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Chapter 8 · 5 hours

Pressure

SEE past exam questions

Past questions and answers

57 questions set from this chapter, 3 of them more than once; 1 is most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 31 exams
  • Asked 3 times
  • 2080 Chaitra · Madhesh · 1 mark
  • 2081 Shrawan · Grade increment · 1 mark
  • 2079 Chaitra · Koshi · 1 mark

Write Pascal's law.

Answer

Pascal's law: Pressure applied on an enclosed fluid is transmitted equally and undiminished in all directions throughout the fluid.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Asked 2 times
  • 2081 Shrawan · Grade increment · 1 mark
  • 2082 Chaitra · Karnali · 1 mark

On which principle is hydraulic press based?
(a) Pascal's law (b) Archimedes principle (c) Law of flotation (d) Principle of baro-meter

Answer

(a) Pascal's law. A hydraulic press transmits the pressure applied on the small piston equally to the large piston through the liquid, so a small effort lifts a large load.

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  • Asked 2 times
  • 2082 Chaitra · Lumbini · 1 mark
  • 2082 Chaitra · Sudurpashchim · 1 mark

State the law of pressure that can be verified by the given figure.
Figure: A closed vessel filled with liquid, with pistons on four openings: a piston at the top (A) with force F, a piston B on the left with force F/2, a piston D on the right, and a piston C at the bottom with force 2F; marked to verify pressure transmission in a liquid.

Answer

The figure verifies Pascal's law: pressure applied on an enclosed liquid is transmitted equally and undiminished in all directions through the liquid. Each piston feels the same pressure P=FAP=\frac{F}{A}, so the force on each piston is proportional to its area.

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  • 2080 Chaitra · Lumbini · 1 mark

What is the upthrust exerted by the water to the body given in the figure.
(a) 0.2 N (b) 2 N (c) 20 N (d) 200 N
Figure: A beaker of water in which a body labelled A with mass 200 g is shown immersed (floating/submerged in the water).

Similar questions: Upthrust on 300 g body (2080 Chaitra · Sudurpashchim)

Answer

(b) 2 N. The body A floats or is suspended in water, so the upthrust equals its weight: U=mg=0.2 kg×10=2U = mg = 0.2\text{ kg}\times 10 = 2 N.

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  • 2080 Chaitra · Sudurpashchim · 1 mark

What is the upthrust exerted by water in the given body?
(a) 0.03 N (b) 0.3 N (c) 3 N (d) 30 N
Figure: A beaker of water in which a body A labelled 300 g is shown immersed (fully submerged, floating/suspended in the water).

Similar questions: Upthrust on 200 g body (2080 Chaitra · Lumbini)

Answer

(c) 3 N. The body A floats or is suspended in water, so the upthrust equals its weight: U=mg=0.3 kg×10=3U = mg = 0.3\text{ kg}\times 10 = 3 N.

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  • CDC model paper 2080, Science and Technology · 1 mark
  • SEE 2080 model question, Set 3, Science and Technology · 1 mark
  • 2081 Chaitra · Gandaki · 1 mark

State Archimedes' principle.

Answer

Archimedes' principle: When a body is partly or wholly immersed in a fluid, it experiences an upward force (upthrust) equal to the weight of the fluid displaced by it.

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  • 2081 Chaitra · Koshi · 1 mark
  • SEE 2080 model question, Set 3, Science and Technology · 1 mark

When a ship enters a river from a sea, it is found to sink more than before. What is the reason for this? (a) Due to the temperature of the sea water being more than that of river. (b) Due to the density of sea water is more than that of river water. (c) Due to the temperature of sea water being less than that of river water. (d) Due to the density of sea water being less than that of river water.

Answer

(b) Due to the density of sea water is more than that of river water. Sea water is denser, so it gives the same upthrust with less displaced volume. River water is less dense, so the ship must displace more water to get the same upthrust and therefore sinks deeper.

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  • SEE 2080 model question, Set 1, Science and Technology · 1 mark
  • 2079 Chaitra · Lumbini · 1 mark

What is upthrust?

Answer

Upthrust is the upward force exerted by a fluid (liquid or gas) on a body that is partly or fully immersed in it. It is also called buoyant force, and its SI unit is the newton (N).

