CBSE Class 10 Physics Hydrostatics MCQs Set 30

Download CBSE MCQs for Class 10 Physics: Hydrostatics

Review structured MCQ sets for Class 10 Physics Hydrostatics. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Hydrostatics

Navigate directly to the 50 objective questions for Hydrostatics using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: Which of the following is/are units of pressure?
(a) pascal
(b) dyne cm-2
(c) N m-2
(d) all of the above
Show Answer & Explanation

Answer: (d) all of the above

Explanation:
1. Pressure = force/area, so its SI unit is N m-2, which is called the pascal.
2. In CGS the unit is dyne cm-2.
3. So all the three listed are units of pressure.

Teacher's Note:
1. 1 Pa = 1 N m-2.
2. 1 Pa = 10 dyne cm-2.

Question 2: Which of the following is used to measure atmospheric pressure accurately?
(a) mercury barometer
(b) manometer
(c) Fortin's barometer
(d) Torricelli barometer
Show Answer & Explanation

Answer: (c) Fortin's barometer

Explanation:
1. A Fortin's barometer has an adjustable mercury reservoir with a fixed zero-mark, so it gives an accurate reading.
2. A manometer measures gas pressure differences, not the atmospheric pressure.

Teacher's Note:
1. Fortin's barometer is a refined mercury barometer.

Question 3: A ship floating on water is in equilibrium. Then
(a) its centre of gravity and centre of buoyancy lie on the same vertical line.
(b) the volume of the immersed part of the ship is not equal to the volume of water displaced.
(c) the weight of the ship is not equal to the upthrust acting on it.
(d) its centre of gravity and centre of buoyancy lie on the same horizontal line.
Show Answer & Explanation

Answer: (a) its centre of gravity and centre of buoyancy lie on the same vertical line.

Explanation:
1. In equilibrium the weight equals the upthrust and the volume immersed equals the volume of water displaced.
2. The weight (through G) and the upthrust (through B) must be collinear, so G and B lie on the same vertical line.

Teacher's Note:
1. Equal and opposite forces must act along the same line.

Question 4: Match Column A with Column B.
(A) Neutral equilibrium (B) Stable equilibrium (C) Unstable equilibrium
(P) the metacentre is above the centre of gravity (Q) the metacentre coincides with the centre of gravity (R) the metacentre is below the centre of gravity
(a) A → Q; B → P; C → R
(b) A → R; B → P; C → Q
(c) A → P; B → R; C → Q
(d) A → R; B → Q; C → P
Show Answer & Explanation

Answer: (a) A → Q; B → P; C → R

Explanation:
1. Stable: the metacentre is above G, so a restoring couple acts (P).
2. Neutral: M coincides with G (Q).
3. Unstable: M is below G, so the ship topples further (R).

Teacher's Note:
1. Metacentre above the centre of gravity gives stability.

Question 5: Arrange the steps of the activity that shows that the atmosphere exerts pressure.
(A) Heat the can until steam comes out forcibly from its mouth.
(B) Take a tin can half filled with water, place it on a stand and heat it with a burner.
(C) Observe that the can gets crushed.
(D) Stop heating and immediately close the mouth of the can tightly with an airtight stopper.
(E) Remove the can from the stand and pour cold water over it.
(a) D E C A B
(b) A D E B C
(c) B A E D C
(d) B A D E C
Show Answer & Explanation

Answer: (d) B A D E C

Explanation:
1. Heat the water in the can (B), until steam drives out the air (A).
2. Stop heating and seal the can (D), then cool it (E).
3. The steam condenses, the inside pressure drops and the atmosphere crushes the can (C).

Teacher's Note:
1. Atmospheric pressure acts on the outside of the can.

Question 6: Assertion (A): The volume of an air bubble increases as it rises from the bottom to the surface of a water tank.
Reason (R): At constant temperature, the pressure of a gas is directly proportional to the volume it occupies.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) Both A and R are false.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. The pressure on the bubble falls as it rises, so its volume increases (Boyle's law); A is true.
2. At constant temperature pressure is inversely proportional to volume, so R is false.

Teacher's Note:
1. Boyle's law: PV = constant.

