Consider a completely full cylindrical water tank of height 1.6~m$1.6\mathrm{~m}$ and cross-sectional area 0.5~m²$0.5\mathrm{~m}^2$ . It has a small hole in its side at a height 90~cm$90\mathrm{~cm}$ from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50~kg$50\mathrm{~kg}$ is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is: (g = 10~m/s²)$(\mathrm{g} = 10\mathrm{~m/s}^2)$
A.3~m/s$3\mathrm{~m/s}$
B.5~m/s$5\mathrm{~m/s}$
C.2~m/s$2\mathrm{~m/s}$
D.4~m/s$4\mathrm{~m/s}$
Solution & Explanation
Related Formula
Bernoulli's Principle between the top surface (1) and the hole (2):
P₁ + (1)/(2)ρ v₁² + ρ g h₁ = P₂ + (1)/(2)ρ v₂² + ρ g h₂$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$
Core Logic
Let the top surface be point 1, and the opening of the hole be point 2.
Height of tank, H = 1.6~m$H = 1.6\mathrm{~m}$ (thus height of point 1, h₁ = 1.6~m$h_1 = 1.6\mathrm{~m}$)
Height of hole, h₂ = 90~cm = 0.9~m$h_2 = 90\mathrm{~cm} = 0.9\mathrm{~m}$ from the bottom.
Depth of the hole below top surface, h = H - h₂ = 1.6 - 0.9 = 0.7~m$h = H - h_2 = 1.6 - 0.9 = 0.7\mathrm{~m}$.
Since area of hole is negligibly small compared to tank area, the velocity of fluid surface at the top is negligible (v₁ ≈ 0$v_1 \approx 0$).
Pressure at point 1 (top surface):
P₁ = P₀ + (Mg)/(A)$$P_1 = P_0 + \frac{Mg}{A}$$
where M = 50~kg$M = 50\mathrm{~kg}$, A = 0.5~m²$A = 0.5\mathrm{~m}^2$, P₀$P_0$ is atmospheric pressure.
Pressure at point 2 (outside the hole):
P₂ = P₀$P_2 = P_0$
Step 1: Applying Bernoulli's Equation
Substitute these values into Bernoulli's equation with reference datum at the hole (h₂ = 0$h_2 = 0$):
P₁ + ρ g h = P₂ + (1)/(2)ρ v₂²$$P_1 + \rho g h = P_2 + \frac{1}{2}\rho v_2^2$$(P₀ + (Mg)/(A)) + ρ g h = P₀ + (1)/(2)ρ v₂²$$\left(P_0 + \frac{Mg}{A}\right) + \rho g h = P_0 + \frac{1}{2}\rho v_2^2$$(Mg)/(A) + ρ g h = (1)/(2)ρ v₂²$$\frac{Mg}{A} + \rho g h = \frac{1}{2}\rho v_2^2$$
Step 2: Substitution and Calculation
Substitute M = 50~kg$M = 50\mathrm{~kg}$, g = 10~m/s²$g = 10\mathrm{~m/s}^2$, A = 0.5~m²$A = 0.5\mathrm{~m}^2$, ρ = 10³~kg/m³$\rho = 10^3\mathrm{~kg/m}^3$, and h = 0.7~m$h = 0.7\mathrm{~m}$:
Keywords:#efflux velocity cylindrical tank#Bernoulli's equation JEE Main#Fluid Mechanics JEE Main 2025#Torricelli's law with load
More Fluid Mechanics 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.
