# Class 9 Physics Chapter 7 Numerical Problems 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 =  wx ρ / (w1 – w2

A = 2πR2

F1 / a = F2 / 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)

Solution:

Volume of wooden block = V = 40 cm x 10 cm x 5cm = 2000 cm3

V = 2000 x 1/1000000  m3

V = 0.002 m3

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

#### (iii) A gold bar of mass 0.2 kg. The density of gold is 19300 kgm – 3.(1.04×10 – 5 m3)

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 – 4m3

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 – 5m3

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 – 5m3

#### 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)

Solution:

Density of air = ρ = 1.3 kgm – 3

Volume of room = V = 8 m x 5 m x 4 m = 160 m3

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 cm2 ?  (5×105 Nm – 2)

Solution:

Force = F = 75 N

Contact Area A = 1.5 cm2 = 1.5 x  x   m2 = 1.5 x 10 – 4 m2

Pressure under the thumb = P =?

P =  F / A

P  = 75 / 1.5 x 104

P = 5 x 105 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.  (2×105 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 m2

Pressure under the thumb = P =?

= F/A

= 20 / 1x 10-4

P =  2 x 105 Nm – 2

#### (ii) Density of the wood. (1778 Nm – 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 cm2

A = 56.25 x 1/100 x 1/100 m2 = 56.25 x 10 – 4 m2

Pressure under the thumb = P =?

P = F/A = mg/A

P  = 1x 10 / 56.25 x 10-4

P = 0.1778 x 104

P = 1778 Nm – 2

Volume = V = 20 cm x 7.5 cm x 7.5 cm = 1125 cm3

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

V = 1125 x 10 – 6 m3

V = 1.125 x 10 – 3 m3

Density =  Mass / Volume

ρ = 1/ 1.125 x 10-3 = 0.8888 x 103 = 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 cm3)

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 cm3

Volume of the glass = Mass / Density

Volume = 306 / 2.55

V = 120 cm3

Volume of the cavity = 125 cm3 – 120 cm3

V = 5 cm3

#### 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)

Solution:

Weight of object in air = w1 = 18 N

Weight of object immersed in water = w2 = 11.4 N

Density of glass = ρ = 1000 kgm – 3

Density of the object = D =?

Nature of the material =?

D =  wx ρ / (w1 – w2

D =   18 / 18-11.4 x 1000

D = 18/6.6 x 1000 = 2.727 x 103 = 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 cm3 , 340 cm3)

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 cm3

Volume = Mass / Density

V = 306 / 0.9  = 340 cm3

#### 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 = 2πR2 = 2 x 3.14 x (0.15)2

A = 0.1413 m2

Weight of the car = w = F2 = 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 = 2πR2 = 2 x 3.14 x (0.015)2

A = 1.413 x 10 – 3  m2

Force = F1 = ?

F1 / a = F2 / A

F= F2 x a / A

F= 200000 N x 1.413 x 10 – 3 / 0.1413

F= 200000 N x 0.01

F= 200 N

#### 7.12 A steel wire of cross-sectional area 2×10 – 5  m2 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.  (2×1011 Nm – 2)

Solution:

Cross-sectional area = A = 2 x 10– 5 m2

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 1011 Nm – 2

Class 9 Physics Full Book Numerical Problems with Solution

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