Question 7: A body of density 0.5 g cm-3 floats in a liquid of density 2 g cm-3. The maximum weight that can be placed on the body before it sinks is ______ the weight of the object.
(a) twice
(b) thrice
(c) equal to
(d) four times
Show Answer & Explanation

Answer: (b) thrice

Explanation:
1. Let the volume be V. The weight of the body is W = V × 0.5 × g.
2. When fully immersed the upthrust is V × 2 × g = 4W, and it must equal W + Wextra.
3. So Wextra = 3W (thrice).

Teacher's Note:
1. Maximum load is when the body is just fully immersed.

Question 8: A container is filled with a liquid of density 5 g cm-3. What is the pressure at a depth of 20 cm from the surface (including the atmospheric pressure of 105 Pa)? (g = 10 m s-2)
(a) 1.1 × 105 Pa
(b) 2.1 × 105 Pa
(c) 8 × 104 Pa
(d) 7 × 104 Pa
Show Answer & Explanation

Answer: (a) 1.1 × 105 Pa

Explanation:
1. Density = 5000 kg m-3, so hρg = 0.2 × 5000 × 10 = 104 Pa.
2. Total pressure = 105 + 104 = 1.1 × 105 Pa.

Teacher's Note:
1. P = Patm + hρg.

Question 9: An object floats in a liquid with 65% of its volume above the surface. If the density of the object is 0.8 g cm-3, what is the density of the liquid (in g cm-3)?
(a) 1.6
(b) 2.28
(c) 3.68
(d) 1.84
Show Answer & Explanation

Answer: (b) 2.28

Explanation:
1. The immersed fraction = 35% = 0.35.
2. For floating, Vimmersed ρl = V ρobject, so ρl = 0.8/0.35 = 2.28 g cm-3.

Teacher's Note:
1. Fraction immersed = ρobject/ρliquid.

Question 10: The angle of contact of a liquid in a capillary tube is 135°. The level of the liquid in the tube is ______ the level of the liquid in the container.
(a) above
(b) below
(c) equal to
(d) above or below
Show Answer & Explanation

Answer: (b) below

Explanation:
1. An obtuse angle of contact means the liquid does not wet the tube, so the meniscus is convex and the liquid is depressed.
2. The level in the tube is below the level outside.

Teacher's Note:
1. Acute angle: rise. Obtuse angle: depression.

Question 11: A pipe of negligible mass is fixed to a building to fill an overhead tank. Its vertical portion is completely filled with water. The pressure of water leaking out from a hole 100 cm above the bottom changes from 5 × 104 Pa to 2 × 104 Pa in 2 minutes. What is the speed of the centre of gravity of the water? (g = 10 m s-2, ρ = 1000 kg m-3)
(a) 0.125 cm s-1
(b) 1.25 cm s-1
(c) 0.125 m s-1
(d) 1.25 m s-1
Show Answer & Explanation

Answer: (b) 1.25 cm s-1

Explanation:
1. The pressure at the hole is ρgh with h the depth of water above the hole, so h falls from 5 m to 2 m, a drop of 3 m of the water level.
2. The centre of gravity of the column falls by half of that, 1.5 m, in 120 s.
3. Speed = 1.5/120 = 0.0125 m s-1 = 1.25 cm s-1.

Teacher's Note:
1. For a uniform column the centre of gravity is at half its height.

Question 12: The left limb of a manometer is connected to a gas cylinder and the mercury level in the right limb rises by 2 cm at normal atmospheric pressure. A vacuum pump then reduces the pressure of the gas by 2 mm of Hg every 4 seconds. By how much does the mercury level in the right limb shift in 10 minutes from the instant the pump is connected?
(a) 26 cm
(b) 13 cm
(c) 15 cm
(d) 9 cm
Show Answer & Explanation

Answer: (c) 15 cm

Explanation:
1. In 10 min = 600 s the pressure falls by (600/4) × 2 mm = 300 mm = 30 cm of Hg.
2. A change of 30 cm in pressure difference is shared equally between the two limbs, so the level in the right limb shifts by 15 cm.

Teacher's Note:
1. In a manometer the difference in levels equals the pressure difference; each limb shifts half.

Question 13: Liquids A, B and C of the same density and surface tension rise in three identical capillary tubes. The angles of contact are 30°, 35° and 40° respectively, and the capillary rises are hA, hB and hC. Which relation is correct?
(a) hA > hB > hC
(b) hA < hB < hC
(c) hA = hB = hC
(d) hA > hC > hB
Show Answer & Explanation

Answer: (a) hA > hB > hC

Explanation:
1. h = 2T cosθ/(rρg), so h ∝ cosθ.
2. cos30° > cos35° > cos40°, so hA > hB > hC.

