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}$$
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
More Mechanical Properties of Fluids Previous-Year Questions — Page 5
Qjee_main_2025_24_jan_morningSurface Energy
The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-
A. 15 J
B. 10 J
C. 20 J
D. 5 J
Solution
Related Formula
The work done W$W$ in expanding surface area with surface tension S$S$ is given by:
W = S · Δ A$$W = S \cdot \Delta A$$
Core Logic
By volume conservation during splitting :
(4)/(3)π R³ = n ((4)/(3)π r³) r = Rn1/3$$\frac{4}{3}\pi R^{3} = n \left(\frac{4}{3}\pi r^{3}\right) \implies r = \frac{R}{n^{1/3}}$$
Work scales scaling-wise linearly with the key multiplier index (n1/3 - 1)$(n^{1/3} - 1)$. Taking ratios between targets directly avoids calculations of constants.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q3jee_main_2025_24_jan_morningExcess Pressure and Surface Tension
An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m³$1000\text{ kg/m}^{3}$ If the pressure inside the bubble is 2100 N/m² greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g=10 m/s²)$g=10\text{ m/s}^{2})$
A. 0.02
B. 0.1
C. 0.25
D. 0.04
Solution
Related Formula
The absolute pressure inside an air bubble submerged in a liquid is given by :
Total inside pressure accounts for both the hydrostatic pressure of the fluid column (ρ gh$\rho gh$) and the spherical geometry boundary constraint pressure ((2T)/(R)$\frac{2T}{R}$).
A 400g$400\mathrm{g}$ solid cube having an edge of length 10cm$10\mathrm{cm}$ floats in water. How much volume of the cube is outside the water?
(Given: density of water = 1000kgm⁻³$= 1000\mathrm{kg}\mathrm{m}^{-3}$ )
A.1400cm³$1400\mathrm{cm}^3$
B.4000cm³$4000\mathrm{cm}^3$
C.400cm³$400\mathrm{cm}^3$
D.600~cm³$600~\mathrm{cm}^3$
Solution
Related Formula
By the law of flotation, the weight of a floating body must exactly balance the buoyant force exerted by the displaced fluid volume:
M · g = ρfluid · Vsubmerged · g$$M \cdot g = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g$$
Core Logic
Given parameters:
Mass of the cube, M = 400 g = 0.4 kg$M = 400 \text{ g} = 0.4 \text{ kg}$
Total volume of the cube, Vtotal = (10 cm)³ = 1000 cm³ = 10⁻³ m³$V_{\text{total}} = (10 \text{ cm})^3 = 1000 \text{ cm}^3 = 10^{-3} \text{ m}^3$
Density of water, ρwater = 1000 kg/m³$\rho_{\text{water}} = 1000 \text{ kg/m}^3$
Equating weight to buoyant force to find the submerged volume Vd$V_d$ :
The fraction of a floating body's volume that is submerged equals the ratio of the body's density to the fluid's density: VsubmergedVtotal = ρbodyρfluid$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$. Here, the cube's effective density is 0.4 g/cm³$0.4 \text{ g/cm}^3$, meaning 40%$40\%$ is submerged and 60%$60\%$ stays outside.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Qjee_main_2025_29_jan_morningPascal\'s Law
In a hydraulic lift, the surface area of the input piston is 6cm²$6\mathrm{cm}^2$ and that of the output piston is 1500cm²$1500\mathrm{cm}^2$ . If 100N$100\mathrm{N}$ force is applied to the input piston to raise the output piston by 20cm$20\mathrm{cm}$ , then the work done is ________ kJ.
A plane is in level flight at constant speed and each of its two wings has an area of 40~m²$40\mathrm{~m}^2$. If the speed of the air is 180~km/h$180\mathrm{~km/h}$ over the lower wing surface and 252~km/h$252\mathrm{~km/h}$ over the upper wing surface, the mass of the plane is _______ kg$\mathrm{kg}$. (Take air density to be 1~kg m⁻³$1\mathrm{~kg\ m}^{-3}$ and g = 10~ms⁻²$g = 10\mathrm{~ms}^{-2}$)
Numerical Answer.Answer: 9600 to 9600
Solution
Related Formula
Bernoulli's pressure balance equation for aerofoils:
Δ P = P₁ - P₂ = (1)/(2)ρ(v₂² - v₁²)$$\Delta P = P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$
Dynamic Lift force balancing plane weight:
Flift = Δ P · Atotal = mg$$F_{\text{lift}} = \Delta P \cdot A_{\text{total}} = mg$$
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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