Bernoulli's Theorem — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Bernoulli's Theorem MCQs with step-by-step solutions (28 questions). Part of Mechanical Properties of Fluids. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Bernoulli's Theorem · easy · theory
Bernoulli's equation for the steady flow of an ideal fluid states that the following quantity is constant along a streamline:
A. $P + \dfrac{1}{2}\rho v + \rho g h$
B. $P + \rho v^2 + \rho g h$
C. $P + \dfrac{1}{2}\rho v^2 + \rho g h$ ✓ Correct
D. $P \times \dfrac{1}{2}\rho v^2 \times \rho g h$
Solution: The three terms are the pressure energy, kinetic energy and potential energy per unit volume. Their sum is conserved along a streamline for an ideal fluid.
Q2 — Bernoulli's Theorem · easy · theory
Bernoulli's theorem is essentially a statement of the conservation of:
A. Energy ✓ Correct
B. Momentum
C. Angular momentum
D. Mass
Solution: Each term in the equation is an energy per unit volume; the theorem says the total mechanical energy per unit volume of the fluid stays constant along a streamline.
Q3 — Bernoulli's Theorem · easy · numerical
An open water tank has a small orifice $5\text{ m}$ below the free surface. Taking $g = 10\text{ m/s}^2$, the velocity of efflux through the orifice is:
A. $10\text{ m/s}$ ✓ Correct
B. $20\text{ m/s}$
C. $5\text{ m/s}$
D. $15\text{ m/s}$
Solution: By Torricelli's law $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\text{ m/s}$.
Q4 — Bernoulli's Theorem · easy · numerical
A tank is filled with water to a height of $20\text{ m}$. The speed of water escaping from a small hole at the bottom is ($g = 10\text{ m/s}^2$):
A. $20\text{ m/s}$ ✓ Correct
B. $40\text{ m/s}$
C. $200\text{ m/s}$
D. $10\text{ m/s}$
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\text{ m/s}$ — the same speed a body would reach falling freely through that height.
Q5 — Bernoulli's Theorem · hard · numerical
Air streams past an aeroplane wing at $70\text{ m/s}$ over the top surface and $60\text{ m/s}$ past the bottom surface. If the air density is $1.3\text{ kg/m}^3$ and the wing area is $20\text{ m}^2$, the dynamic lift is:
A. $4225\text{ N}$
B. $33800\text{ N}$
C. $8450\text{ N}$
D. $16900\text{ N}$ ✓ Correct
Solution: $\Delta P = \dfrac{1}{2}\rho(v_{top}^2 - v_{bot}^2) = \dfrac{1}{2}(1.3)(4900 - 3600) = 845\text{ Pa}$. Lift $= \Delta P \times A = 845 \times 20 = 16900\text{ N}$.
Q6 — Bernoulli's Theorem · hard · numerical
Water flows through a horizontal tube whose cross-sections at $P_1$ and $P_2$ are $A$ and $2A$. If the speed at $P_1$ is $4\text{ m/s}$, the pressure difference $(P_1 - P_2)$ for water ($\rho = 1000\text{ kg/m}^3$) is:
A. $-8000\text{ Pa}$
B. $+8000\text{ Pa}$
C. $+6000\text{ Pa}$
D. $-6000\text{ Pa}$ ✓ Correct
Solution: Continuity gives $v_2 = 2\text{ m/s}$. For a horizontal tube, $P_1 - P_2 = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = \dfrac{1}{2}(1000)(4 - 16) = -6000\text{ Pa}$ — the wider, slower section is at higher pressure.
Q7 — Bernoulli's Theorem · medium · theory
Bernoulli's theorem in its simple form is NOT applicable when the fluid is:
A. In steady streamline flow
B. Ideal and incompressible
C. Viscous and turbulent ✓ Correct
D. Non-viscous
Solution: The derivation assumes a non-viscous, incompressible fluid in steady streamline flow. Viscosity dissipates mechanical energy as heat, so the sum of the three terms no longer stays constant.
Q8 — Bernoulli's Theorem · easy · theory
According to Bernoulli's principle, at a point in a horizontal pipe where the flow speed is high, the pressure is:
A. Equal to atmospheric pressure
B. Low ✓ Correct
C. High
D. Unchanged
Solution: With the $\rho g h$ term constant in a horizontal pipe, $P + \dfrac{1}{2}\rho v^2$ is fixed, so an increase in $v$ must be paid for by a decrease in $P$.
Q9 — Bernoulli's Theorem · easy · theory
A Venturi meter is used to measure:
A. The surface tension of a liquid
B. The viscosity of a liquid
C. The rate of flow of a liquid through a pipe ✓ Correct
D. The atmospheric pressure
Solution: It has a constriction where the speed rises and the pressure falls. Measuring the pressure drop and applying Bernoulli plus continuity gives the flow rate.