Teacher's Note:
1. A smaller angle of contact gives a greater rise.

Question 14: Which of the following is a unit of thrust?
(a) dyne
(b) newton
(c) kg
(d) both (a) and (b)
Show Answer & Explanation

Answer: (d) both (a) and (b)

Explanation:
1. Thrust is a force, so it is measured in newton (SI) or dyne (CGS).
2. The kilogram is a unit of mass.

Teacher's Note:
1. 1 N = 105 dyne.

Question 15: A barometer in which no liquid is used is a/an
(a) aneroid barometer
(b) Torricellian barometer
(c) Fortin's barometer
(d) mercury barometer
Show Answer & Explanation

Answer: (a) aneroid barometer

Explanation:
1. The aneroid barometer uses a sealed metal box with flexible walls and no liquid.
2. The others use mercury.

Teacher's Note:
1. Aneroid means without liquid.

Question 16: A ship on the surface of water is in unstable equilibrium. Then
(a) its metacentre lies between the centre of gravity and the centre of buoyancy.
(b) its centre of gravity lies between the metacentre and the centre of buoyancy.
(c) its metacentre and centre of gravity coincide.
(d) its centre of gravity is below the metacentre.
Show Answer & Explanation

Answer: (a) its metacentre lies between the centre of gravity and the centre of buoyancy.

Explanation:
1. In unstable equilibrium the metacentre M is below the centre of gravity G.
2. Since B is below M, the order from the bottom is B, M, G, so M lies between B and G.

Teacher's Note:
1. Unstable: M is below G.

Question 17: Match Column A with Column B.
(A) Kinetic energy of molecules is zero (B) Fall in atmospheric pressure (C) Hydraulic press (D) Upthrust
(P) Pascal's law (Q) absolute zero temperature (R) volume of liquid displaced by a solid (S) depression
(a) A → R; B → S; C → P; D → Q
(b) A → Q; B → S; C → P; D → R
(c) A → Q; B → P; C → S; D → R
(d) A → P; B → Q; C → S; D → R
Show Answer & Explanation

Answer: (b) A → Q; B → S; C → P; D → R

Explanation:
1. Molecular kinetic energy is zero at absolute zero (Q).
2. A fall in atmospheric pressure is called a depression (S).
3. A hydraulic press works on Pascal's law (P).
4. Upthrust equals the weight of the volume of liquid displaced by the solid (R).

Teacher's Note:
1. Archimedes' principle links upthrust to displaced volume.

Question 18: Arrange the steps to find the relative density of a liquid using Archimedes' principle.
(A) Find the apparent loss of weight of the body in the liquid.
(B) Determine the weight of the body in air using a spring balance.
(C) Find the apparent loss of weight of the body in water.
(D) Immerse the body in the given liquid and determine its weight.
(E) Immerse the body in water and determine its weight.
(F) Find the relative density as the ratio of the loss of weight in the liquid to that in water.
(a) E F B D C A
(b) B D A E C F
(c) B D E F A C
(d) A E C F B D
Show Answer & Explanation

Answer: (b) B D A E C F

Explanation:
1. Weigh in air (B), weigh in the liquid (D) and find the loss (A).
2. Weigh in water (E) and find the loss (C).
3. Finally take the ratio (F). The order is B, D, A, E, C, F.

Teacher's Note:
1. Relative density = loss of weight in liquid / loss of weight in water.

Question 19: Assertion (A): The cohesive forces in solids are very strong.
Reason (R): The intermolecular space in solids is very small.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) Both A and R are false.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Molecules in solids are packed closely, so the intermolecular forces are strong; A is true and R is true.
2. The small intermolecular space is the reason the forces are strong, so R explains A.

Teacher's Note:
1. Force between molecules grows as they get closer.

Question 20: A barometer has some dry air trapped above the mercury in the glass tube. At normal atmospheric pressure, the height of the mercury column is
(a) < 76 cm
(b) > 76 cm
(c) = 76 cm
(d) = 100 cm
Show Answer & Explanation

Answer: (a) < 76 cm

Explanation:
1. The trapped air exerts a pressure on the mercury, so Patm = hρg + Pair.
2. Hence the mercury column is shorter than 76 cm.

