Properties Of Matter Ques 19
- Two identical cylindrical vessels with their bases at the same level each contain a liquid of density $\rho$. The height of the liquid in one vessel is $h_1$ and in the other is $h_2$. The area of either base is $A$. What is the work done by gravity in equalising the levels when the two vessels are connected?
(1981, 4 M)
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Answer:
Correct Answer: 19.$\frac{\rho A g}{4}\left(h_1-h_2\right)^2$
Solution:
- Let $h$ be the level in equilibrium. Equating the volumes, we have
$\begin{aligned} & A h_1+A h_2 & =2 A h \\ \therefore & h & =\left(\frac{h_1+h_2}{2}\right)\end{aligned}$
Work done by gravity $=U_i-U_f$
$ \begin{aligned} W & =\left(m_1 g \frac{h_1}{2}+m_2 g \frac{h_2}{2}\right)-\left(m_1+m_2\right) g \frac{h}{2} \\ & =\frac{A h_1 \rho g h_1}{2}+\frac{A h_2 \rho g h_2}{2}-\left[A h_1 \rho+A h_2 \rho\right] g\left(\frac{h_1+h_2}{4}\right) \end{aligned} $
Simplifying this, we get
$ W=\frac{\rho A g}{4}\left(h_1-h_2\right)^2 $