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}$$
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
More Mechanical Properties of Fluids Previous-Year Questions — Page 2
Q45jee_main_2026_23_january_eveningViscosity and Terminal Velocity
A small metallic sphere of diameter 2 mm and density 10.5 g/cm³$10.5 \, \mathrm{g/cm}^{3}$ is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm³$1.5 \, \mathrm{g/cm}^{3}$ respectively. The terminal velocity attained by the sphere is ____ cm/s.
(π = (22)/(7) and g = 10 m/s²$\pi = \frac{22}{7} \text{ and } g = 10 \, \mathrm{m/s^2}$)
Watch the units closely. Poise is the CGS unit for viscosity, so converting all values to CGS (cm, g, s) makes the calculation direct and prevents factor-of-10 errors common when mixing SI and CGS.
Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s coalesce to form a bigger drop. The terminal velocity of bigger drop is ____ cm/s.
When N$N$ identical drops coalesce, the new radius is N1/3$N^{1/3}$ times the old radius. Consequently, the new terminal velocity scales precisely as N2/3$N^{2/3}$. For N=64$N=64$, 642/3 = 16$64^{2/3} = 16$. Just multiply initial v$v$ by 16.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q37jee_main_2026_24_january_eveningBuoyancy and Archimedes Principle
A cubical block of density ρb = 600 kg/m³$\rho_{\mathrm{b}} = 600 \, \mathrm{kg/m^{3}}$ floats in a liquid of density ρₑ = 900 kg/m³$\rho_{\mathrm{e}} = 900 \, \mathrm{kg/m^{3}}$ . If the height of block is H = 8.0 cm$H = 8.0 \, \mathrm{cm}$ then height of the submerged part is ____ cm.
A.7.3$7.3$
B.4.3$4.3$
C.6.3$6.3$
D.5.3$5.3$
Solution
Related Formula
Fbuoyancy = Mg$$F_{\text{buoyancy}} = Mg$$ρliquid Vsubmerged g = ρblock Vtotal g$$\rho_{\text{liquid}} V_{\text{submerged}} g = \rho_{\text{block}} V_{\text{total}} g$$
Core Logic
For a floating block, the weight of the block equals the buoyant force exerted by the liquid.
M g = Fb$M g = F_b$
ρb × A × H × g = ρₑ × A × h × g$$\rho_b \times A \times H \times g = \rho_e \times A \times h \times g$$
Submerged fraction is exactly the ratio of densities: (h)/(H) = ρobjectρfluid$\frac{h}{H} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q50jee_main_2026_24_january_eveningSurface Tension and Work Done
A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 - x) μJ of work is done in blowing it further to make its diameter 14 cm, then the value of x is ____.
(π = 2 2 / 7)$(\pi = 2 2 / 7)$
Numerical Answer.Answer: 11304 to 11304
Solution
Related Formula
W = Δ U = S · Δ A$$W = \Delta U = S \cdot \Delta A$$
For a soap bubble with two surfaces:
Δ A = 2 × 4π (r₂² - r₁²)$$\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)$$
A soap bubble has two free surfaces; always multiply area changes by 2 (W = 8π S Δ r²$W = 8\pi S \Delta r^2$). Failing to account for both surfaces yields exactly half the answer, acting as a common trap.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q4jee_main_2025_02_april_eveningSurface Tension and Surface Energy
Two water drops each of radius r$r$ coalesce to form a bigger drop. If T$T$ is the surface tension, the surface energy released in this process is:
A.4π r² T [2 - 2(2)/(3)]$4\pi r^2 T \left[2 - 2^{\frac{2}{3}}\right]$
B.4π r² T [2 - 2(1)/(3)]$4\pi r^2 T \left[2 - 2^{\frac{1}{3}}\right]$
C.4π r² T [1 + √(2)]$4\pi r^2 T \left[1 + \sqrt{2}\right]$
D.4π r² T [√(2) - 1]$4\pi r^2 T \left[\sqrt{2} - 1\right]$
Solution
Related Formula
Surface Energy:
U = T · A = T · (4π R²)$$U = T \cdot A = T \cdot (4\pi R^2)$$
Conservation of Volume during coalescence of drops:
Δ E = Uᵢ - Uf = T(Aᵢ - Af)$$\Delta E = U_i - U_f = T(A_i - A_f)$$Δ E = T ( 8π r² - 4π r² 22/3 ) = 4π r² T [ 2 - 22/3 ]$$\Delta E = T \left( 8\pi r^2 - 4\pi r^2 2^{2/3} \right) = 4\pi r^2 T \left[ 2 - 2^{2/3} \right]$$
Pattern Recognition
Sees: Coalescence of N$N$ identical drops.
Trap: Forgetting to conserve volume first, or confusing initial and final surface areas.
Shortcut: Energy released when N$N$ drops coalesce into one big drop is:
Δ E = 4π r² T [ N - N2/3 ]$$\Delta E = 4\pi r^2 T \left[ N - N^{2/3} \right]$$
Here, substituting N = 2$N = 2$ directly gives 4π r² T [ 2 - 22/3 ]$4\pi r^2 T \left[ 2 - 2^{2/3} \right]$.
Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
More Mechanical Properties of Fluids Questions — jee_main_2025_03_april_evening
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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