Class 9 Physics Chapter 7 numerical problems are according to the new syllabus of the Punjab Board. Chapter 7 is related to the properties of matter in which you will study the working principles of how matter is converted into different substances using multiple phenomenons.

## Class 9 Physics Chapter 7 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

Density = mass/volume

P = F / A

D = w_{1 }x ρ / (w_{1} – w_{2})

A = 2πR^{2}

F_{1} / a = F_{2} / A

Y = FL / A△L

### Numerical Problems Chapter 7 – Properties of Matter

**7.1 A wooden block measuring 40 cm x 10 cm x 5 cm has a mass 850 g. Find the density of 3 wood. **** (425 kgm **^{– 3})

^{– 3})

**Solution: **

Volume of wooden block = V = 40 cm x 10 cm x 5cm = 2000 cm^{3}

V = 2000 x 1/1000000 m^{3}

V = 0.002 m^{3}

Mass = m = 850 g = = 0.85 kg

Density of wood = ρ =?

Density = mass/volume

ρ = 0.85 / 0.02

ρ = 425 kgm^{– 3}

**7.2 How much would be the volume of ice formed by freezing 1 litre of water? ****(1.09 litre)**

**Solution: **

Volume of water = 1 litre

Volume of ice = ?

1 litre of water = 1 kg mass and density = 1000 kg ^{– 3}

Since the density of ice is 0.92 times of liquid water, therefore,

Volume of ice = mass/density

Volume = 1000 / 920

V = 1.09 litre

**7.3 Calculate the volume of the following objects:**

** (i) An iron sphere of mass 5 kg, the density of iron is 8200 kgm **^{– 3}. (6.1×10 ^{– 4 }m^{3})

^{– 3}. (6.1×10

^{– 4 }m

^{3})

** (ii) 200 g of lead shot having density 11300 kgm **^{– 3}. **(1.77×10 **^{– 5 }m^{3})

^{– 3}.

^{– 5 }m

^{3})

**(iii) A gold bar of mass 0.2 kg. The density of gold is 19300 kgm **^{– 3}. **(1.04×10 **^{– 5 }m^{3})

^{– 3}.

^{– 5 }m

^{3})

**Solution: **

Mass of iron sphere = m = 5 kg

Density of iron = ρ = 8200 kgm ^{– 3}

Volume of iron sphere = V =?

Volume = Mass / Density

Volume = 5 / 8200

Volume = 0.00060975 = 6.0975 x 10 ^{– 4}

Volume = 6.1 x 10 ^{– 4}m^{3}

Mass of lead shot = m = 200 g = kg = 0.2 kg

Density of lead = r = 11300 kgm ^{– 3}

Volume of lead shot = V =?

Volume = Mass / Density

Volume = 0.2 / 11300

Volume = 0.000017699 = 1.76699 x 10 ^{– 5}

Volume = 1.77 x 10 ^{– 5}m^{3}

Mass of gold bar = m = 0.2 kg

Density of gold = ρ = 19300 kgm ^{– 3}

Volume of gold bar = V =?

Volume = Mass / Density

Volume = 0.2 / 19300

Volume = 0.000010362 = 1.0362 x 10 ^{– 5}

Volume = 1.04 x 10 ^{– 5}m^{3}

**7.4 The density of air is 1.3 kgm **^{– 3}. Find the mass of air in a room measuring 8m x 5m x 4m. **(208 kg)**

^{– 3}. Find the mass of air in a room measuring 8m x 5m x 4m.

**Solution: **

Density of air = ρ = 1.3 kgm ^{– 3}

Volume of room = V = 8 m x 5 m x 4 m = 160 m^{3}

Mass of air = m =?

Mass of air = Density of air x volume of the room

Mass of air = 1.3 x 160

Mass of air = 208 kg

**7.5 A student presses her palm by her thumb with a force of 75 N. How much would be the pressure under her thumb having contact area 1.5 cm**^{2} ? ** (5×10**^{5} Nm ^{– 2})

^{2}?

^{5}Nm

^{– 2})

**Solution: **

Force = F = 75 N

Contact Area A = 1.5 cm^{2} = 1.5 x x m^{2} = 1.5 x 10 ^{– 4 }m^{2}

Pressure under the thumb = P =?

P = F / A

P = 75 / 1.5 x 10^{4}

P = 5 x 10^{5} Nm ^{– 2}

**7.6 The head of a pin is a square of side 10 mm. Find the pressure on it due to a force of 20 N. ****(2×10**^{5} Nm ^{– 2})

^{5}Nm

^{– 2})

**Solution: **

Force = F = 20 N

Area of head of a pin = A = 10mm x 10mm = 10/10 cm x 10/10 cm

A = 1/100 m x 1/100 m

A = 10 ^{– 4 }m^{2}

Pressure under the thumb = P =?

P *= F/A** *

*P * = 20 / 1x 10^{-4}

P = 2 x 10^{5} Nm ^{– 2}

**7.7 A uniform rectangular block of wood 20 cm x 7.5 cm x 7.5 cm and of mass 1000g stands on a horizontal surface with its longest edge vertical. Find**

** (i) The pressure exerted by the block on the surface**

** (ii) Density of the wood.** ** (1778 Nm **^{– 2 } , 889 kgm ^{– 3})

^{– 2 }, 889 kgm

^{– 3})

**Solution: **

Length of the smallest side of the block = 7.5 cm

Mass of the block = m = 1000g = 1kg

Pressure exerted by the block = P =?

Density of wood = ρ =?

Since the smallest edge of the block is rested on the horizontal surface. Therefore, the area of the block will be:

Area = A = 7.5 cm x 7.5 cm = 56.25 cm^{2}

A = 56.25 x 1/100 x 1/100 m^{2} = 56.25 x 10 ^{– 4 }m^{2}

Pressure under the thumb = P =?