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  • 2082 Chaitra · Bagmati · 1 mark

On which principle is the given instrument based? (i) Archimedes' principle (ii) Pascal's law (iii) law of gravitation (iv) Snell's law
Figure: A spherical vessel with several cylindrical pipes/pistons attached at the top, bottom, left and right, with the sphere filled with fluid (Pascal's law apparatus).

Answer

(ii) Pascal's law. The fluid-filled sphere with pistons shows that pressure applied at one piston is transmitted equally in all directions through the fluid.

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  • 2080 Grade increment · 1 mark

On the basis of the given table choose the correct fact.
Piston APiston B
Area = 25 cm2^2Area = 40 cm2^2
(a) Load of piston A can be lifted by applying less effort of piston B. (b) Load of piston B can be lifted by applying less effort of piston A. (c) Load of piston A can be lifted by applying more effort of piston B. (d) Effort applied on piston B will be equal to the load lifted in piston A.
Figure: Table giving piston areas: Piston A = 25 cm^2, Piston B = 40 cm^2.

Answer

(b) Load of piston B can be lifted by applying less effort on piston A. Piston B has the larger area (40 cm2^2) and piston A the smaller (25 cm2^2). Pressure is equal in both, so FB=FA×4025F_B = F_A\times\frac{40}{25}, which means the force on B is greater than the effort on A.

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  • 2080 Chaitra · Lumbini · 2 marks

By using a hydraulic lift a load of 2500 N needs to be lifted. If cross sectional area of the small piston is 110\frac{1}{10} of the big piston, calculate the force required to apply on small piston.

Answer

Given: Load (force on big piston) F2=2500F_2 = 2500 N; A1=110A2A_1 = \frac{1}{10}A_2 (small piston area A1A_1, big piston A2A_2).

Formula (Pascal's law):

F1A1=F2A2\frac{F_1}{A_1}=\frac{F_2}{A_2}

Substitution:

F1=F2×A1A2=2500×110F_1=F_2\times\frac{A_1}{A_2}=2500\times\frac{1}{10}

Result: F1=250F_1 = 250 N.

The force required on the small piston is 250 N.

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  • 2081 Chaitra · Bagmati · 1 mark

If 500 N force is applied on the piston A, how much force is produced on piston B?
(a) 875 N (b) 975 N (c) 1075 N (d) 1175 N
Figure: Hydraulic U-tube with liquid: piston A has area 200 cm² (500 N applied downward on it), piston B on the right has area 350 cm².

Answer

(a) 875 N.

Given: F1=500F_1 = 500 N, A1=200A_1 = 200 cm2^2, A2=350A_2 = 350 cm2^2. By Pascal's law F1A1=F2A2\frac{F_1}{A_1}=\frac{F_2}{A_2}, so

F2=500×350200=875 NF_2=\frac{500\times 350}{200}=875\text{ N}

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  • 2081 Chaitra · Sudurpashchim · 1 mark

In the adjoining figure, how much weight can be balanced on Piston B?
(i) 100 N (ii) 150 N (iii) 200 N (iv) 250 N
Figure: Hydraulic press (U-tube with two pistons holding liquid). Piston A on the right has area 2 cm^2 and carries a load of 30 N; piston B on the left has area 10 cm^2 and its load is unknown (?).

Answer

(ii) 150 N.

Given: FA=30F_A = 30 N, AA=2A_A = 2 cm2^2, AB=10A_B = 10 cm2^2. By Pascal's law FBAB=FAAA\frac{F_B}{A_B}=\frac{F_A}{A_A}, so

FB=30×102=150 NF_B=\frac{30\times 10}{2}=150\text{ N}

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  • 2082 Asar · Grade increment · 2 marks

In the figure, a sample of hydraulic machine is constructed by using syringes is shown. How much force is produced in piston B?
Figure: Two syringes joined by a tube: small syringe with piston A, force 20 N applied, area = 0.5 cm^2; large syringe piston B with area = 0.0005 m^2.

Answer

Given: FA=20F_A = 20 N, AA=0.5 cm2=0.5×10−4 m2=5×10−5 m2A_A = 0.5\text{ cm}^2 = 0.5\times10^{-4}\text{ m}^2 = 5\times10^{-5}\text{ m}^2, AB=0.0005 m2=5×10−4 m2A_B = 0.0005\text{ m}^2 = 5\times10^{-4}\text{ m}^2.