Q10 — Bernoulli's Theorem · easy · theory
In a Venturi meter, the pressure of the liquid at the throat compared with that in the wider portion is:
A. Zero
B. The same
C. Higher
D. Lower ✓ Correct
Solution: The throat has a smaller area, so continuity makes the speed higher there, and Bernoulli then requires the pressure to be lower.
Q11 — Bernoulli's Theorem · medium · theory
A filter pump or atomizer (spray gun) works on the principle that:
A. Viscous drag pushes the liquid up the tube
B. Surface tension draws the liquid upward
C. A fast-moving stream of air produces a region of reduced pressure ✓ Correct
D. A fast-moving stream of air produces a region of increased pressure
Solution: Air blown rapidly across the top of the tube lowers the pressure there, so atmospheric pressure on the liquid surface pushes the liquid up and into the air stream.
Q12 — Bernoulli's Theorem · easy · theory
When a strip of paper is held below the lips and air is blown horizontally over its upper surface, the paper rises because:
A. The pressure above the strip falls below the pressure beneath it ✓ Correct
B. The blown air pushes the paper upward directly
C. The pressure above the strip rises above the pressure beneath it
D. The weight of the paper decreases
Solution: Fast-moving air above the paper has reduced pressure, while the still air beneath remains at atmospheric pressure. The resulting upward pressure difference lifts the strip.
Q13 — Bernoulli's Theorem · medium · theory
A spinning cricket ball swerves in flight. This phenomenon is known as:
A. The Raman effect
B. The Doppler effect
C. The Coriolis effect
D. The Magnus effect ✓ Correct
Solution: Spin drags air faster on one side and slower on the other. The speed difference creates a pressure difference across the ball, and the resulting sideways force curves its path.
Q14 — Bernoulli's Theorem · medium · theory
Two ships moving parallel and close to each other tend to be pulled together because:
A. The water between them flows faster, lowering the pressure there ✓ Correct
B. Of the surface tension of the sea water
C. Of the gravitational attraction between the ships
D. The water between them flows slower, lowering the pressure there
Solution: The gap between the hulls acts like a constriction, so the water speeds up and its pressure drops. The higher pressure on the outer sides then pushes the ships towards each other.
Q15 — Bernoulli's Theorem · medium · theory
Each term in Bernoulli's equation has the dimensions of:
A. Power per unit area
B. Energy per unit volume ✓ Correct
C. Force per unit length
D. Energy per unit mass
Solution: The terms $P$, $\dfrac{1}{2}\rho v^2$ and $\rho g h$ all carry the dimensions $[M^1L^{-1}T^{-2}]$, which is pressure, and equivalently energy per unit volume.
Q16 — Bernoulli's Theorem · easy · theory
For a fluid flowing through a horizontal pipe, Bernoulli's equation reduces to:
A. $\dfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$
B. $P = \text{constant}$
C. $P + \dfrac{1}{2}\rho v^2 = \text{constant}$ ✓ Correct
D. $P + \rho g h = \text{constant}$
Solution: A horizontal pipe keeps $h$ the same at every section, so the $\rho g h$ term is common to both sides and cancels out.
Q17 — Bernoulli's Theorem · medium · theory
The speed of efflux from a small hole at depth $h$ below the surface of a liquid is independent of:
A. Both the depth and gravity
B. The depth $h$ of the hole
C. The acceleration due to gravity
D. The density of the liquid ✓ Correct
Solution: Torricelli's result $v = \sqrt{2gh}$ contains only $g$ and $h$. The density cancels, so mercury and water escape at the same speed from the same depth.
Q18 — Bernoulli's Theorem · easy · numerical
Water escapes through a small hole $45\text{ m}$ below the free surface of a large tank. The efflux speed is ($g = 10\text{ m/s}^2$):
A. $15\text{ m/s}$
B. $90\text{ m/s}$
C. $45\text{ m/s}$
D. $30\text{ m/s}$ ✓ Correct
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 45} = \sqrt{900} = 30\text{ m/s}$.
Q19 — Bernoulli's Theorem · hard · numerical
Water ($\rho = 1000\text{ kg/m}^3$) flows through a horizontal pipe at $2\text{ m/s}$ in the wide section and $6\text{ m/s}$ in the narrow section. The pressure difference between the wide and narrow sections is:
A. $16000\text{ Pa}$ ✓ Correct
B. $32000\text{ Pa}$
C. $8000\text{ Pa}$
D. $4000\text{ Pa}$
Solution: $P_1 - P_2 = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = \dfrac{1}{2}(1000)(36 - 4) = 500 \times 32 = 16000\text{ Pa}$.
