First slide

Work

Question

Two identical cylindrical vessels with their bases at same level each contains a liquid of density $ρ$ . The height of the liquid in one vessel is h₁ and that in the other vessel is h₂. The area of either base is A. The work done by gravity in equalizing the levels when the two vessels are connected, is

Moderate

A

$(h_{1} - h_{2}) gρ$
B

$(h_{1} - h_{2}) gAρ$
C

$\frac{1}{2} {(h_{1} - h_{2})}^{2} gAρ$
D

$\frac{1}{4} {(h_{1} - h_{2})}^{2} gAρ$

Solution

If h is the common height when they are connected, by conservation of mass
${ρA}_{1} h_{1} + {ρA}_{2} h_{2} = ρh (A_{1} + A_{2})$
$h = (h_{1} + h_{2}) / 2$ [as $A_{1} = A_{2} = A$ given]
As (h₁/2) and (h₂/2) are heights of initial centre of gravity of liquid in two vessels., the initial potential energy of the system
$U_{i} = (h_{1} Aρ) g \frac{h_{1}}{2} + (h_{2} Aρ) \frac{h_{2}}{2} = ρgA \frac{(h_{1}^{2} + h_{2}^{2})}{2}$ ...(i)
When vessels are connected the height of centre of gravity of liquid in each vessel will be h/2,
i.e. ( $\frac{(h_{1} + h_{2})}{4}$ [as $h = (h_{1} + h_{2}) / 2]$
Final potential energy of the system
$U_{F} = [\frac{(h_{1} + h_{2})}{2} Aρ] g (\frac{h_{1} + h_{2}}{4})$
$= Aρg [\frac{{(h_{1} + h_{2})}^{2}}{4}]$ …(ii)
Work done by gravity
$W = U_{i} - U_{f} = \frac{1}{4} ρgA [2 (h_{1}^{2} + h_{2}^{2}) - {(h_{1} + h_{2})}^{2}]$
$= \frac{1}{4} ρgA {(h_{1} ~ h_{2})}^{2}$

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