two identical cylindrical vessels with their bases at the same level each contain a liquid of density…

two identical cylindrical vessels with their bases at the same level each contain a liquid of density $0.801\\text{ g/cm}^3$. the area of each base is $4.77\\text{ cm}^2$, but in one vessel the liquid height is $0.915\\text{ m}$ and in the other it is $1.37\\text{ m}$. find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

two identical cylindrical vessels with their bases at the same level each contain a liquid of density $0.801\\text{ g/cm}^3$. the area of each base is $4.77\\text{ cm}^2$, but in one vessel the liquid height is $0.915\\text{ m}$ and in the other it is $1.37\\text{ m}$. find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

Answer

Explanation:

Step1: Convert given values to SI units

$$ \rho = 0.801 \text{ g/cm}^3 = 801 \text{ kg/m}^3 $$ $$ A = 4.77 \text{ cm}^2 = 4.77 \times 10^{-4} \text{ m}^2 $$ $$ h_1 = 0.915 \text{ m}, \quad h_2 = 1.37 \text{ m} $$

Step2: Calculate the final equalized height

$$ h_f = \frac{h_1 + h_2}{2} = \frac{0.915 + 1.37}{2} = 1.1425 \text{ m} $$

Step3: Determine the mass of liquid moved

$$ \Delta h = h_2 - h_f = 1.37 - 1.1425 = 0.2275 \text{ m} $$ $$ m = \rho A \Delta h = 801 \times (4.77 \times 10^{-4}) \times 0.2275 \approx 0.08692 \text{ kg} $$

Step4: Calculate the change in center of mass

$$ \Delta y = \frac{h_2 + h_f}{2} - \frac{h_1 + h_f}{2} = \frac{h_2 - h_1}{2} = 0.2275 \text{ m} $$

Step5: Calculate work done by gravity

$$ W = m g \Delta y = \rho A \Delta h \cdot g \cdot \Delta y = \rho A g \frac{(h_2 - h_1)^2}{4} $$ $$ W = 801 \times (4.77 \times 10^{-4}) \times 9.81 \times \frac{(1.37 - 0.915)^2}{4} $$

Step6: Compute the final numerical value

$$ W = 0.3748 \times \frac{0.207025}{4} \approx 0.01939 \text{ J} $$

Answer:

0.0194 J