9th Class Physics · Punjab Board · Chapter 6

Chapter 6: Mechanical Properties of MatterNumericals Solved Step by Step

All 11 numericals of 9th Class Physics Chapter 6, Mechanical Properties of Matter, solved step by step: Given Data, To Find, Formula, Calculation and Result, the way you write them in the exam. Punjab Board new syllabus.

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Chapter 6· Mechanical Properties of Matter

Numericals Solved Step by Step

Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.

Numerical 6.1

A spring is stretched 20 mm by a load of 40 N. Calculate the spring constant. If an object causes an extension of 16 mm, what will be its weight?

Given Data

x₁ = 20 mm = 0.02 m
F₁ = 40 N
x₂ = 16 mm = 0.016 m

To Find

Spring constant k
Weight F₂ for 16 mm

Formulas

k = F ÷ x

F = kx

Calculation

k = F₁ ÷ x₁ = 40 ÷ 0.02 = 2000 N m⁻¹.
F₂ = k × x₂ = 2000 × 0.016 = 32 N.

Result

k = 2000 N m⁻¹; weight = 32 N

Numerical 6.2

The mass of 5 litres of milk is 4.5 kg. Find its density in SI units.

Given Data

m = 4.5 kg
V = 5 litres = 5 × 10⁻³ m³

To Find

Density ρ

Formula

ρ = m ÷ V

Calculation

1 litre = 10⁻³ m³, so V = 5 × 10⁻³ m³.
ρ = 4.5 ÷ (5 × 10⁻³) = 900 kg m⁻³.

Result

ρ = 900 kg m⁻³

Note: 1000 litres = 1 m³.

Numerical 6.3

When a solid of mass 60 g is lowered into a measuring cylinder, the level of water rises from 40 cm³ to 44 cm³. Calculate the density of the solid.

Given Data

m = 60 g = 0.06 kg
V₁ = 40 cm³
V₂ = 44 cm³

To Find

Density ρ

Formulas

V = V₂ − V₁

ρ = m ÷ V

Calculation

V = 44 − 40 = 4 cm³ = 4 × 10⁻⁶ m³.
ρ = 0.06 ÷ (4 × 10⁻⁶) = 15 000 kg m⁻³.

Result

ρ = 1.5 × 10⁴ kg m⁻³ (15 g cm⁻³)

Numerical 6.4

A block of density 8 × 10³ kg m⁻³ has a volume of 60 cm³. Find its mass.

Given Data

ρ = 8 × 10³ kg m⁻³
V = 60 cm³ = 60 × 10⁻⁶ m³

To Find

Mass m

Formula

m = ρV

Calculation

m = 8 × 10³ × 60 × 10⁻⁶ = 0.48 kg.

Result

m = 0.48 kg (480 g)

Numerical 6.5

A brick measures 5 cm × 10 cm × 20 cm. If its mass is 5 kg, calculate the maximum and minimum pressure the brick can exert on a horizontal surface. (g = 10 m s⁻²)

Smallest face 5 cm × 10 cm, largest face 10 cm × 20 cm.

Given Data

m = 5 kg, so F = mg = 50 N
smallest face = 5 × 10 = 50 cm² = 50 × 10⁻⁴ m²
largest face = 10 × 20 = 200 cm² = 200 × 10⁻⁴ m²

To Find

Maximum pressure
Minimum pressure

Formula

P = F ÷ A

Calculation

Maximum pressure, on the smallest face: P = 50 ÷ (50 × 10⁻⁴) = 1.0 × 10⁴ Pa.
Minimum pressure, on the largest face: P = 50 ÷ (200 × 10⁻⁴) = 2.5 × 10³ Pa.

Result

Maximum 1.0 × 10⁴ Pa; minimum 2.5 × 10³ Pa

Numerical 6.6

What will be the height of the column in a barometer at sea level if mercury is replaced by water of density 1000 kg m⁻³? (Density of mercury = 13.6 × 10³ kg m⁻³; the mercury column is 0.76 m.)