Formula (Pascal's law):

FBAB=FAAA\frac{F_B}{A_B}=\frac{F_A}{A_A}

Substitution:

FB=20×5×10−45×10−5=20×10=200 NF_B=\frac{20\times 5\times10^{-4}}{5\times10^{-5}}=20\times 10=200\text{ N}

The force produced on piston B is 200 N.

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  • 2082 Chaitra · Koshi · 1 mark

How much force is produced in piston Q in the given figure?
(a) 80 N (b) 100 N (c) 400 N (d) 500 N
Figure: Hydraulic press: two connected pistons over a liquid. Piston P (area 20 cm^2) carries a 200 N force; piston Q (area 50 cm^2) produces force F2.

Answer

(d) 500 N.

Given: FP=200F_P = 200 N, AP=20A_P = 20 cm2^2, AQ=50A_Q = 50 cm2^2. By Pascal's law

FQ=FP×AQAP=200×5020=500 NF_Q=\frac{F_P\times A_Q}{A_P}=\frac{200\times 50}{20}=500\text{ N}

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  • 2080 Chaitra · Bagmati · 1 mark

What will be the weight of the water displaced by a body of 400 g floating in water? (i) 0.4 N (ii) 4 N (iii) 40 N (iv) 400 N

Answer

(ii) 4 N.

A floating body displaces water whose weight equals the body's own weight (law of floatation). Weight =mg=0.4 kg×10 m/s2=4= mg = 0.4\text{ kg}\times 10\text{ m/s}^2 = 4 N, so the displaced water weighs about 4 N.

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  • 2080 Chaitra · Bagmati · 2 marks

Two dams are being made for the irrigation of field from a river. Which one is long lasting in between the base of dam of 6 ft. and the base dam of 8 ft. wider? Give scientific reason.

Answer

The dam with the 8 ft wide base will be longer lasting.

Reason: Pressure of water increases with depth (P=dghP = dgh), so the water exerts the greatest pressure and force at the bottom of the dam. A wider base is stronger and can withstand this large pressure, so the dam does not break. A 6 ft base is weaker against the high pressure at the bottom.

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  • 2080 Grade increment · 2 marks

Draw a labelled diagram that proves Archimede's principle.

Answer

Apparatus: spring balance, solid body, eureka (overflow) can filled to the spout, small beaker, and a balance to weigh the beaker.

          (W1)                         (W2)
     Spring balance                Spring balance
           |                              |
        [solid body]                  [solid body]
        in air                        in water
                                     _______________
                                    | eureka can    |
                                    |   [body]      |
                                    | ~~~~~~~~~~~~~ |___
                                    | water         |   \ spout
                                    |_______________|    \
                                                         [beaker]
                                                      displaced water (W3)
                                                            |
                                                         balance

Observation: Weight of body in air =W1= W_1. Weight of body submerged in water =W2= W_2. Loss of weight (upthrust) =W1−W2= W_1 - W_2. The displaced water is collected in the beaker. Let the weight of the empty beaker be WeW_e and the weight of the beaker with displaced water be W4W_4. Then the weight of displaced water is W3=W4−WeW_3 = W_4 - W_e.

We find W1−W2=W3W_1 - W_2 = W_3, i.e. the upthrust equals the weight of the displaced water. This proves Archimedes' principle.

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  • 2080 Chaitra · Gandaki · 2 marks

The object having more density sinks into the liquid having less density. Is this statement absolutely true? Justify with suitable example.

Answer

No, the statement is not absolutely true. A denser object sinks only if its average density is greater than that of the liquid, and the average density depends on the shape of the object, not only the material.

Example: A solid iron needle is made of iron, which is denser than water. It sinks, but a steel ship made hollow floats, because its average density (including the air inside) is less than that of water. Hence a denser material does not always sink in a less dense liquid: what matters is the average density of the whole object.

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  • 2080 Chaitra · Karnali · 1 mark

A fact and two statements about Pascal's law are given below:
Fact: The pressure given on an enclosed liquid is equally transmitted in all the directions by the liquid.
Statements:
a) The law is applicable in functioning of a submarine. b) The law is applicable in braking system of four wheelers.
(i) Fact and both the statements are correct. (ii) Fact is correct but both the statements are incorrect. (iii) Fact and statement 1 are correct but statement 2 is incorrect. (iv) Fact and statement 2 are correct but statement 1 is incorrect.

