Question

Two identical cylindrical vessels with their bases at the same level, each contains a liquid of density $13 \times 10^3 \,kg/m^3$ . The area of each base is $4.00\, cm^2$ , but in one vessel, the liquid height is $0.854 \,m$ and in the other it is $1.560\, m$ . Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

• A. $0.0635 \,J$
• B. $0.635 \,J$
• C. $6.35 \,J$
• D. $63.5 \,J$

Answer 1

Answer: (B)

$0.635 \,J$

Answer 2

Given, $\rho = 1.3 \times 10^3\,kg/m^3$
$A = 4.00\,cm^2 = \frac{4.00}{100\times 100}$
$= 4.00 \times 10^{-4} \,m^2$
$h_1 = 0.854\,m$
$h_2 = 1.560\,m$
Now, work done by gravity $=$ loss in $PE$
$= \frac{4\rho_g}{4} (h_1 - h_2)^2$
$= \frac{4.00 \times 10^{-4} \times 1.3 \times 10^3 \times 10(1.560 - 0.854)}{4}$
$= 0.635\,J$