Q20 — Bernoulli's Theorem · hard · numerical
In a Venturi meter, water enters a section of area $10\text{ cm}^2$ at $2\text{ m/s}$ and passes through a throat of area $5\text{ cm}^2$. The pressure drop at the throat is ($\rho = 1000\text{ kg/m}^3$):
A. $6000\text{ Pa}$ ✓ Correct
B. $2000\text{ Pa}$
C. $3000\text{ Pa}$
D. $12000\text{ Pa}$
Solution: Continuity gives $v_2 = 4\text{ m/s}$. Then $\Delta P = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = 500(16 - 4) = 6000\text{ Pa}$.
Q21 — Bernoulli's Theorem · easy · numerical
A tank holds water to a depth of $1.25\text{ m}$. The speed of water emerging from a small hole at the bottom is ($g = 10\text{ m/s}^2$):
A. $2.5\text{ m/s}$
B. $5\text{ m/s}$ ✓ Correct
C. $12.5\text{ m/s}$
D. $25\text{ m/s}$
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.25} = \sqrt{25} = 5\text{ m/s}$.
Q22 — Bernoulli's Theorem · easy · numerical
Water escapes from a hole $80\text{ m}$ below the surface of a large tank. The efflux speed is ($g = 10\text{ m/s}^2$):
A. $20\text{ m/s}$
B. $1600\text{ m/s}$
C. $40\text{ m/s}$ ✓ Correct
D. $80\text{ m/s}$
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 80} = \sqrt{1600} = 40\text{ m/s}$.
Q23 — Bernoulli's Theorem · easy · numerical
A tank holds water to a depth of $0.8\text{ m}$. Water emerges from a small hole at the bottom with speed ($g = 10\text{ m/s}^2$):
A. $4\text{ m/s}$ ✓ Correct
B. $8\text{ m/s}$
C. $16\text{ m/s}$
D. $2\text{ m/s}$
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.8} = \sqrt{16} = 4\text{ m/s}$.
Q24 — Bernoulli's Theorem · hard · numerical
Water ($\rho = 1000\text{ kg/m}^3$) flows through a horizontal pipe at $3\text{ m/s}$ in the wide section and $5\text{ m/s}$ in the narrow section. The pressure difference between them is:
A. $4000\text{ Pa}$
B. $8000\text{ Pa}$ ✓ Correct
C. $16000\text{ Pa}$
D. $2000\text{ Pa}$
Solution: $\Delta P = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = 500(25 - 9) = 500 \times 16 = 8000\text{ Pa}$.
Q25 — Bernoulli's Theorem · hard · numerical
Air of density $1.2\text{ kg/m}^3$ flows at $50\text{ m/s}$ over the top and $40\text{ m/s}$ under a wing of area $10\text{ m}^2$. The dynamic lift is:
A. $2700\text{ N}$
B. $5400\text{ N}$ ✓ Correct
C. $10800\text{ N}$
D. $540\text{ N}$
Solution: $\Delta P = \dfrac{1}{2}(1.2)(2500 - 1600) = 0.6 \times 900 = 540\text{ Pa}$, so lift $= 540 \times 10 = 5400\text{ N}$.
Q26 — Bernoulli's Theorem · hard · numerical
In a Venturi meter, water enters a section of area $8\text{ cm}^2$ at $3\text{ m/s}$ and passes through a throat of area $4\text{ cm}^2$. The pressure drop at the throat is ($\rho = 1000\text{ kg/m}^3$):
A. $4500\text{ Pa}$
B. $6750\text{ Pa}$
C. $13500\text{ Pa}$ ✓ Correct
D. $27000\text{ Pa}$
Solution: Continuity gives $v_2 = 6\text{ m/s}$. Then $\Delta P = \dfrac{1}{2}(1000)(36 - 9) = 500 \times 27 = 13500\text{ Pa}$.
Q27 — Bernoulli's Theorem · easy · numerical
Water escapes from a small hole $3.2\text{ m}$ below the free surface of a tank. The efflux speed is ($g = 10\text{ m/s}^2$):
A. $6.4\text{ m/s}$
B. $16\text{ m/s}$
C. $8\text{ m/s}$ ✓ Correct
D. $4\text{ m/s}$
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 3.2} = \sqrt{64} = 8\text{ m/s}$.
Q28 — Bernoulli's Theorem · medium · numerical
The kinetic energy per unit volume of water ($\rho = 1000\text{ kg/m}^3$) flowing at $10\text{ m/s}$ is:
A. $10000\text{ J/m}^3$
B. $50000\text{ J/m}^3$ ✓ Correct
C. $100000\text{ J/m}^3$
D. $5000\text{ J/m}^3$
Solution: The kinetic term in Bernoulli's equation is $\dfrac{1}{2}\rho v^2 = \dfrac{1}{2}(1000)(100) = 50000\text{ J/m}^3$.