Answer

(iv) Fact and statement 2 are correct but statement 1 is incorrect.

The fact is Pascal's law. The braking system of four wheelers (hydraulic brake) works on Pascal's law, so statement b is correct. A submarine works on Archimedes' principle and the law of floatation, not Pascal's law, so statement a is incorrect.

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  • 2080 Chaitra · Karnali · 2 marks

The weight of displaced water is equal to the weight of the floating body on the water. Justify it with diagram.

Answer

When a body floats, it is in equilibrium: its weight (downward) is balanced by the upthrust (upward). By Archimedes' principle, upthrust equals the weight of the water displaced.

  Upthrust (U)
      ^
  ~~~~|~~~~~~~~~~~~ water surface
  |  [ body ]   |
  |      |       |
  |      v       |
  |  Weight (W)  |

For a floating body, U=WU = W (weight of body). Also UU = weight of displaced water.

Therefore, weight of displaced water = weight of the floating body. This is the law of floatation.

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  • 2080 Chaitra · Koshi · 2 marks

Differentiate between Archimedes' principle and law of floatation in two points.

Answer

Archimedes' principleLaw of floatation
It states that the upthrust on a body immersed in a fluid equals the weight of the fluid displaced.It states that a floating body displaces fluid whose weight equals the weight of the body.
It applies to bodies that are partly or fully immersed (sinking, floating or suspended).It applies only to floating bodies, where upthrust equals the weight of the body.

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  • 2081 Chaitra · Lumbini · 1 mark

In the given figure, what is the upthrust given by liquid on the immersed body.
(a) 65 N (b) 50 N (c) 35N (d) 15N
Figure: Two spring balances. In the first the body hangs in air and the reading is 50 N. In the second the body is immersed in liquid in an overflow vessel, and the liquid that overflows into a small beaker is labelled 15 N.

Answer

(d) 15 N. By Archimedes' principle, upthrust equals the weight of the liquid displaced, which is 15 N.

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  • 2082 Chaitra · Bagmati · 1 mark

What is the weight of the given object in water?
Figure: Two spring-balance set-ups. Left: a solid hanging in air from a spring balance reading 16 N. Right: the same solid immersed in water in a overflow (eureka) can, with the displaced water collected in a small beaker; the spring balance reading is marked 5 N.

Answer

From the figure, the spring balance reading with the solid immersed in water is 5 N. So the weight of the object in water is 5 N. (Loss of weight, i.e. upthrust, =16−5=11= 16-5 = 11 N, which equals the weight of the displaced water.)

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  • 2082 Chaitra · Sudurpashchim · 1 mark

In the given figure, what is the weight of an object in liquid? (i) 5 N (ii) 25 N (iii) 30 N (iv) 35 N
Figure: Two spring balances: the first shows the object's weight in air as 30 N; the second shows the object suspended in a liquid in an overflow vessel, with 5 N of liquid displaced and collected in a beaker.

Answer

(ii) 25 N. Upthrust = weight of displaced liquid = 5 N.

Weight in liquid = weight in air −- upthrust =30−5=25= 30 - 5 = 25 N.

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  • 2081 Chaitra · Bagmati · 2 marks

In the given figure, what does 16 N represent? What should be the weight of the object to float on the water?
Figure: A spring balance holds an object that is lowered into an overflow can of water; the displaced water collects in a vessel on a weighing scale reading 16 N.

Answer

The 16 N represents the weight of the displaced water, which by Archimedes' principle equals the upthrust acting on the object.

For the object to float on water, by the law of floatation its weight must be equal to the weight of the displaced water, i.e. 16 N.

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  • 2081 Chaitra · Gandaki · 2 marks

Study the given diagram and answer the questions asked below.
(i) How much upthrust is experienced by the stone?
(ii) What is the mass of stone if its weight in air is 7.35 N?
Figure: A spring balance holds a stone suspended in a eureka can full of water; displaced water flows out of the spout into a small beaker labelled 5 N (weight of displaced water).

Answer

(i) Upthrust = weight of displaced water (Archimedes' principle) == 5 N.

(ii) Given: weight in air W=7.35W = 7.35 N, g=9.8g = 9.8 m/s2^2.

m=Wg=7.359.8=0.75 kgm=\frac{W}{g}=\frac{7.35}{9.8}=0.75\text{ kg}

The mass of the stone is 0.75 kg (750 g).

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  • 2081 Chaitra · Karnali · 2 marks

