Fluid Pressure & Density — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Fluid Pressure & Density MCQs with step-by-step solutions (29 questions). Part of Mechanical Properties of Fluids. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Fluid Pressure & Density · easy · numerical
The gauge pressure at a depth of $10\text{ m}$ below the surface of water is ($\rho = 1000\text{ kg/m}^3$, $g = 10\text{ m/s}^2$):
A. $10^6\text{ Pa}$
B. $10^4\text{ Pa}$
C. $10^5\text{ Pa}$ ✓ Correct
D. $2 \times 10^5\text{ Pa}$
Solution: Gauge pressure is the excess over atmospheric: $P = \rho g h = 1000 \times 10 \times 10 = 10^5\text{ Pa}$, which is about one atmosphere.
Q2 — Fluid Pressure & Density · easy · theory
The pressure exerted by a liquid at a point inside it depends on:
A. The total weight of the liquid in the vessel
B. The depth of the point and the density of the liquid ✓ Correct
C. The area of the base of the vessel
D. The shape of the containing vessel
Solution: From $P = P_0 + \rho g h$, only depth and density matter. This is the hydrostatic paradox: vessels of very different shapes holding different amounts of liquid show the same pressure at the same depth.
Q3 — Fluid Pressure & Density · medium · numerical
A block of ice of density $900\text{ kg/m}^3$ floats in water of density $1000\text{ kg/m}^3$. The percentage of its volume that remains submerged is:
A. $50\%$
B. $10\%$
C. $100\%$
D. $90\%$ ✓ Correct
Solution: For a floating body, weight equals buoyant force: $V\rho_b g = V_{sub}\rho_l g$. So $\dfrac{V_{sub}}{V} = \dfrac{\rho_b}{\rho_l} = \dfrac{900}{1000} = 0.9$, i.e. $90\%$.
Q4 — Fluid Pressure & Density · easy · theory
Archimedes' principle states that the buoyant force acting on a body immersed in a fluid is equal to:
A. The weight of the body
B. The weight of the fluid displaced by the body ✓ Correct
C. The volume of the fluid displaced
D. The density of the fluid times the volume of the body
Solution: The upthrust equals the weight of the displaced fluid, $F_B = V_{disp}\,\rho_{fluid}\,g$, and acts vertically upward through the centre of buoyancy.
Q5 — Fluid Pressure & Density · medium · numerical
A body weighs $50\text{ N}$ in air and $40\text{ N}$ when fully immersed in water. Its relative density is:
A. $1.25$
B. $5$ ✓ Correct
C. $0.8$
D. $4$
Solution: Loss of weight $= 50 - 40 = 10\text{ N}$ is the buoyant force. Relative density $= \dfrac{\text{weight in air}}{\text{loss of weight in water}} = \dfrac{50}{10} = 5$.
Q6 — Fluid Pressure & Density · medium · numerical
A U-tube contains water and kerosene separated by mercury. If the mercury levels in both limbs are equal, the ratio of the heights of the water and kerosene columns $h_w / h_k$ is ($\rho_w = 1.0\text{ g/cm}^3$, $\rho_k = 0.8\text{ g/cm}^3$):
A. $1.25$
B. $0.64$
C. $0.8$ ✓ Correct
D. $1.0$
Solution: Equal mercury levels means equal pressure at the interface: $\rho_w g h_w = \rho_k g h_k$, so $\dfrac{h_w}{h_k} = \dfrac{\rho_k}{\rho_w} = \dfrac{0.8}{1.0} = 0.8$.
Q7 — Fluid Pressure & Density · medium · numerical
A barometer reads $76\text{ cm}$ with mercury. If it is filled instead with a liquid of half the density of mercury, the height of the liquid column supported by the same atmospheric pressure is:
A. $152\text{ cm}$ ✓ Correct
B. $76\text{ cm}$
C. $38\text{ cm}$
D. $19\text{ cm}$
Solution: Atmospheric pressure is fixed, and $P = \rho g h$ means $h \propto \dfrac{1}{\rho}$. Halving the density doubles the column height to $152\text{ cm}$.
