open liquid column is the addition of atmospheric pressure over the surface of the liquid and the pressure due to liquid above that point.
Complete step by step answer:
According to Ideal gas law, the product of pressure and volume is constant.
Therefore, we can write,
\[\Rightarrow{P_1}{V_1} = {P_2}{V_2}\]
Here, \[{V_1}\] is the volume of the arm A at atmospheric pressure \[{P_1}\] and \[{V_2}\] is the volume of the arm A at pressure \[{P_2}\].
The initial volume is V and the final volume is \[\dfrac{V}{2}\]. The pressure \[{P_1}\] is the atmospheric pressure P.
Therefore, the above equation becomes,
\[\Rightarrow PV = {P_2}\dfrac{V}{2}\]
\[ \Rightarrow {p_2} = 2P\]
We know that the pressure below the height H is,
\[ \Rightarrow P = H\rho g\]
Therefore,
\[\Rightarrow {p_2} = 2H\rho g\]
Here, \[\rho \] is the density of the liquid and g is the acceleration due to gravity.
Let the height of the mercury column is x. The pressure below the height x is the sum of atmospheric pressure and the pressure due to the mercury column above it. We have determined the pressure at the arm A which is 2P.
Therefore,
\[\Rightarrow 2P = {P_0} + x\rho g\]
We have given, the atmospheric pressure is H cm of mercury. Therefore, the atmospheric pressure is,
\[\Rightarrow{P_0} = H\rho g\]
Therefore, the pressure at arm A is,
\[ \Rightarrow 2H\rho g = H\rho g + x\rho g\]
\[ \Rightarrow 2H\rho g = \rho g\left( {H + x} \right)\]
\[ \Rightarrow 2H = \left( {H + x} \right)\]
\[ \Rightarrow\therefore x = h\]
So, the correct answer is option (A).
Note:The pressure inside an open liquid column is the addition of atmospheric pressure over the surface of the liquid and the pressure due to liquid above that point.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![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)
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)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)