CBSE Class 10 Physics Thrust and Pressure MCQs Set 26

Answer: (b) 2.67 N m\( ^{-2} \)

Explanation:
1. Weight = 60 \( \times \) 10 = 600 N.
2. Pressure = 600 / 225 = 2.67 N m\( ^{-2} \).

Teacher's Note:
1. Use the weight (a force in newton), not the mass, in the pressure formula.
2. The unusually large area is what makes the pressure so small here.

Answer: (a) 5.1 \( \times \) 10\( ^{-2} \) m\( ^3 \)

Explanation:
1. Upthrust = 1400 - 900 = 500 N.
2. Upthrust = \( \rho \)Vg, so V = 500 / (998 \( \times \) 9.8) = 5.1 \( \times \) 10\( ^{-2} \) m\( ^3 \).

Teacher's Note:
1. When fully submerged the volume of displaced water equals the volume of the body.
2. Use the density of the liquid, not that of the rock.

Answer: (d) 1 : 1

Explanation:
1. Pressure in a liquid at a given depth is P = \( \rho \)gh.
2. It does not depend on the area of the surface.
3. Both membranes are at the same depth, so the ratio is 1 : 1.

Teacher's Note:
1. Pressure does not depend on the area, but the thrust does.
2. The thrusts here would be in the ratio 2 : 3.

Answer: (c) (x - y) : x

Explanation:
1. R.D. of the liquid = (x - y)/(x - z).
2. R.D. of the body = x/(x - z).
3. Ratio = (x - y)/x, i.e. (x - y) : x.

Teacher's Note:
1. Divide the loss in the liquid by the loss in water to get the relative density of the liquid.

Answer: (d) 7830 kg

Explanation:
1. Volume of the raft = 9 m\( ^3 \), so its mass = 150 \( \times \) 9 = 1350 kg.
2. Maximum water displaced = 9 \( \times \) 1020 = 9180 kg.
3. Load = 9180 - 1350 = 7830 kg.

Teacher's Note:
1. At the limit the raft is fully submerged.
2. Load + raft weight = upthrust.

Answer: (b) 2 N

Explanation:
1. Loss of weight = 10 - 8 = 2 N.
2. This loss equals the upthrust.
3. By Archimedes' principle the weight of the water displaced is 2 N.

Teacher's Note:
1. Loss of weight in a fluid = upthrust = weight of fluid displaced.

Answer: (c) 2 \( \times \) 10\( ^4 \) cm\( ^2 \)

Explanation:
1. F\( _1 \) = 10\( ^7 \) dyne = 10\( ^7 \times 10^{-5} \) = 100 N.
2. F\( _2 \) = 2000 \( \times \) 10 = 20000 N.
3. By Pascal's law F\( _1 \)/A\( _1 \) = F\( _2 \)/A\( _2 \), so A\( _2 \) = 100 \( \times \) 20000/100 = 2 \( \times \) 10\( ^4 \) cm\( ^2 \).

Teacher's Note:
1. 1 dyne = 10\( ^{-5} \) N.
2. The ratio of areas can be used directly, so there is no need to convert cm\( ^2 \) to m\( ^2 \).

Answer: (a) 10

Explanation:
1. Loss of weight = 32 - 28.8 = 3.2 gf.
2. R.D. = 32/3.2 = 10.

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

Answer: (d) weight of the liquid displaced by the object

Explanation:
1. By Archimedes' principle the upthrust on a body immersed in a fluid equals the weight of the fluid it displaces.
2. F\( _B \) = \( \rho \)\( _L \)V\( _L \)g.

Teacher's Note:
1. Option (c) is a mass, not a force, so it cannot be a force on its own.
2. The weight of the object has no direct role in the upthrust.

Answer: (a) 16,000 N

Explanation:
1. A floating body displaces fluid equal to its own weight.
2. The weight of the boat is 16,000 N, which does not change.
3. So it displaces 16,000 N of salt water as well, but a smaller volume.

Teacher's Note:
1. The weight of fluid displaced by a floating body is fixed by the body's weight.
2. The volume displaced changes with the liquid density.

Answer: (b) 1 : 1

Explanation:
1. The same force of 10 N acts on both surfaces.
2. Thrust equals the perpendicular force, so the ratio is 10 : 10 = 1 : 1.
3. The pressures would be in the ratio 2 : 1.

Teacher's Note:
1. Thrust is a force and pressure is force per unit area.
2. Read carefully which of the two the question asks for.

Answer: (d) 0.69

Explanation:
1. Fraction inside = 0.25/0.81 = 0.309.
2. Fraction above the surface = 1 - 0.309 = 0.69.

Teacher's Note:
1. Do not mix up the fraction inside with the fraction outside.

Answer: (d) 185 cm

Explanation:
1. Same pressure: 13.6 \( \times \) 74 = 5.44 \( \times \) h.
2. h = 13.6 \( \times \) 74 / 5.44 = 185 cm.

Teacher's Note:
1. Height varies inversely with density.
2. The liquid is 2.5 times less dense than mercury, so the column is 2.5 times taller.

Answer: (c) \( \tfrac{6}{7} \)V\( \rho \)g

Explanation:
1. Upthrust = V\( \rho \)g.
2. Weight of the ball = V(\( \rho \)/7)g.
3. Force needed = upthrust - weight = (6/7)V\( \rho \)g.

Teacher's Note:
1. The applied force must make up for the difference between upthrust and weight.

