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

The Correct Option is B

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$