CBSE Class 10 Physics Thrust and Pressure MCQs Set 19

Answer: (b) 8/9 g cm\( ^{-3} \)

Explanation:
1. In water the body is 2/3 immersed, so its density = 2/3 g cm\( ^{-3} \).
2. In the other liquid: \( \rho_s \) = (3/4)\( \rho_L \).
3. \( \rho_L \) = (2/3)/(3/4) = 8/9 g cm\( ^{-3} \).

Teacher's Note:
1. Find the density of the body first using the fraction in water.

Answer: (c) increases

Explanation:
1. Upthrust = \( \rho \)\( _L \)Vg.
2. For a fully submerged rod V is fixed, so upthrust is proportional to the density of the liquid.
3. It increases.

Teacher's Note:
1. Same body, denser liquid, larger upthrust.

Answer: (d) manometer

Explanation:
1. A manometer uses a liquid column to measure fluid pressure.
2. A barometer measures atmospheric pressure, a hydrometer measures the density of liquids and an odometer measures distance.

Teacher's Note:
1. Match each instrument with the quantity it measures.

Answer: (d) 12 N

Explanation:
1. For a fully submerged body, F\( _B \) is proportional to the liquid density.
2. The density becomes 3 times, so F\( _B \) = 3 \( \times \) 4 = 12 N.

Teacher's Note:
1. The cylinder's own density is a distractor here.

Answer: (d) 1 kg m\( ^{-1} \) s\( ^{-2} \)

Explanation:
1. 1 Pa = 1 N/m\( ^2 \).
2. 1 N = 1 kg m s\( ^{-2} \), so 1 Pa = kg m s\( ^{-2} \) / m\( ^2 \) = kg m\( ^{-1} \) s\( ^{-2} \).

Teacher's Note:
1. Always reduce a derived unit to base units to test a given option.
2. Option (a) is the unit of force per length, not per area.

Answer: (c) 9 kg

Explanation:
1. Density of the liquid = 0.8 \( \times \) 1 kg per litre = 0.8 kg/L.
2. Mass of liquid = 10 \( \times \) 0.8 = 8 kg.
3. Total mass = 8 + 1 = 9 kg.

Teacher's Note:
1. Do not forget to add the mass of the bucket.
2. 1 litre of water has a mass of 1 kg.

Answer: (c) increases in size

Explanation:
1. Pressure in water decreases as the bubble rises.
2. The bubble's volume is inversely related to the surrounding pressure (Boyle's law), so it expands.

Teacher's Note:
1. Bubbles from the bottom of a pond are small and grow as they approach the surface.
2. Pressure depends only on depth, not on the size of the bubble.

Answer: (c) decreases

Explanation:
1. The density of the water increases after the salt is added.
2. For the same weight a smaller volume of the denser liquid supplies the same upthrust.
3. So the object rises a little and the immersed volume decreases.

Teacher's Note:
1. Objects float higher in sea water than in river water.
2. The object does not sink.

Answer: (a) 1.47 cm

Explanation:
1. Equal pressures give \( \rho \)\( _{Hg} \)h\( _{Hg} \) = \( \rho \)\( _w \)h\( _w \).
2. h\( _{Hg} \) = 1 \( \times \) 20 / 13.6 = 1.47 cm.

Teacher's Note:
1. Equal pressure means the denser liquid has the shorter column.

Answer: (c) it has a high density

Explanation:
1. The height of the column is h = P/\( \rho \)g.
2. Mercury is very dense, so the column is only about 76 cm.
3. Water would need a column of about 10.3 m.

Teacher's Note:
1. Height is inversely proportional to density at a given pressure.
2. 760 mm of Hg = 1 atm.

Answer: (b) decrease the pressure on the road

Explanation:
1. More wheels give a larger total area of contact with the road.
2. For the same weight, a larger area means a smaller pressure.

Teacher's Note:
1. Heavy vehicles use many wheels or broad tyres to avoid damaging roads.
2. The same idea explains the wide foundations of buildings.

Answer: (b) increases of pressure are transmitted equally to all parts of a fluid

Explanation:
1. Hydraulic machines work on Pascal's law.
2. A pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of the container.

Teacher's Note:
1. A small force on a small piston gives a large force on a large piston.
2. Option (c) explains the variation of pressure with depth, which is a different idea.

Answer: (c) \( \dfrac{D}{d} \)

Explanation:
1. Weight = upthrust gives DVg = dvg.
2. So v/V = D/d.

Teacher's Note:
1. This ratio is less than 1 only if D < d, which is the condition for floating.

Answer: (b) a unit of pressure

Explanation:
1. One atmosphere is the pressure exerted by the air column at sea level.
2. 1 atm = 1.013 \( \times \) 10\( ^5 \) Pa, so it is a unit of pressure.

Teacher's Note:
1. Do not confuse the atmosphere (the air envelope) with the atm (a unit).
2. 1 bar = 10\( ^5 \) Pa is another common unit.

