Two cylindrical vessels of equal cross sectional area of 2mathrm~m^2$2\mathrm{~m}^{2}$ contain water upto height 10mathrm~m$10\mathrm{~m}$ and 6mathrm~m$6\mathrm{~m}$, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is : (Density of water is 10^3mathrm~kg/m^3$10^{3}\mathrm{~kg/m}^{3}$ and g=10mathrm~m/s^2$g=10\mathrm{~m/s}^{2}$)

### Related Formula The gravitational potential energy U$U$ of a liquid column of mass m$m$ and height h$h$ is evaluated relative to its bottom by placing its total mass at its center of mass (h/2$h/2$): U = m g left(frach2right) = (rho A h) g left(frach2right) = frac12 rho A g h^2$$U = m g \left(\frac{h}{2}\right) = (\rho A h) g \left(\frac{h}{2}\right) = \frac{1}{2} \rho A g h^2$$ Work done by the force of gravity (W$W$) equals the negative change in potential energy: W = -Delta U = U_i - U_f$$W = -\Delta U = U_i - U_f$$ ### Core Logic Since the vessels are identical and connected at the bottom, water flows from the higher column to the lower one until their final heights equalize at: h_f = frac10 + 62 = 8mathrm~m$$h_f = \frac{10 + 6}{2} = 8\mathrm{~m}$$ ### Step 1: Calculate Initial Potential Energy (U_i$U_i$) Let the reference level U=0$U=0$ be at the bottom: U_i = U_1 + U_2 = frac12 rho A g h_1^2 + frac12 rho A g h_2^2$$U_i = U_1 + U_2 = \frac{1}{2} \rho A g h_1^2 + \frac{1}{2} \rho A g h_2^2$$ U_i = frac12 rho A g left(10^2 + 6^2right) = frac12 rho A g (100 + 36) = 68 rho A g$$U_i = \frac{1}{2} \rho A g \left(10^2 + 6^2\right) = \frac{1}{2} \rho A g (100 + 36) = 68 \rho A g$$

### Step 2: Calculate Final Potential Energy (U_f$U_f$) Both vessels equalize to h_f = 8mathrm~m$h_f = 8\mathrm{~m}$: U_f = 2 times left[ frac12 rho A g h_f^2 right] = rho A g (8^2) = 64 rho A g$$U_f = 2 \times \left[ \frac{1}{2} \rho A g h_f^2 \right] = \rho A g (8^2) = 64 \rho A g$$ ### Step 3: Work Done by Gravity (W$W$) W = U_i - U_f = 68 rho A g - 64 rho A g = 4 rho A g$$W = U_i - U_f = 68 \rho A g - 64 \rho A g = 4 \rho A g$$ Substitute the given values ( ho = 10^3mathrm~kg/m^3$ ho = 10^3\mathrm{~kg/m}^3$, A = 2mathrm~m^2$A = 2\mathrm{~m}^2$, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s}^2$): W = 4 times 10^3 times 2 times 10 = 8 times 10^4mathrm~J$$W = 4 \times 10^3 \times 2 \times 10 = 8 \times 10^4\mathrm{~J}$$ ### Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $ ### Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $\Delta U = \frac{1}{4} \rho A g (h_1 - h_2)^2. Applying this directly: $. Applying this directly: $Delta U = frac14 times 10^3 times 2 times 10 times (10 - 6)^2 = 5000 times 16 = 8 times 10^4mathrm~J$\Delta U = \frac{1}{4} \times 10^3 \times 2 \times 10 \times (10 - 6)^2 = 5000 \times 16 = 8 \times 10^4\mathrm{~J}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids