Answer
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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.
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