Answer: (d) 10\( ^{-3} \)

Explanation:
1. 1 g cm\( ^{-3} \) = 10\( ^{-3} \) kg / 10\( ^{-6} \) m\( ^3 \) = 1000 kg m\( ^{-3} \).
2. So 1 kg m\( ^{-3} \) = 10\( ^{-3} \) g cm\( ^{-3} \).
3. The SI unit is therefore 10\( ^{-3} \) times the CGS unit.

Teacher's Note:
1. The larger numerical value (1000) goes with the smaller unit.
2. Water has density 1 g cm\( ^{-3} \) = 1000 kg m\( ^{-3} \).

Answer: (a) both parts will sink

Explanation:
1. Sinking depends on density, which is the same throughout a homogeneous object.
2. Each part has the same density as the whole, which is greater than that of water.
3. So both parts sink.

Teacher's Note:
1. Size does not decide floating or sinking; density does.

Answer: (c) 100 N m\( ^{-2} \)

Explanation:
1. 1 bar = 10\( ^5 \) N m\( ^{-2} \).
2. 1 millibar = 10\( ^{-3} \) bar = 10\( ^{-3} \times 10^5 \) = 100 N m\( ^{-2} \).

Teacher's Note:
1. Milli means one thousandth.
2. Weather reports often give pressure in millibar (1 mbar = 1 hPa).

Answer: (b) 50 N

Explanation:
1. Thrust is the force acting perpendicular to the surface.
2. The whole 50 N acts perpendicular to the surface, so the thrust is 50 N.
3. The small area affects the pressure, not the thrust.

Teacher's Note:
1. Thrust does not depend on the area.
2. Pressure = thrust / area, which is very large for a tiny tip.

Answer: (c) decreases

Explanation:
1. Atmospheric pressure is due to the weight of the air above.
2. At higher altitude there is less air above, so the pressure decreases.

Teacher's Note:
1. The decrease is not uniform because air density also falls with height.
2. This is why mountaineers carry oxygen.

Answer: (a) 45 \( \times \) 10\( ^{-4} \) N

Explanation:
1. Contact area = 3 cm \( \times \) 3 cm = 9 cm\( ^2 \) = 9 \( \times \) 10\( ^{-4} \) m\( ^2 \).
2. Weight = P \( \times \) A = 5 \( \times \) 9 \( \times \) 10\( ^{-4} \) = 45 \( \times \) 10\( ^{-4} \) N.

Teacher's Note:
1. A cube rests on one face, so the area is side squared.
2. Convert the area to m\( ^2 \) before using pascal.

Answer: (a) pressure is the same at all points at the same level

Explanation:
1. Pressure in a liquid at rest depends on depth: P = P\( _0 \) + \( \rho \)gh.
2. Points at the same level have the same depth, so they have the same pressure.
3. Pressure at a point acts equally in all directions and also includes the atmospheric pressure on the surface.

Teacher's Note:
1. This is the content of Pascal's law for pressure at equal heights.
2. Pressure increases with depth, so option (b) is false.

Answer: (c) remain the same

Explanation:
1. The floating ice displaces water equal to its own weight.
2. Melted ice gives the same mass of water, which occupies exactly the displaced volume.
3. So the level does not change.

Teacher's Note:
1. This works because the ice weight equals the weight of water displaced.

Answer: (b) maximum when breadth and height form the base

Explanation:
1. The weight is the same in all cases.
2. Pressure = weight / base area, so it is largest for the smallest base.
3. Breadth \( \times \) height = 40 \( \times \) 20 = 800 cm\( ^2 \), smaller than 60 \( \times \) 40 = 2400 cm\( ^2 \) and 60 \( \times \) 20 = 1200 cm\( ^2 \).

Teacher's Note:
1. For a fixed force, pressure is inversely proportional to area.
2. Always compare the three possible base areas.

Answer: (b) 2 \( \times \) 10\( ^5 \) Pa

Explanation:
1. Force = weight = 40 \( \times \) 10 = 400 N.
2. Total area = 2 \( \times \) 10 cm\( ^2 \) = 20 cm\( ^2 \) = 2 \( \times \) 10\( ^{-3} \) m\( ^2 \).
3. Pressure = 400 / (2 \( \times \) 10\( ^{-3} \)) = 2 \( \times \) 10\( ^5 \) Pa.

Teacher's Note:
1. Convert cm\( ^2 \) to m\( ^2 \) with 1 cm\( ^2 \) = 10\( ^{-4} \) m\( ^2 \).
2. Remember that two femurs share the load, so the areas add.

Answer: (c) 4 g

Explanation:
1. Mass of liquid displaced = 2 \( \times \) 8 = 16 g.
2. Apparent weight = 20 - 16 = 4 g.

Teacher's Note:
1. The scale is calibrated in grams, so the upthrust can also be treated in gram-weight.
2. Loss of weight = mass of liquid displaced.

Answer: (d) it has no unit

Explanation:
1. Relative density is the ratio of two densities.
2. The units cancel, so it is a pure number.

Teacher's Note:
1. Relative density is also called specific gravity.
2. It tells how many times a substance is denser than water.

Answer: (b) 10 N

Explanation:
1. Volume of the ball = 4 / 4000 = 10\( ^{-3} \) m\( ^3 \).
2. Buoyant force = 1000 \( \times \) 10\( ^{-3} \) \( \times \) 10 = 10 N.

Teacher's Note:
1. The ball displaces its own volume of water.
2. Cross-check: weight = 40 N and relative density 4 gives upthrust 40/4 = 10 N.