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 = w1 x 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
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 m3)
(ii) 200 g of lead shot having density 11300 kgm – 3. (1.77×10 – 5 m3)
(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 =?
P = F/A
P = 20 / 1x 10-4
P = 2 x 105 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)
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 = w1 x 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
F1 = F2 x a / A
F1 = 200000 N x 1.413 x 10 – 3 / 0.1413
F1 = 200000 N x 0.01
F1 = 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
