CBSE Class 9 - Science - Chapter 10 - Floatation & Archimedes’ Principle
Floatation & Archimedes’ Principle
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| Italian Stamp in honour of Archimedes credits: St. Andrews Univ |
1. The force acting per unit area of the object is known as pressure.
The SI unit of Pressure is N/m2 or Pascal (Pa)
2. Density or Mass density is the ratio of mass per unit volume.
The SI unit of density is kg/m3
A density of an object is a key factor that decides whether it’ll float or sink in a liquid.
Q3: The weight of a man is 690N which contains 5.3 x 10-3 m3 of blood.
Find (a) blood's weight. Take density of blood is 1060 kg/m3
(b) express it as a percentage of body weight.
Answer: (a) Since m = ρ x V
and W = mg
∴ W = (ρV)g = 1060 x 5.3 x 10-3 x 9.80
⇒ W = 54N
(b) % of blood weight wrt Body's weight = (54N/690N) x 100 = 7.8%
4. Pressure in Liquids
- Fluids (gases and liquids) exert pressure due to their weight.
- The greater the height of a fluid above a surface, the greater the pressure exerted by the fluid on that surface.
- Pressure Increases with the depth.
- At any point inside a fluid, pressure is exerted in all directions. The pressure on each side is perpendicular to the surface, no matter what its shape.
- Pressure at any point in fluids (P) = hdg,
where h is the height/depth, d is a density of liquids and g is acceleration due to gravity.
a. same as P
b. smaller than P
c. larger than P
d. cannot be determined.
Answer: Correct answer is b. The Pressure at any point is = hdg. For the given situation h and g are fixed. So fluid with less density will apply less pressure.
Q5: What do you mean by Thrust or Buoyant force?
Answer: The upward force exerted by a liquid when a body is immersed in the liquid is called thrust or Buoyant Force.
Note: When in equilibrium the buoyant force is balanced by the weight of an object or the force of gravity acting on it.
Q6: What are the factors the buoyant force depends on?
Answer: 1. The volume of the object immersed in the liquid.
2. The density of the liquid.
(Note: Buoyant force does not change with depth).
Q7 (NCERT):Why does an object float or sink when placed on the surface of the water?
Answer: If the density of an object is higher than the density of the liquid, then it sinks in the liquid. This is because the buoyant force acting on the object is less than the force of gravity. In case, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity.
Q8: The following figure shows four containers containing a liquid at the same height. Rank them according to the pressure exerted by the liquid on the bottom of the containers in ascending order.
b. 4. 3, 1, 2
c. 3, 4, 1, 2
d. same pressure in each case.
Answer: (d)
Q9: What are Plimsoll Lines?
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| Plimsoll Lines |
Answer: While loading a ship, care is taken it should not be loaded too much. Its load depends on the density of the sea in which it floats. The ship would sink if its load exceeds. To indicate the depth up to which the ship can sink in a sea of different densities under different climatic conditions, several lines are marked on the sides of the ship. It ensures stability and safety of the ship and these lines are called Plimsoll lines (name after sailor Captain Plimsoll) For details, check the wiki.
Q10: Define 1 Pascal.
Answer: When a force of 1 newton acts normally on an area of cross-section 1 m2, then the pressure experienced by the surface is said to one pascal.
Q11(CBSE 2011): State Archimedes’ Principle. Based on this principle, write its two applications.
Answer: When a body is partially or wholly immersed in a fluid, it experiences an upthrust and apparently loses its weight which is equal to the weight of the fluid displaced by the immersed part of the body.
Applications based on Archimedes’ principle are:
- designing of ships and submarines
- Lactometers, which are used to determine the purity of a sample of milk
- hydrometers used for determining the density of liquids
Q12:Mass of a rectangular bar of an iron piece is 320 g. Its dimensions are 2 × 2 × 10 cm3. What is its specific gravity? Will the bar float or sink in water?
Answer: Relative density is also called as specific gravity. It is the ratio of the density of a substance to the density of water (i.e. 1 g per cubic centimetre).
∴ Sp. Gravity of iron = Density of Iron/Density of Water
Density of Iron = mass / volume = 320g/(2 × 2 × 10 cm3) = 320 / 40 = 8 g/cm3
Density of water = 1 g/cm3
∴ Sp. Gravity of iron = 8/1 = 8 ... (answer)
Since the density of the bar is more than the density of water, the bar will sink.
Q13(NCERT): Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer: A sheet of paper will fall slower than the one which is crumpled into a ball because air offers resistance to the falling objects. The paper in sheet form has larger surface area than that of a crumpled paper ball. Thus the sheet will experience more pressure (air resistance) and falls slower.
Q14(NCERT): In what direction does the buoyant force on an object immersed in a liquid act?
Answer: The buoyant force on an object immersed in a liquid always acts vertically upwards.
