## 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

#### (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 – 4m3

(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 – 5m3

(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 – 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 = 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

#### (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