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$h = 0.7\mathrm{~m}$:
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.
