Friday, November 25, 2016

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

Floatation
7 Simple Numerical You Must Know

CBSE Class 9 - 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


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