Q15(NCERT): Why does a block of plastic released under water come up to the surface of the water?
Answer: The upward buoyant force acting on the block is greater than the downward gravitational force due to its weight. Due to this the block of plastic released under water comes up to the surface of the water.
Q16(NCERT): The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm-3, will the substance float or sink?
Answer: Given, mass of the substance (m) = 50g
volume of the substance (v) = 20 cm3
Density of the substance = m/v = 50/20 = 2.5 g/cm3
Since the density of a substance is more than the density of water, the substance will sink.
Q17(NCERT): The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm-3? What will be the mass of the water displaced by this packet?
Answer: Given, mass of the packet (m) = 500g
volume of the packet = 350 cm3
∴ density of the packet = m/v = 500/250 = 1.428 g cm-3
density of water = 1g cm-3
Since the density of the packet is higher than density of water, it will sink.
mass of water displaced by the packet = volume of the packet × density of water
= 350 cm3 × 1 g cm-3 = 350g
Q18(NCERT): Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer: Pressure = Force/Area
⇒ Pressure is inversely proportional to the surface area on which the force acts.
Since the contact surface area of a thin string, strap is very small, it exerts larger pressure on the shoulder.
Q19(NCERT): You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer: The actual mass shall be more than 42kg. When we weigh, we experience upward thrust (buoyant force) due to air. Thus weighing machine shows our apparent weight.
Apparent weight = True weight - buoyant force.
Q20: As you climb up a mountain, your ears “pop” because of the changes in atmospheric pressure. In which direction, outward or inward, does your eardrum move when you climb up?
Answer: Outward. Since the internal pressure of the body is higher than atmospheric pressure.
Q21: Is the Archimedes principle applicable to gases?
Answer: Yes. The Archimedes principle is applicable to fluids (liquids + gases).
Q22: Why do sewing needles and all pins have a sharp pointed tip?
Answer: Pressure = Force/Area
⇒ Pressure is inversely proportional to the surface area on which the force acts.When we pin up the
papers or sew the clothes, we apply force over a very small area which leads to a larger pressure. That's why sewing needles and all pins have pointed tips.
Q23(CBSE 2011): Loaded test-tube placed in pure milk sinks to a certain mark (M). Now some water is mixed with the milk. Will the test tube sink more or less? Explain.
Answer: Density of milk is higher than the density of water. When some water is mixed with milk, then the mixture will have less density than pure milk. It will exert less amount of buoyant force. Consequently, test tube will sink more.
Q24(CBSE Proficiency Test): The weight of an empty glass is 1 N. When completely filled with water, the glass weighs 4 N. If you now throw the water away and fill the glass partly with sand, the weight becomes 4 N. If the rest of the glass is now filled up with water, the weight becomes 6 N. What is the relative density of sand?
Answer: Given, weight of empty glass = 1N
Weight of glass filled with water = 4N
⇒ Weight of water filled in glass = 4 - 1 = 3N
Since Weight = mg (let g = 10 m/s2)
⇒ mass of water filled in glass = 3/10 = 0.3 kg
Density of water = 1000kg/m3
Since Density = mass/ volume
⇒ volume of water 3N or empty glass = mass/density = 0.3/1000 = 3.0 × 10-4 m3 ...(I)
Weight of glass + partly filled sand = 4N
⇒ Weight of partly filled sand = 4 -1 = 3N
mass of sand in the glass = 3/10 = 0.3kg
Weight of glass + sand + water = 6N
⇒ Weight of water in glass = 6 - 4 = 2N
From (I) , Volume of water occupied by 2N water in glass = 2.0 × 10-4 m3 ...(II)
⇒ volume occupied by sand = (I) - (II) = 1.0 × 10-4 m3
⇒ density of sand = mass/volume = 0.3kg/ 1.0 × 10-4 = 3000 kg m-3
relative density of sand = density of sand/density of water = 3000/1000 = 3 ... (answer)
Alternate Method:
There is one more formula to calculate the relative density.
R.D. = Weight of sand in air / (Weight of sand in air - Weight of sand in water)
{NOTE: In this particular question we are taking formula based on sand, but in other problems, we take the mass/weight of the solvent and not the solute}
Weight of empty glass = 1N
Weight of glass filled with water = 4N
⇒ Weight of water filled in glass = 4 - 1 = 3N ...(I)
Weight of sand in glass = 4N
⇒ Weight of sand filled in glass = 4 - 1 = 3N
Weight of water and sand mixed in glass = 6N
⇒ Weight of water and sand filled in glass = 6 - 1 = 5N
Weight of sand in air = 3N [from (I)]
Weight of sand in water = Weight of water and sand in water - Weight of water
= 5N - 3N = 2N
R.D. = 3N / (3N - 2N)
= 3N / 1N
= 3
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