Q8 — Fluid Pressure & Density · easy · theory
In a connected vessel containing a liquid at rest, the pressure at all points lying in the same horizontal plane is:
A. Greater in the narrower limb
B. Dependent on the cross-sectional area
C. Greater in the wider limb
D. The same ✓ Correct
Solution: Since pressure varies only with depth in a static fluid, all points at the same horizontal level in the same continuous liquid are at equal pressure, whatever the shape of the limb.
Q9 — Fluid Pressure & Density · easy · theory
The relation between absolute pressure $P_{abs}$, gauge pressure $P_g$ and atmospheric pressure $P_0$ is:
A. $P_{abs} = P_g - P_0$
B. $P_{abs} = P_g + P_0$ ✓ Correct
C. $P_{abs} = P_g \times P_0$
D. $P_{abs} = P_0 - P_g$
Solution: A gauge reads pressure relative to the atmosphere, so the true (absolute) pressure is the gauge reading plus atmospheric pressure.
Q10 — Fluid Pressure & Density · easy · theory
The SI unit of pressure is the pascal, which is equivalent to:
A. $\text{N}\cdot\text{m}$
B. $\text{N}\cdot\text{m}^2$
C. $\text{N/m}^2$ ✓ Correct
D. $\text{N/m}$
Solution: Pressure is force per unit area, so $1\text{ Pa} = 1\text{ N/m}^2$.
Q11 — Fluid Pressure & Density · easy · numerical
The pressure due to a water column at the bottom of a tank $2\text{ m}$ deep is ($\rho = 1000\text{ kg/m}^3$, $g = 9.8\text{ m/s}^2$):
A. $19600\text{ Pa}$ ✓ Correct
B. $39200\text{ Pa}$
C. $2000\text{ Pa}$
D. $9800\text{ Pa}$
Solution: $P = \rho g h = 1000 \times 9.8 \times 2 = 19600\text{ Pa}$.
Q12 — Fluid Pressure & Density · medium · numerical
Two liquids of densities $\rho_1$ and $\rho_2$ are mixed in equal volumes. The density of the mixture is:
A. $\sqrt{\rho_1\rho_2}$
B. $\dfrac{\rho_1 + \rho_2}{2}$ ✓ Correct
C. $\rho_1 + \rho_2$
D. $\dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}$
Solution: Take each volume as $V$. Total mass $= (\rho_1 + \rho_2)V$ and total volume $= 2V$, so $\rho_{mix} = \dfrac{\rho_1 + \rho_2}{2}$ — the arithmetic mean.
Q13 — Fluid Pressure & Density · hard · numerical
Two liquids of densities $\rho_1$ and $\rho_2$ are mixed in equal masses. The density of the mixture is:
A. $\dfrac{\rho_1\rho_2}{\rho_1 + \rho_2}$
B. $\sqrt{\rho_1\rho_2}$
C. $\dfrac{\rho_1 + \rho_2}{2}$
D. $\dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}$ ✓ Correct
Solution: Take each mass as $m$. Total volume $= \dfrac{m}{\rho_1} + \dfrac{m}{\rho_2}$ and total mass $= 2m$, giving $\rho_{mix} = \dfrac{2\rho_1\rho_2}{\rho_1 + \rho_2}$ — the harmonic mean.
Q14 — Fluid Pressure & Density · easy · theory
The relative density (specific gravity) of a substance is:
A. A dimensionless ratio with no unit ✓ Correct
B. Measured in $\text{kg/m}^3$
C. Measured in $\text{N/m}^2$
D. Measured in $\text{g/cm}^3$
Solution: Relative density is the ratio of the density of a substance to the density of water, so the units cancel and it is a pure number.
