Numerical Problems

 

 

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

                                                                       = 2000 x  x  x  m³

= 0.002 m³

                                         Mass = m = 850 g =  = 0.85 kg

Density of wood = r = ?

Density =  mass / volume = 0.85 / 0.02 = 425 kg ^{– 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 density of ice is 0.92 times of the liquid water therefore,

Volume of ice   =  mass / density

=  1000 / 920

Volume of ice = 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 = r = 8200 kgm ^{– 3}

Volume of iron sphere = V =?

Volume = Mass / Density

Volume = 5 / 8200

=  0.00060975 = 6.0975 x 10 ^{– 4}

Volume = 6.1 x 10 ^{– 4}m³

 

(ii)                 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

= 0.000017699 = 1.76699 x 10 ^{– 5}

Volume = 1.77 x 10 ^{– 5}m³

 

(iii)                Mass of gold bar = m = 0.2 kg

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

Volume of gold bar = V =?

Volume = Mass / Density

Volume =  0.2 / 19300

= 0.000010362 = 1.0362 x 10 ^{– 5}

Volume = 1.04 x 10 ^{– 5}m³

 

 

 

 

 

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 = r = 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 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 = 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 =

= 1/100 m x 1/100 m

= 10 ^{– 4} m²

Pressure under the thumb = P = ?

P = F/A 

P  = 20 / 1x 10^-4  =  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 r = ?

 

Calculations: (i) since the smallest edge of the block is rested on the horizontal surface. Therefore, area of the block will be:

                   Area = A = 7.5 cm x 7.5 cm = 56.25 cm²

= 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   = 0.1778 x 10⁴ = 1778 Nm ^{– 2}

(ii)                 Volume = V = 20 cm x 7.5 cm x 7.5 cm = 1125 cm³

= 1125 x 1/100 m x 1/100 m x 1/100 m = 1125 x 10 ^{– 6} m³

Or     V = 1.125 x 10 ^{– 3} m³

Density =  Mass / Volume

Density  = 1/ 1.125 x 10^-3 = 0.8888 x 10³ = 888.8 kgm ^{– 3}

                       Density = 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 = r = 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  = 120 cm³

Volume of the cavity = 125 cm³ – 120 cm³ = 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 = r = 1000 kgm ^{– 3}

• Density of the object = D = ?
• Nature of the material = ?

                                  D =  w₁ / w₁ – w₂  x p

D =   18 / 18-11.4 x 1000

= 18/6.6 x 1000 = 2.727 x 10³ = 2727 kgm ^{– 3}

• The density of aluminium is 2700 kgm ^{– 3} , the above 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       or          m = w / g = 3.06 / 10 = 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

 

(b)                     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

= 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 the 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  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}