Answer: (b) 1.8 \( \times \) 10\( ^4 \) N m\( ^{-2} \)

Explanation:
1. Each layer is 1 m deep. Water: 1000 \( \times \) 10 \( \times \) 1 = 10000 Pa.
2. Oil (800 kg m\( ^{-3} \)): 800 \( \times \) 10 \( \times \) 1 = 8000 Pa.
3. Total = 18000 Pa = 1.8 \( \times \) 10\( ^4 \) N m\( ^{-2} \).

Teacher's Note:
1. Add the pressure of each layer separately.
2. Convert 0.80 g cm\( ^{-3} \) to 800 kg m\( ^{-3} \).

Answer: (d) decreases

Explanation:
1. The upthrust acts opposite to the weight.
2. Apparent weight = weight - upthrust, which is less than the weight in air.
3. So the reading decreases.

Teacher's Note:
1. The reading shows the apparent weight.
2. The spring balance reads the true weight only in a vacuum or, nearly, in air.

Answer: (b) 78 kg

Explanation:
1. Volume = 3 \( \times \) 4 \( \times \) 5 = 60 m\( ^3 \).
2. Mass = density \( \times \) volume = 1.3 \( \times \) 60 = 78 kg.

Teacher's Note:
1. Mass = density \( \times \) volume.
2. Air is light, but a whole room still holds tens of kilograms of it.

Answer: (d) 1.03 \( \times \) 10\( ^7 \) Pa

Explanation:
1. Gauge pressure = \( \rho \)gh.
2. = 1.03 \( \times \) 10\( ^3 \) \( \times \) 10 \( \times \) 1000 = 1.03 \( \times \) 10\( ^7 \) Pa.

Teacher's Note:
1. Gauge pressure excludes the atmosphere.
2. Deep-sea submarines are built with very strong hulls for this reason.

Answer: (a) add the atmospheric pressure

Explanation:
1. Gauge pressure is measured above atmospheric pressure.
2. So absolute pressure = gauge pressure + atmospheric pressure.

Teacher's Note:
1. Adding 273 converts Celsius to kelvin and has nothing to do with pressure.
2. For a pressure below atmospheric, subtract the vacuum pressure instead.

Answer: (a) relative density

Explanation:
1. Relative density = density of substance / density of water.
2. Hence density of substance = relative density \( \times \) density of water.

Teacher's Note:
1. Mass and volume alone cannot give density without dividing one by the other.

Answer: (c) 2.01 \( \times \) 10\( ^5 \) Pa

Explanation:
1. Gauge pressure due to water = \( \rho \)gh = 1000 \( \times \) 10 \( \times \) 10 = 1.0 \( \times \) 10\( ^5 \) Pa.
2. Absolute pressure = atmospheric + gauge = 1.01 \( \times \) 10\( ^5 \) + 1.0 \( \times \) 10\( ^5 \) = 2.01 \( \times \) 10\( ^5 \) Pa.

Teacher's Note:
1. Absolute pressure includes the atmosphere acting on the lake surface.
2. About every 10 m of water adds one atmosphere.

Answer: (d) lactometer

Explanation:
1. A lactometer floats in milk and its depth of sinking shows the density of the sample.
2. Adulterated milk has a different density from pure milk.

Teacher's Note:
1. A hydrometer is the general instrument for liquid density, while a lactometer is specifically for milk.
2. Both work on the law of floatation.

Answer: (d) hang at whatever position it is released

Explanation:
1. The density of the solid equals that of water.
2. Upthrust equals weight at every depth, so the net force is zero.
3. It stays where it is released.

Teacher's Note:
1. The solid is in neutral equilibrium.

Answer: (b) increase the pressure created at the tip

Explanation:
1. A pointed tip has a very small area.
2. For the same force, a smaller area gives a larger pressure, so the nail enters the surface easily.

Teacher's Note:
1. Knives, needles and nails all use small areas to create high pressure.
2. The thrust is unchanged; only the pressure rises.

Answer: (c) centre of buoyancy

Explanation:
1. The upthrust acts through the centre of gravity of the displaced liquid.
2. This point is called the centre of buoyancy.

Teacher's Note:
1. It coincides with the geometric centre only for a uniform fully immersed body.

Answer: (a) fall

Explanation:
1. Before melting, the floating system displaces water of weight equal to (ice + lead).
2. After melting, the ice turns into water that fills exactly the volume the ice part displaced, but the lead sinks and displaces only its own small volume.
3. The lead now displaces less water than it did while floating, so the level falls.

Teacher's Note:
1. A floating lead piece displaces more water (by weight) than a sunk piece.
2. Total displaced volume decreases.

Answer: (c) 3/4

Explanation:
1. For a uniform cylinder the immersed fraction of volume equals the fraction of length.
2. The fraction = 750/1000 = 3/4.

Teacher's Note:
1. The cross-sectional area is the same along the length.