If the weight of a cricket ball in water is 2 N and the weight of the water displaced by it is 1 N,
(i) What is the actual weight of the cricket ball?
(ii) On which principle is this experiment based?

Answer

(i) Upthrust = weight of displaced water = 1 N.

Actual weight = weight in water + upthrust =2+1== 2 + 1 = 3 N.

(ii) The experiment is based on Archimedes' principle.

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  • 2080 Chaitra · Madhesh · 1 mark

To measure the pressure of liquid p=dghp = dgh formula, what does hh stand for and what change occurs in pp when hh increases?
(a) Depth, pp increases (b) Depth, pp decreases (c) Height, pp increases (d) Height, pp decreases

Answer

(a) Depth, pp increases. In p=dghp = dgh, hh is the depth of the point below the liquid surface. Pressure is directly proportional to depth, so when hh increases, pp increases.

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  • 2080 Chaitra · Madhesh · 1 mark

Which fundamental units are involved in the unit of pressure?

Answer

Pressure =forcearea=\frac{\text{force}}{\text{area}}, so its SI unit is N m−2=kg m−1s−2\text{N m}^{-2}=\text{kg m}^{-1}\text{s}^{-2}. The fundamental units involved are kilogram (kg), metre (m) and second (s).

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  • 2080 Chaitra · Sudurpashchim · 2 marks

Consider you have to use air in breaking system of a four wheeler and in an immersed submarine to float it on water. What changes will you make in air pressure for them?

Answer

  • Braking system of a four wheeler: The air pressure is increased (compressed air is used in air brakes). By Pascal's law this high pressure is transmitted equally to the brake mechanism, giving a large braking force.
  • Submarine to float: Compressed air at high pressure is pumped into the ballast tanks to force the water out. This reduces the weight of the submarine, so its average density decreases and it rises and floats.

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  • 2081 Chaitra · Madhesh · 1 mark

When aluminum foil is placed on water in a beaker, it floats. When the same piece of foil is squeezed into a spherical shape, it sinks. What could be the reason for this? (a) The surface area decreases, upthrust decreases and sink. (b) Weight increases and sinks (c) Surface area increases and sinks (d) Density of foil increases and sinks

Answer

(a) The surface area decreases, upthrust decreases and sink. A flat foil spreads out and displaces a large volume of water, so the upthrust equals its weight and it floats. When squeezed into a ball it displaces much less water, so the upthrust becomes less than its weight and it sinks.

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  • 2081 Chaitra · Madhesh · 1 mark

Define one Pascal pressure.

Answer

One pascal is the pressure produced when a force of 1 newton acts perpendicularly on an area of 1 square metre: 1 Pa=1 N m−21\text{ Pa} = 1\text{ N m}^{-2}.

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  • 2081 Chaitra · Sudurpashchim · 1 mark

Name the force applied by water in upward direction to the body immersed in it.

Answer

The upward force is called upthrust (also called buoyant force or buoyancy).

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  • 2081 Chaitra · Koshi · 1 mark

Write the relation between up-thrust and density of liquid.

Answer

Upthrust is directly proportional to the density of the liquid: U=VdgU = V d g, so for the same volume immersed, a denser liquid gives a larger upthrust.

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  • 2082 Chaitra · Gandaki · 2 marks

Why does a stone sink when dropped in water? Explain on the basis of the law of floatation.

Answer

According to the law of floatation, a body floats only when the weight of the liquid it displaces equals its own weight. A stone is denser than water, so when fully immersed the weight of the water it displaces (the upthrust) is less than the weight of the stone. The net force is downward, so the stone sinks.

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  • 2082 Chaitra · Karnali · 1 mark

W1W_1, W2W_2 and W3W_3 are the weights of an object on weighing by a spring balance in air, water and oil respectively. What is the correct ascending order of given weights?
(a) W2<W3<W1W_2 < W_3 < W_1 (b) W1<W2<W3W_1 < W_2 < W_3 (c) W3<W2<W1W_3 < W_2 < W_1 (d) W2<W1<W3W_2 < W_1 < W_3

Answer

(a) W2<W3<W1W_2 < W_3 < W_1. In air there is almost no upthrust, so the weight is greatest (W1W_1). Water is denser than oil, so it gives a larger upthrust, making the apparent weight in water (W2W_2) the least and in oil (W3W_3) in between.

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  • 2082 Chaitra · Koshi · 1 mark

Write the principle which explains why a ship floats in water.