P = F/A = mg/A

P* * = 1x 10 / 56.25 x 10^{-4}

P = 0.1778 x 10^{4}

P = 1778 Nm ^{– 2}

Volume = V = 20 cm x 7.5 cm x 7.5 cm = 1125 cm^{3}

V = 1125 x 1/100 m x 1/100 m x 1/100 m

V = 1125 x 10 ^{– 6 }m^{3}

V = 1.125 x 10 ^{– 3 }m^{3}

Density = Mass / Volume

ρ = 1/ 1.125 x 10^{-3} = 0.8888 x 10^{3} = 888.8 kgm ^{– 3}

ρ = 889 kgm ^{– 3}

**7.8 A cube of glass of 5 cm side and mass 306 g, has a cavity inside it. If the density of glass is 2.55 gcm **^{– 3}. Find the volume of the cavity. **(5 cm**^{3})

^{– 3}. Find the volume of the cavity.

^{3})

**Solution: **

Size of the cube = 7.5 cm

Mass of the cube = m = 306 g

Density of glass = ρ = 2.55 kgm ^{– 3}

Volume of the cavity = V =?

Volume of the whole cube = 5 cm x 5 cm x 5 cm = 125 cm^{3}

Volume of the glass = Mass / Density

Volume = 306 / 2.55

V = 120 cm^{3}

Volume of the cavity = 125 cm^{3} – 120 cm^{3}

V = 5 cm^{3}

**7.9 An object has weight 18 N in air. Its weight is found to be 11.4 N when immersed in water. Calculate its density. Can you guess the material of the object?**** (2727 kgm **^{– 3}, Aluminium)

^{– 3}, Aluminium)

**Solution: **

Weight of object in air = w_{1} = 18 N

Weight of object immersed in water = w_{2} = 11.4 N

Density of glass = ρ = 1000 kgm ^{– 3}

Density of the object = D =?

Nature of the material =?

D = w_{1 }x ρ / (w_{1} – w_{2})

D = 18 / 18-11.4 x 1000

D = 18/6.6 x 1000 = 2.727 x 10^{3} = 2727 kgm ^{– 3}

The density of aluminium is 2700 kgm ^{– 3}, and the calculated value of density is 2727 kgm^{– 3} nearest to the density of aluminium, so the material of the object is aluminium.

**7.10 A solid block of wood of density 0.6 gcm **^{– 3 }weighs 3.06 N in air. Determine (a) volume of the block (b) the volume of the block immersed when placed freely in a liquid of density 0.9 gcm ^{– 3 }?**(510 cm**^{3} , 340 cm^{3})

^{– 3 }weighs 3.06 N in air. Determine (a) volume of the block (b) the volume of the block immersed when placed freely in a liquid of density 0.9 gcm

^{– 3 }?

^{3}, 340 cm

^{3})

**Solution: **

Density of wood = D = 0.6 gcm ^{– 3}

Weight of the wooden block = w= 3.06 N

Since w = mg

m = w / g

m = 3.06 / 10

m = 0.306 kg = 306 g

Density of liquid = D = 0.9 gcm ^{– 3}

Volume of the block V =?

Volume of the block immersed in a liquid V =?

Density = Mass / Volume

Volume = Mass / Volume

V = 306 / 0.6 = 510 cm^{3}

Volume = Mass / Density

V = 306 / 0.9 = 340 cm^{3}

**7.11 The diameter of the piston of a hydraulic press is 30 cm. How much force is required to lift a car weighing 20 000 N on its piston if the diameter of the piston of the pump is 3 cm? (200 N)**

Solution: Diameter = D = 30 cm

Radius of the piston = R = D / 2 = 30 / 2 = 15 cm = 15 / 100 m = 0.15 m

Area of the piston = A = 2πR^{2} = 2 x 3.14 x (0.15)^{2}

A = 0.1413 m^{2}

Weight of the car = w = F_{2} = 20000 N

Diameter of the piston = d = 3 cm

Radius of the piston = R = D/2 = 3/2 = 1.5 cm = 1.5/100 m = 0.015 m

Area of the piston = A = 2πR^{2} = 2 x 3.14 x (0.015)^{2}

A = 1.413 x 10 ^{– 3 } m^{2}

Force = F_{1} = ?

F_{1} / a = F_{2} / A

F_{1 }= F_{2} x a / A

F_{1 }= 200000 N x 1.413 x 10 ^{– 3 }/ 0.1413

F_{1 }= 200000 N x 0.01

F_{1 }= 200 N

**7.12 A steel wire of cross-sectional area 2×10 **^{– 5 } m^{2} is stretched through 2 mm by a force of 4000 N. Find Young’s modulus of the wire. The length of the wire is 2 m. (2×10^{11} Nm ^{– 2})

^{– 5 }m

^{2}is stretched through 2 mm by a force of 4000 N. Find Young’s modulus of the wire. The length of the wire is 2 m. (2×10

^{11}Nm

^{– 2})

**Solution: **

Cross-sectional area = A = 2 x 10^{– 5} m^{2}

Extension = △L = 2 mm = 2 x 10^{-3} m = 0.002 m

Force = F = 4000 N

Length of the wire = L = 1m

Y = FL / A△L

Y = 4000×2 / 2×10^{– 3} x 0.002 = 8000 / 0.004 x 10 ^{– 5}

Y = 8000 / 0.004 x 10 ^{– 5}

Y = 2,000,000 x 10 ^{– 5} = 2 x 10^{11} Nm ^{– 2}

**Class 9 Physics Full Book Numerical Problems with Solution**