CBSE Class 9 - Science - Floatation 7 Simple Numerical You Must Know

Floatation
7 Simple Numerical You Must Know

Q1: A force of 200 N acts on a surface of area 10 cm². Calculate thrust and pressure. How much the pressure changes if the area of contact increases to 50 cm².

Answer:
Force = 200 N
Area = 10cm² = 10 x 10^-4m²
Pressure (P) = Force ÷ Area
P1 = 200 ÷ 10 x 10^-4 = 2 x 10⁵ Pa

New area = 50cm² = 50 x 10^-4m²
P2 = 200 ÷ 50 x 10^-4 = 4 x 10⁴ Pa

Change in Pressure = P1 - P2 = 2 x 10⁵   - 4 x 10⁴
                                = 1.6 x 10⁵ Pa



Q2: A cuboid room has dimensions 50m x 15m x 3.5m. What is the mass of the air enclosed in the room if the density of air = 1.30 kg/m³.


Answer: Volume of room (V) = 50m x 15m x 3.5m = 2625m³
Density of air (d) = 1.30 kg/m³
Mass of air = V x d = 2625 x 1.30 = 3412.5 kg





Q3: Relative density of silver is 10.8. If the density of water is 10³kg/m³, find density of silver.

Answer:
Relative Density (RD) of silver = 10.8
Density of Water = 10³kg/m³

RD of Silver = Density of Silver ÷ Density of Water
⇒ Density of Silver = RD of Silver x Density of Water
= 10.8 x 10³
= 1.08 x 10⁴kg/m³



Q4: A ball of mass 4kg and density 4000 kgm³ is completely immersed in water. Find the buoyant force acting on ball if the density of water is 10³kg/m³. (Take g = 10ms^-2)

Answer:
Volume of water displaced = Volume of ball = mass ÷ density
= 4 ÷ 4000 = 10 ^-3m³.

Buoyant force = Weight of water displaced
= Mass of water displaced x g
= Volume of water displaced x density of water x g
= 10 ^-3m³ x 10³kg/m³ x 10
= 10 N



Q5(NCERT): The volume of 50g of a substance is 20 cm³. If the density of water is 1g/cm³, will the substance float or sink?

Answer:

Given, mass (m) = 50g
volume (V) = 20 cm³
Density ofsubstance = m / V = 50/20 = 2.5 g/cm³
Since density of substance is greater than that of water, the substance will sink.



Q6: A solid weighs 80g in air, 64g in water. Calculate the relative density of solid.

Answer:
Weight in air = 80g
Weight in water = 64g

                         
                             Weight in air                   Weight in air
 Relative Density = --------------------------   = ---------------------------------
       Loss of Weight in water        Weight in air - Weight in water

= 80 / (80 -64)
= 80/16
= 5


Q7: A solid block of density 6000 kg/m³ weighs 0.6 kg in air. It is completely immersed in water of density 1000 kg/m³. Calculate the apparent weight of the block in water and the buoyant force. (Take g = 10ms^-2)

Answer: Density (d) = 6000 kg m³
mass of block (m) = 0.6 kg

Volume (V) = m/d = 0.6/ 6000 = 10^-4m³

Buoyant force (Fb) = weight of water displace = V x d x g
= 10^-4 x 1000 x 10
= 1 N
Apparent Weight = Weight in Air - Fb
= 0.6 x 10 - 1N
= 6 - 1 = 5N


See also

✎See also: Floatation and Archimedes Principle (Q & A)