Q15 — Fluid Pressure & Density · medium · numerical
A block floats in water with one-fourth of its volume above the surface. The density of the block is:
A. $1000\text{ kg/m}^3$
B. $250\text{ kg/m}^3$
C. $750\text{ kg/m}^3$ ✓ Correct
D. $500\text{ kg/m}^3$
Solution: Three-fourths of the volume is submerged, so $\dfrac{\rho_b}{\rho_w} = \dfrac{3}{4}$, giving $\rho_b = 0.75 \times 1000 = 750\text{ kg/m}^3$.
Q16 — Fluid Pressure & Density · medium · theory
A body floats in a liquid. The centre of buoyancy is located at the:
A. Topmost point of the body
B. Centre of gravity of the body
C. Centre of gravity of the displaced liquid ✓ Correct
D. Point of contact with the liquid surface
Solution: The buoyant force is the resultant of pressure forces on the immersed surface, and it acts through the centre of gravity of the displaced liquid, called the centre of buoyancy.
Q17 — Fluid Pressure & Density · easy · theory
A manometer is an instrument used to measure:
A. The density of a liquid
B. The surface tension of a liquid
C. The viscosity of a liquid
D. The pressure difference between a gas and the atmosphere ✓ Correct
Solution: A manometer balances the unknown pressure against a liquid column; the height difference between the limbs gives the pressure difference directly as $\rho g h$.
Q18 — Fluid Pressure & Density · easy · numerical
The gauge pressure at a depth of $20\text{ m}$ in water is ($\rho = 1000\text{ kg/m}^3$, $g = 10\text{ m/s}^2$):
A. $10^5\text{ Pa}$
B. $2 \times 10^5\text{ Pa}$ ✓ Correct
C. $4 \times 10^5\text{ Pa}$
D. $2 \times 10^4\text{ Pa}$
Solution: $P = \rho g h = 1000 \times 10 \times 20 = 2 \times 10^5\text{ Pa}$.
Q19 — Fluid Pressure & Density · medium · numerical
A body weighs $80\text{ N}$ in air and $60\text{ N}$ when fully immersed in water. Its relative density is:
A. $0.75$
B. $4$ ✓ Correct
C. $1.33$
D. $3$
Solution: Loss of weight $= 80 - 60 = 20\text{ N}$. Relative density $= \dfrac{80}{20} = 4$.
Q20 — Fluid Pressure & Density · medium · numerical
A block floats in water with two-fifths of its volume above the surface. Its density is:
A. $400\text{ kg/m}^3$
B. $800\text{ kg/m}^3$
C. $1000\text{ kg/m}^3$
D. $600\text{ kg/m}^3$ ✓ Correct
Solution: Three-fifths is submerged, so $\rho_b = \dfrac{3}{5} \times 1000 = 600\text{ kg/m}^3$.
Q21 — Fluid Pressure & Density · medium · numerical
A tank of base area $0.5\text{ m}^2$ holds water to a depth of $2\text{ m}$. The force exerted by the water on the base is ($\rho = 1000\text{ kg/m}^3$, $g = 10\text{ m/s}^2$):
A. $10^4\text{ N}$ ✓ Correct
B. $2 \times 10^4\text{ N}$
C. $4 \times 10^4\text{ N}$
D. $5 \times 10^3\text{ N}$
Solution: $P = \rho g h = 1000 \times 10 \times 2 = 2 \times 10^4\text{ Pa}$, so $F = PA = 2 \times 10^4 \times 0.5 = 10^4\text{ N}$.
Q22 — Fluid Pressure & Density · medium · numerical
A mercury barometer reads $75\text{ cm}$. The atmospheric pressure is approximately ($\rho_{Hg} = 13600\text{ kg/m}^3$, $g = 9.8\text{ m/s}^2$):
A. $1.0 \times 10^5\text{ Pa}$ ✓ Correct
B. $7.5 \times 10^5\text{ Pa}$
C. $1.36 \times 10^5\text{ Pa}$
D. $1.0 \times 10^4\text{ Pa}$
Solution: $P = \rho g h = 13600 \times 9.8 \times 0.75 \approx 9.996 \times 10^4 \approx 1.0 \times 10^5\text{ Pa}$.