Answer

A ship floats because of the law of floatation (based on Archimedes' principle): the weight of water displaced by the ship equals the weight of the ship, because its hollow shape displaces a large volume of water.

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  • 2082 Chaitra · Lumbini · 1 mark

Why does an empty plastic bottle forcefully pushed into the water come up when it is released?
(a) Due to up thrust exerted by the bottle on water. (b) Due to up thrust exerted by the water on plastic bottle. (c) Due to pressure of water on plastic bottle. (d) Due to pressure of plastic bottle on water.

Answer

(b) Due to up thrust exerted by the water on plastic bottle. When fully pushed in, the bottle displaces a volume of water whose weight is much greater than the weight of the bottle, so the upward upthrust exceeds its weight and it rises when released.

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  • 2082 Chaitra · Madhesh · 1 mark

On which principle is the submarine based on?
(a) Pascal's law (b) Archimedes' principle (c) law of flotation (d) Newton's law

Answer

(b) Archimedes' principle. A submarine floats or sinks by taking in water or expelling it from ballast tanks, which changes its weight relative to the upthrust (the weight of water displaced), as in the law of floatation.

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  • 2081 Chaitra · Lumbini · 1 mark

Write an instrument based on Pascal's law.

Answer

A hydraulic press (also hydraulic lift or hydraulic brake) is based on Pascal's law.

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  • 2082 Chaitra · Madhesh · 1 mark

Name an instrument which works on the basis of atmospheric pressure.

Answer

A barometer works on the basis of atmospheric pressure. (A syringe or a drinking straw also works because of atmospheric pressure.)

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  • 2079 Chaitra · Gandaki · 1 mark

On which principle is a hydraulic garage lift based?

Answer

A hydraulic garage lift works on Pascal's law: pressure applied to an enclosed liquid is transmitted equally in all directions.

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  • 2079 Chaitra · Gandaki · 1 mark

How much is the specific gravity of gold?

Answer

The specific gravity (relative density) of gold is about 19.3. It has no unit, since it is a ratio of densities.

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  • 2079 Chaitra · Karnali · 1 mark

Name the person who invented the mercury barometer.

Answer

The mercury barometer was invented by Evangelista Torricelli, an Italian scientist, in 1643.

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  • 2079 Chaitra · Sudurpashchim · 1 mark

What change occurs in the upthrust when the density of the liquid increases?

Answer

Upthrust increases when the density of the liquid increases, because upthrust =Vdg= V d g is directly proportional to the density dd of the liquid.

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  • 2079 Chaitra · Madhesh · 1 mark

It is easier to lift a stone in water than in air. Why?

Answer

In water the stone experiences an upward force (upthrust) that acts against its weight, so its apparent weight is less and it is easier to lift.

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  • 2079 Chaitra · Madhesh · 2 marks

Write any two factors affecting liquid pressure.

Answer

The pressure at a point in a liquid, P=h d gP = h\,d\,g, depends on:

  1. Depth (hh): pressure increases as the depth below the liquid surface increases.
  2. Density of the liquid (dd): a denser liquid exerts more pressure at the same depth.

(The value of gg also matters, but it is constant on the earth's surface. The shape or size of the container and the amount of liquid do not affect it.)

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  • 2079 Chaitra · Sudurpashchim · 2 marks

Explain the working method of a syringe.

Answer

A syringe works on atmospheric pressure.

  • Drawing the liquid in: when the piston is pulled out, the volume inside the barrel increases and the pressure inside falls below atmospheric pressure. The atmospheric pressure acting on the liquid outside pushes it through the needle into the barrel.
  • Pushing the liquid out: when the piston is pushed in, the pressure inside the barrel rises above atmospheric pressure, so the liquid is forced out through the needle.

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  • 2079 Chaitra · Madhesh · 3 marks

A mercury barometer measures 760 mm of Hg atmospheric pressure at sea level. If the barometer is filled with water, what will be the height of the water raised inside the barometer at that place? Which of the two liquids is better to use as a barometric liquid? Explain.

