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JEE Advanced · Physics · 13. Thermodynamics

Paragraph :
\(P_{14-16}\) : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius \(R\) and height \(L_0\). The cylinder is open at its bottom and has a small hole at its top. A piston of mass \(M\) is held at a distance \(L\) from the top surface as shown in the figure. The atmospheric pressure is \(p_0\).

Question :
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is \(\rho\). In equilibrium, the height \(H\) of the water column in the cylinder satisfies

  1. A \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)+L_0 p_0=0\)
  2. B \(\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)-L_0 p_0=0\)
  3. C \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0\)
  4. D \(\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)+L_0 p_0=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0\)

Step-by-step Solution

Detailed explanation

\(
\begin{gathered}
p_1=p_2 \\
p_0+\rho g\left(L_0-H\right)=p
\end{gathered}
\)
Now, applying \(p_1 V_1=p_2 V_2\)
for the air inside the cylinder, we have
\(
p_0\left(L_0\right)=p\left(L_0-H\right)
\)

\(\therefore \quad p=\frac{p_0 L_0}{L_0-H}\)
Substituting in Eq. (i), we have
\(
p_0+\rho g\left(L_0-H\right)=\frac{p_0 L_0}{L_0-H}
\)
or \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-p_0 L_0=0\)
\(\therefore\) Option (c) is correct.
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