Teacher's Note:
1. A good barometer has a vacuum above the mercury.

Question 21: An object floats on a liquid of density 5 g cm-3. The extra force needed to immerse the object completely is four times the weight of the object. What is the density of the object (in g cm-3)?
(a) 2
(b) 1
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Full upthrust = W + 4W = 5W, i.e. Vρlg = 5Vρog.
2. ρo = 5/5 = 1 g cm-3.

Teacher's Note:
1. Extra force = full upthrust - weight.

Question 22: The pressure at the bottom of a swimming pool is 2.1 × 105 Pa and the atmospheric pressure is 105 Pa. What is the depth of the pool? (ρ = 1000 kg m-3, g = 10 m s-2)
(a) 5 m
(b) 11 m
(c) 15 m
(d) 20 m
Show Answer & Explanation

Answer: (b) 11 m

Explanation:
1. hρg = 2.1 × 105 - 105 = 1.1 × 105 Pa.
2. h = 1.1 × 105/(1000 × 10) = 11 m.

Teacher's Note:
1. Subtract the atmospheric pressure first.

Question 23: The density of an object floating on a liquid of density 2 g cm-3 is 0.6 g cm-3. What percentage of the volume of the object is outside the liquid?
(a) 70%
(b) 30%
(c) 35%
(d) 65%
Show Answer & Explanation

Answer: (a) 70%

Explanation:
1. Immersed fraction = 0.6/2 = 30%.
2. Volume outside = 100% - 30% = 70%.

Teacher's Note:
1. Fraction immersed = ρobject/ρliquid.

Question 24: The angle of contact for a liquid in a capillary tube is 40°. The level of the liquid in the tube is ______ the level in the container.
(a) equal to
(b) below
(c) above
(d) above or below
Show Answer & Explanation

Answer: (c) above

Explanation:
1. An acute angle of contact means the liquid wets the tube and the meniscus is concave.
2. The liquid rises in the tube, so the level is above that in the container.

Teacher's Note:
1. Acute angle: capillary rise.

Question 25: A pipe of negligible mass is fixed to a building to fill an overhead tank. Its vertical portion is completely filled with water, and the pressure of water leaking from a hole 50 cm above its bottom changes from 6 × 104 Pa to x × 104 Pa in 1 minute. If the average speed of the centre of gravity of the water is 2 cm s-1, find x. (g = 10 m s-2, ρ = 1000 kg m-3)
(a) 6
(b) 3.6
(c) 2.4
(d) 3.25
Show Answer & Explanation

Answer: (b) 3.6

Explanation:
1. In 60 s the centre of gravity falls 2 cm s-1 × 60 s = 1.2 m, so the water level falls 2.4 m.
2. The pressure at the hole falls by 2.4 m × 1000 × 10 = 2.4 × 104 Pa.
3. x = 6 - 2.4 = 3.6.

Teacher's Note:
1. The level falls twice as much as the centre of gravity.

Question 26: The left limb of a manometer is connected to a gas cylinder at normal atmospheric pressure, and the mercury level in the right limb is 1 cm lower. A vacuum pump is then connected, and the mercury level in the right limb falls a further 3 cm in 8 minutes. What is the rate of decrease in the pressure of the gas?
(a) 125 cm of Hg s-1
(b) 76 cm of Hg s-1
(c) 0.125 mm of Hg s-1
(d) 12.5 mm of Hg s-1
Show Answer & Explanation

Answer: (c) 0.125 mm of Hg s-1

Explanation:
1. A fall of 3 cm in the right limb is matched by a 3 cm rise in the left limb, so the pressure difference changes by 6 cm = 60 mm of Hg.
2. Rate = 60 mm/(8 × 60 s) = 0.125 mm of Hg s-1.

Teacher's Note:
1. The change in level difference equals twice the shift of one limb.

Question 27: Soaps and detergents are added to water for washing clothes because
(a) they prevent wetting.
(b) they decrease the angle of contact.
(c) they increase the surface tension.
(d) they increase the angle of contact.
Show Answer & Explanation

Answer: (b) they decrease the angle of contact.

Explanation:
1. Soaps lower the surface tension and reduce the angle of contact, so water spreads and wets the cloth fibres better and penetrates dirt.

Teacher's Note:
1. Detergents are wetting agents.

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Multiple Choice Questions (MCQs) for Class 10 Physics Hydrostatics

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