Answer

The atmospheric pressure supports a column of either liquid, so hwdwg=hHgdHggh_w d_w g = h_{Hg} d_{Hg} g.

hw=hHg dHgdw=0.76×136001000≈10.34 mh_w=\frac{h_{Hg}\,d_{Hg}}{d_w}=\frac{0.76\times 13600}{1000}\approx 10.34\ \text{m}

The water column would rise to about 10.34 m (about 10336 mm).

Better liquid: mercury is better, because:

  • Its density is very high, so the column is only 760 mm. A water barometer would need a tube over 10 m long, which is impractical to make and read.
  • Mercury has a very low vapour pressure, so there is almost no vapour in the vacuum above it, and it does not wet the glass. Water evaporates easily, which would give wrong readings.

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  • 2079 Chaitra · Bagmati · 3 marks

A solid object is immersed fully in three different liquid media (water, salt solution and mercury) turn by turn, and the displaced mass is found as given in the table.
LiquidDisplaced mass (kg)
A1
B13.6
C1.5
(a) Identify mercury among them. (b) If each liquid has the same volume in a similar container, displaced mass of which liquid has more volume? Write the reason on the basis of Archimedes' principle.

Answer

(a) Mercury is the densest liquid (density 13.6 g/cm313.6\ \text{g/cm}^3), so it gives the largest displaced mass. Mercury is B (13.6 kg).

(b) The solid is fully immersed in each liquid, so it displaces the same volume of liquid every time, equal to its own volume. So the displaced liquids have equal volume; none has more. By Archimedes' principle the upthrust equals the weight of the displaced liquid, and the displaced mass is m=Vdm = V d. With VV the same, the displaced mass is larger for the liquid with greater density, which is why mercury (B) shows 13.6 kg and the others show less.

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  • 2079 Chaitra · Koshi · 4 marks

The figure shows a mercury barometer. 'B' has some fixed reading.
Figure: Mercury barometer with a glass tube, points A and B marked on the tube, mercury in the dish and atmospheric pressure arrow.
(a) What will be the value of reading 'B' when this device is taken to sea level? (b) The level of mercury changes when this device is taken at a higher altitude from sea level. Why? (c) Explain any two reasons for not using water instead of mercury in this device.

Answer

(a) At sea level the atmospheric pressure is normal, so the reading at B (the height of the mercury column) will be 760 mm (76 cm) of Hg.

(b) Atmospheric pressure decreases with altitude because the air above is thinner. The weaker atmospheric pressure supports a shorter mercury column, so the mercury level falls.

(c) Water is not used in place of mercury because:

  1. Water is much less dense than mercury, so the column would be about 10.3 m high. Such a long tube is not practical to build, carry or read.
  2. Water evaporates easily and its vapour fills the space above the column, giving a wrong reading. Water also wets the glass and is colourless, so it is hard to read.

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  • 2079 Chaitra · Gandaki · 4 marks

The densities of some materials are given in the table.
MaterialDensity (g/cm3^3)
A19.1
B11
C1
D0.92
(a) Which material floats on water among them? (b) Mass of which material will be more when the volumes of the materials A and D are the same? (c) Which of A and D displaces more water when they are kept in water and have the same volume? (d) Write the name of material D.

Answer

(a) A material floats on water if its density is less than 1 g/cm31\ \text{g/cm}^3. Only D (0.92) floats.

(b) Mass == volume ×\times density. For equal volumes the material with greater density has more mass, so A (19.1) has more mass.

(c) A displaces more water. A sinks and is fully immersed, so it displaces water equal to its whole volume. D floats with only part of it immersed, so it displaces less water.

(d) Material D, with density 0.92 g/cm30.92\ \text{g/cm}^3, is ice.

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  • CDC model paper 2080, Science and Technology · 1 mark

Why do objects float more easily in sea water than in river? (a) Density of sea water is more than that of river. (b) Density of sea water is less than that of river. (c) Temperature of sea water is more than that of river. (d) Temperature of sea water is less than that of river.

Answer

(a) Density of sea water is more than that of river. Sea water is denser, so it gives a larger upthrust for the same volume displaced, and objects float more easily.

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  • SEE 2080 model question, Set 2, Science and Technology · 1 mark

Write the relation between upthrust and weight of displaced liquid.

Answer