Given Data

ρ(water) = 1000 kg m⁻³
ρ(mercury) = 13.6 × 10³ kg m⁻³
h(mercury) = 0.76 m

To Find

Height of the water column

Formula

ρ₁gh₁ = ρ₂gh₂

Calculation

The same atmospheric pressure holds up both columns: 1000 × g × h = 13 600 × g × 0.76.
h = 13 600 × 0.76 ÷ 1000 = 10.34 m.

Result

h ≈ 10.3 m

Numerical 6.7

In the hydraulic brake system of a car, the force exerted normally on its piston of cross-sectional area 5 cm² is 500 N. What will be the pressure transferred to the brake oil? What will be the force on the second piston of cross-sectional area 20 cm²?

The same pressure acts on both pistons (Pascal's law).

Given Data

A₁ = 5 cm² = 5 × 10⁻⁴ m²
F₁ = 500 N
A₂ = 20 cm² = 20 × 10⁻⁴ m²

To Find

Pressure P in the oil
Force F₂ on the second piston

Formulas

P = F₁ ÷ A₁

F₂ = P × A₂

Calculation

P = 500 ÷ (5 × 10⁻⁴) = 1.0 × 10⁶ Pa.
F₂ = 1.0 × 10⁶ × 20 × 10⁻⁴ = 2000 N.

Result

P = 1.0 × 10⁶ Pa; F₂ = 2000 N

Numerical 6.8

Find the water pressure on a deep-sea diver at a depth of 10 m, where the density of sea water is 1030 kg m⁻³. (g = 10 m s⁻²)

Given Data

h = 10 m
ρ = 1030 kg m⁻³
g = 10 m s⁻²

To Find

Water pressure P

Formula

P = ρgh

Calculation

P = 1030 × 10 × 10 = 1.03 × 10⁵ Pa.

Result

P = 1.03 × 10⁵ Pa

Note: That is about one more atmosphere; the total pressure on the diver is about twice atmospheric pressure.

Numerical 6.9

The cross-sectional areas of the small and large pistons of a hydraulic press are 10 cm² and 100 cm². What force must be exerted on the small piston to lift a car of weight 4000 N?

F₁ on the small piston, the 4000 N car on the large one.

Given Data

A₁ = 10 cm²
A₂ = 100 cm²
F₂ = 4000 N

To Find

Force F₁ on the small piston

Formula

F₁ ÷ A₁ = F₂ ÷ A₂

Calculation

F₁ = F₂ × A₁ ÷ A₂ = 4000 × 10 ÷ 100.
F₁ = 400 N.

Result

F₁ = 400 N

Note: Only the ratio of the areas matters, so cm² need not be changed to m² here.

Numerical 6.10

In a hot-air balloon the following readings were taken: altitude 150, 500, 800, 1140, 1300, 1500 m; pressure 99.5, 95.7, 92.4, 88.9, 87.2, 85.3 kPa. Draw a graph of pressure against altitude and find (a) the air pressure at sea level (b) the height at which the air pressure is 90 kPa.

The points lie close to a straight line; extend it back to 0 m, and read across from 90 kPa.

Given Data

altitude (m): 150, 500, 800, 1140, 1300, 1500
pressure (kPa): 99.5, 95.7, 92.4, 88.9, 87.2, 85.3

To Find

(a) Pressure at sea level (0 m)
(b) Altitude where the pressure is 90 kPa

Formula

read from the straight-line graph

Calculation

Plot the points and draw the best straight line through them; the pressure falls about 1.06 kPa for every 100 m.
(a) Extend the line back to altitude 0: it meets the pressure axis at about 101 kPa.
(b) Read across from 90 kPa to the line and down: about 1030–1040 m (between the 800 m and 1140 m readings).

Result

(a) about 101 kPa (b) about 1030–1040 m

Note: One solved copy answers (b) with "88.9 kPa at 1140 m", which is just a row of the table. The question asks for 90 kPa.

Numerical 6.11

If the pressure in a hydraulic press is increased by an additional 10 N cm⁻², how much extra load will the output platform support if its cross-sectional area is 50 cm²?