Q23 — Fluid Pressure & Density · easy · numerical
The pressure due to a liquid of density $800\text{ kg/m}^3$ at a depth of $5\text{ m}$ is ($g = 10\text{ m/s}^2$):
A. $160\text{ Pa}$
B. $50000\text{ Pa}$
C. $40000\text{ Pa}$ ✓ Correct
D. $4000\text{ Pa}$
Solution: $P = \rho g h = 800 \times 10 \times 5 = 40000\text{ Pa}$.
Q24 — Fluid Pressure & Density · easy · numerical
A body of volume $2 \times 10^{-3}\text{ m}^3$ is completely immersed in water. The buoyant force on it is ($\rho = 1000\text{ kg/m}^3$, $g = 10\text{ m/s}^2$):
A. $2\text{ N}$
B. $20\text{ N}$ ✓ Correct
C. $0.2\text{ N}$
D. $200\text{ N}$
Solution: $F_B = V\rho g = 2 \times 10^{-3} \times 1000 \times 10 = 20\text{ N}$.
Q25 — Fluid Pressure & Density · medium · numerical
A body weighs $100\text{ N}$ in air and $80\text{ N}$ in water. Its relative density is:
A. $2.5$
B. $4$
C. $5$ ✓ Correct
D. $1.25$
Solution: Loss of weight $= 100 - 80 = 20\text{ N}$, so relative density $= \dfrac{100}{20} = 5$.
Q26 — Fluid Pressure & Density · medium · numerical
A block of density $900\text{ kg/m}^3$ floats in a liquid of density $1200\text{ kg/m}^3$. The fraction of its volume submerged is:
A. $75\%$ ✓ Correct
B. $90\%$
C. $60\%$
D. $25\%$
Solution: $\dfrac{V_{sub}}{V} = \dfrac{\rho_{body}}{\rho_{liquid}} = \dfrac{900}{1200} = 0.75 = 75\%$.
Q27 — Fluid Pressure & Density · hard · numerical
In a U-tube, a $10\text{ cm}$ column of water ($\rho = 1000\text{ kg/m}^3$) balances an $8\text{ cm}$ column of another liquid. The density of that liquid is:
A. $800\text{ kg/m}^3$
B. $1000\text{ kg/m}^3$
C. $1250\text{ kg/m}^3$ ✓ Correct
D. $625\text{ kg/m}^3$
Solution: Equal pressures at the interface give $\rho_1h_1 = \rho_2h_2 \Rightarrow \rho_2 = \dfrac{1000 \times 10}{8} = 1250\text{ kg/m}^3$.
Q28 — Fluid Pressure & Density · medium · numerical
The absolute pressure at a depth of $10\text{ m}$ in water, taking atmospheric pressure as $10^5\text{ Pa}$, is ($\rho = 1000$, $g = 10$):
A. $10^5\text{ Pa}$
B. $3 \times 10^5\text{ Pa}$
C. $2 \times 10^5\text{ Pa}$ ✓ Correct
D. $1.1 \times 10^5\text{ Pa}$
Solution: $P_{abs} = P_0 + \rho g h = 10^5 + (1000)(10)(10) = 10^5 + 10^5 = 2 \times 10^5\text{ Pa}$.
Q29 — Fluid Pressure & Density · medium · numerical
A block floats in water with $20\%$ of its volume above the surface. Its density is:
A. $600\text{ kg/m}^3$
B. $200\text{ kg/m}^3$
C. $1000\text{ kg/m}^3$
D. $800\text{ kg/m}^3$ ✓ Correct
Solution: Eighty per cent is submerged, so $\rho_{body} = 0.8 \times 1000 = 800\text{ kg/m}^3$.