Two cylindrical vessels of equal cross sectional area of 2~m²$2\mathrm{~m}^{2}$ contain water upto height 10~m$10\mathrm{~m}$ and 6~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 10³~kg/m³$10^{3}\mathrm{~kg/m}^{3}$ and g=10~m/s²$g=10\mathrm{~m/s}^{2}$)
A.1 × 10⁵~J$1 \times 10^{5}\mathrm{~J}$
B.4 × 10⁴~J$4 \times 10^{4}\mathrm{~J}$
C.6 × 10⁴~J$6 \times 10^{4}\mathrm{~J}$
D.8 × 10⁴~J$8 \times 10^{4}\mathrm{~J}$
Solution & Explanation
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 ((h)/(2)) = (ρ A h) g ((h)/(2)) = (1)/(2) ρ A g h²$$U = m g \left(\frac{h}{2}\right) = (\rho A h) g \left(\frac{h}{2}\right) = \frac{1}{2} \rho A g h^2$$
Work done by the force of gravity (W$W$) equals the negative change in potential energy:
W = -Δ U = Uᵢ - Uf$$W = -\Delta U = U_i - U_f$$
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:
Step 1: Calculate Initial Potential Energy (Uᵢ$U_i$)
Let the reference level U=0$U=0$ be at the bottom:
Uᵢ = U₁ + U₂ = (1)/(2) ρ A g h₁² + (1)/(2) ρ A g h₂²$$U_i = U_1 + U_2 = \frac{1}{2} \rho A g h_1^2 + \frac{1}{2} \rho A g h_2^2$$Uᵢ = (1)/(2) ρ A g (10² + 6²) = (1)/(2) ρ A g (100 + 36) = 68 ρ A g$$U_i = \frac{1}{2} \rho A g \left(10^2 + 6^2\right) = \frac{1}{2} \rho A g (100 + 36) = 68 \rho A g$$
Work Done by Gravity in Connecting Vessels
Step 2: Calculate Final Potential Energy (Uf$U_f$)
Both vessels equalize to hf = 8~m$h_f = 8\mathrm{~m}$:
Uf = 2 × [ (1)/(2) ρ A g hf² ] = ρ A g (8²) = 64 ρ 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 = Uᵢ - Uf = 68 ρ A g - 64 ρ A g = 4 ρ A g$$W = U_i - U_f = 68 \rho A g - 64 \rho A g = 4 \rho A g$$
Substitute the given values (ho = 10³~kg/m³$
ho = 10^3\mathrm{~kg/m}^3$, A = 2~m²$A = 2\mathrm{~m}^2$, g = 10~m/s²$g = 10\mathrm{~m/s}^2$):
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:
$Δ U = (1)/(4) × 10³ × 2 × 10 × (10 - 6)² = 5000 × 16 = 8 × 10⁴~J$\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}$$
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
More Mechanical Properties of Fluids Previous-Year Questions
Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are 6 cm²$6\text{ cm}^{2}$ and 3 cm²$3\text{ cm}^{2}$ respectively. The rate of flow will be ____ cm³$\text{cm}^{3}$/s. (take g = 10 m/s²$\text{m/s}^{2}$)
The figure illustrates a Venturi meter setup with water flowing through a variable cross-section tube.
In a horizontal Venturi meter, substituting V₂ = V₁ (A₁/A₂)$V_2 = V_1 (A_1/A_2)$ directly into ρ g h = (1)/(2)ρ (V₂² - V₁²)$\rho g h = \frac{1}{2}\rho (V_2^2 - V_1^2)$ is the standard path. Working in CGS vs MKS units requires careful tracking; convert VA$V_A$ to cm/s before multiplying by AA$A_A$ in cm².
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q38jee_main_2026_21_jan_eveningSurface Tension
Surface tension of two liquids (having same densities), T₁$T_{1}$ and T₂$T_{2}$, are measured using capillary rise method utilizing two tubes with inner radii of r₁$r_{1}$ and r₂$r_{2}$ where r₁ > r₂$r_{1} > r_{2}$. The measured liquid heights in these tubes are h₁$h_{1}$ and h₂$h_{2}$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights h₁$h_{1}$ and h₂$h_{2}$ and surface tensions T₁$T_{1}$ and T₂$T_{2}$ satisfy the relation :
A.h₁ < h₂ and T₁ = T₂$h_1 < h_2 \text{ and } T_1 = T_2$
B.h₁ = h₂ and T₁ = T₂$h_1 = h_2 \text{ and } T_1 = T_2$
C.h₁ > h₂ and T₁ = T₂$h_1 > h_2 \text{ and } T_1 = T_2$
D.h₁ > h₂ and T₁ < T₂$h_1 > h_2 \text{ and } T_1 < T_2$
Solution
Related Formula
h = (2T θ)/(ρ g r)$$h = \frac{2T\cos\theta}{\rho g r}$$
Core Logic
Since we are assessing surface tension T$T$ as a property of the liquids, the question intends to compare liquids that are identical (implying T₁ = T₂$T_1 = T_2$ and same density).
Capillary rise formula block for Q38 - JEE Main 2026 Evening
By Jurin's Law, for identical liquids, the height of capillary rise is inversely proportional to the radius of the tube:
h ∝ (1)/(r)$$h \propto \frac{1}{r}$$
Step 1: Final Conclusion
Given r₁ > r₂$r_1 > r_2$, the inverse proportionality directly implies that the liquid will rise less in the wider tube.
Therefore, h₁ < h₂$h_1 < h_2$ and naturally T₁ = T₂$T_1 = T_2$ for the intrinsic property.
Pattern Recognition
Wider tubes (r$r$ large) cause lower capillary rises (h$h$ small). h ∝ 1/r$h \propto 1/r$ is a foundational inverse relationship in capillary action.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q47jee_main_2026_21_jan_eveningTerminal Velocity
The terminal velocity of a metallic ball of radius 6 mm$6 \text{ mm}$ in a viscous fluid is 20 cm/s$20 \text{ cm/s}$. The terminal velocity of another ball of same material and having radius 3 mm$3 \text{ mm}$ in the same fluid will be ________ cm/s.
For balls of the same material falling through the same fluid, all terms in the terminal velocity equation except the radius r$r$ are constant.
Therefore, vT ∝ r²$v_T \propto r^2$.
Terminal velocity is directly proportional to the square of the radius. Halving the radius (6 → 3$6 \to 3$) will drop the terminal velocity by a factor of 1/4$1/4$.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q33jee_main_2026_22_january_morningPascal's Law and Surface Tension
Given below are two statements:
Statement I : Pressure of fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid.
Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension.
In the light of the above statements, choose the correct answer from the options given below:
A.Statement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$
B.Both Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
C.Both Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
D.Statement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
According to Pascal's law, pressure exists at every point in a liquid at rest, not just at boundaries. Thus, Statement I is false.
For interior molecules, net cohesive forces are zero, whereas surface molecules possess excess potential energy leading to surface tension. Thus, Statement II is correct.
Pattern Recognition
Sees: Fluid pressure definition + surface tension origin.
Shortcut: Recall Pascal's law (pressure everywhere) and surface energy excess.
Check: Matches option (4). ✓
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q34jee_main_2026_22_january_eveningCapillarity and Surface Tension
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises up to certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1 % each, then the height of the liquid in the tube will change by ____%.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
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