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Floatation

7 Simple Numerical You Must Know

**Q1: A force of 200 N acts on a surface of area 10 cm**

^{2}. Calculate thrust and pressure. How much the pressure changes if the area of contact increases to 50 cm^{2}.Answer:

Force = 200 N

Area = 10cm

^{2}= 10 x 10

^{-4}m

^{2}

Pressure (P) = Force ÷ Area

P1 = 200 ÷ 10 x 10

^{-4}= 2 x 10

^{5}Pa

New area = 50cm

^{2}= 50 x 10

^{-4}m

^{2}

P2 = 200 ÷ 50 x 10

^{-4}= 4 x 10

^{4}Pa

Change in Pressure = P1 - P2 = 2 x 10

^{5 }- 4 x 10

^{4}

= 1.6 x 10

^{5}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/m**

^{3}.Answer: Volume of room (V) = 50m x 15m x 3.5m = 2625m

^{3}

Density of air (d) = 1.30 kg/m

^{3}

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 10**

^{3}kg/m^{3}, find density of silver.Answer:

Relative Density (RD) of silver = 10.8

Density of Water = 10

^{3}kg/m

^{3}

RD of Silver = Density of Silver ÷ Density of Water

⇒ Density of Silver = RD of Silver x Density of Water

= 10.8 x 10

^{3}

= 1.08 x 10

^{4}kg/m

^{3}

**Q4: A ball of mass 4kg and density 4000 kgm**

^{3}is completely immersed in water. Find the buoyant force acting on ball if the density of water is 10^{3}kg/m^{3}. (Take g = 10ms^{-2})Answer:

Volume of water displaced = Volume of ball = mass ÷ density

= 4 ÷ 4000 = 10

^{-3}m

^{3}.

Buoyant force = Weight of water displaced

= Mass of water displaced x g

= Volume of water displaced x density of water x g

= 10

^{-3}m

^{3}x 10

^{3}kg/m

^{3}x 10

= 10 N

**Q5(NCERT): The volume of 50g of a substance is 20 cm**

^{3}. If the density of water is 1g/cm^{3}, will the substance float or sink?Answer:

Given, mass (m) = 50g

volume (V) = 20 cm

^{3}

Density ofsubstance = m / V = 50/20 = 2.5 g/cm

^{3}

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/m**

^{3}weighs 0.6 kg in air. It is completely immersed in water of density 1000 kg/m^{3}. Calculate the apparent weight of the block in water and the buoyant force. (Take g = 10ms^{-2})**Answer: Density (d) = 6000 kg m**

^{3}

mass of block (m) = 0.6 kg

Volume (V) = m/d = 0.6/ 6000 = 10

^{-4}m

^{3}

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

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