By Archimedes' principle, the upthrust on a body is equal to the weight of the liquid displaced by it.

Upthrust=Weight of displaced liquid=V d g\text{Upthrust} = \text{Weight of displaced liquid} = V\,d\,g

where VV is the volume of liquid displaced, dd its density and gg the acceleration due to gravity.

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  • SEE 2080 model question, Set 2, Science and Technology · 1 mark

On which principle given instrument based on?
(a) Pascal's law (b) Archimedes' Principle (c) Law of Gravitation (d) Newton's law
Figure: A U-shaped tube with two arms of different cross-section, each fitted with a piston, containing liquid (hydraulic press type arrangement).

Answer

(a) Pascal's law. The U-tube with two pistons of different cross-sections is a hydraulic press arrangement, which works by transmitting pressure equally through the liquid.

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  • SEE 2080 model question, Set 1, Science and Technology · 1 mark

If large piston A and small piston B of a hydraulic lift have their cross-sectional area, pressure and force/weight are A1A_1 and A2A_2, P1P_1 and P2P_2 and F1F_1 and F2F_2 respectively, then which of the following relation would be correct according to Pascal's law?
(a) P1>P2P_1 > P_2 (b) A1<A2A_1 < A_2 (c) F1>F2F_1 > F_2 (d) F1<F2F_1 < F_2

Answer

(c) F1>F2F_1 > F_2. By Pascal's law the pressure is the same on both pistons, P1=P2P_1=P_2, so F1A1=F2A2\frac{F_1}{A_1}=\frac{F_2}{A_2}. Since the large piston has A1>A2A_1 > A_2, the force on it is larger, F1>F2F_1 > F_2.

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Questions from SEE 2082 Science and Technology, 1041 Koshi (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Madhesh (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Bagmati (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Gandaki (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Lumbini (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Karnali (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2082 Science and Technology, 1041 Sudurpashchim (Board exam, Chaitra 2082 (scanned PDF, bilingual)), SEE 2081 Science and Technology, 1041 Koshi (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Madhesh (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Bagmati (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Gandaki (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Lumbini (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Karnali (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2081 Science and Technology, 1041 Sudurpashchim (Board exam, Chaitra 2081 (PDF; official answer keys exist on the bidhyamandir page)), SEE 2080 Science and Technology, 1041 Koshi (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Madhesh (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Bagmati (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Gandaki (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Lumbini (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Karnali (Board exam, 2080 (scanned PDF, bilingual)), SEE 2080 Science and Technology, 1041 Sudurpashchim (Board exam, 2080 (scanned PDF, bilingual)), SEE 2082 Science and Technology, GI-1041 (grade increment) (Grade increment, Asar 2083 (PDF)), SEE 2081 Science and Technology, GI-1041 (grade increment) (Grade increment 2081 (8 page images); month taken from the August 2025 post, check), SEE 2080 Science and Technology, GI-1041 (grade increment) (Grade increment 2080 (PDF)), CDC model paper 2080, Science and Technology (with answer key) (Official CDC model paper, pages 2-9; answer key pages 10-15), SEE 2080 model question, Set 1, Science and Technology (Model paper Set 1 (typed PDF, publisher not named, no answers)), SEE 2080 model question, Set 2, Science and Technology (Model paper Set 2 (typed PDF, publisher not named, no answers)), SEE 2080 model question, Set 3, Science and Technology (Model paper Set 3 (typed PDF, publisher not named, no answers)), SEE 2079 Compulsory Science (RE-111) Koshi (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Madhesh (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Bagmati (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Gandaki (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Lumbini (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken), SEE 2079 Compulsory Science (RE-111) Karnali (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken) and SEE 2079 Compulsory Science (RE-111) Sudurpashchim (Board exam, Chaitra 2079 (2023), last year of the old course (scanned PDF, bilingual, 75 marks); only questions that fit the current outline were taken). Answers are written for this site; check them against your class notes.

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