StudentAid
HomeLearn
9th Class10th Class11th Class12th ClassPast papers & testsBrowse everything
For schoolsFor driversFor parentsBook a demo
Learn/ 9th Physics/ Unit 6 /Numericals

Unit 6: Mechanical Properties of Matter — Numericals

9th Class Physics · Unit 6: Mechanical Properties of Matter

6.1.A spring is stretched 20 mm by a load of 40 N. Calculate the value of spring constant. If an object causes an extension of 16 mm, what will be its weight?
Given
Extension x = 20 mm = 0.020 m
Load F = 40 N
Result
Spring constant k = 2 kN m-1
Weight at 16 mm extension W = 32 N
6.2.The mass of 5 litres of milk is 4.5 kg. Find its density in SI units.
Given
Mass m = 4.5 kg
Volume V = 5 litres = 5 × 10-3 m3
Density rho = ?
Formula
Density D = mV = frac{4.5}{5 × 10-3
Working
D = 900 kg m-3
Result
Density D = 0.9 × 103 kg m-3
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
Initial water level V1 = 40 cm3
Final water level V2 = 44 cm3
Mass m = 60 g = 0.06 kg
Density of solid D = ?
Formula
Displaced volume V = V2 - V1
Density D = mV
Working
Displaced volume V = 44 cm3 - 40 cm3 = 4 cm3 = 4 × 10-6 m3
D = frac{0.06}{4 × 10-6 = 15000 kg m-3
Result
Density D = 15 × 103 kg m-3
6.4.A block of density 8 x 10^3 kg m^-3 has a volume 60 cm3. Find its mass.
Given
Density D = 8 × 103 kg m-3
Volume V = 60 cm3 = 60 × 10-6 m3
Mass m = ?
Formula
Mass m = V × D
Working
m = (60 × 10-6)(8 × 103)
Result
Mass m = 0.48 kg
6.5.A brick measures 5 cm x 10 cm x 20 cm. If its mass is 5 kg, calculate the maximum and minimum pressure which the brick can exert on a horizontal surface.
Given
Dimensions 5 cm × 10 cm × 20 cm
Mass of brick m = 5 kg
Maximum and minimum pressure P_{max = ? and P_{min = ?
Formula
Weight F = W = mg
Largest face (for minimum pressure) Area_{max = 20 cm × 10 cm = 200 cm2 = 200 × 10-4 m2
Smallest face (for maximum pressure) Area_{min = 5 cm × 10 cm = 50 cm2 = 50 × 10-4 m2
Maximum pressure P_{max = frac{F}{Area_{min
Minimum pressure P_{min = frac{F}{Area_{max
Working
Force F = (5)(10) therefore g = 10 m s-2, F = 50 N
Maximum pressure P_{max = frac{50}{50 × 10-4 = 10,000 Pa
Result
Maximum pressure P_{max = 1 × 104 Pa
Working
Minimum pressure P_{min = frac{50}{200 × 10-4 = 2500 Pa
Result
Minimum pressure P_{min = 25 × 102 Pa
6.6.What will be the height of the column in barometer at sea level if mercury is replaced by water of density 1000 kg m^-3, where density of mercury is 13.6 x 10^3 kg m^-3?
Given
Density of water rhow = 1000 kg m-3
Density of mercury rhom = 13.6 × 103 kg m-3
Result
Height of water column h = 10.3 m
6.7.Suppose in the hydraulic brake system of a car, the force exerted normally on its piston of cross-sectional area of 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 area of cross-section 20 cm2?
Given
Smaller piston area a = 5 cm2 = 5 × 10-4 m2
Larger piston area A = 20 cm2 = 20 × 10-4 m2
Force on smaller piston F1 = 500 N
Pressure and force on larger piston P1 = ? and F2 = ?
Formula
Pressure P1 = F1a
Force on larger piston F2 = Aa × F1
Working
Pressure P1 = frac{500}{5 × 10-4 = 1 × 106 Pa
Force on larger piston F2 = frac{20 × 10-4{5 × 10-4 × 500
Result
Force on larger piston F2 = 2000 N
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.
Given
Depth h = 10 m
Density of sea water rho = 1030 kg m-3
Gravitational acceleration g = 10 m s-2
Water pressure P = ?
Formula
Pressure P = rho g h
Working
P = (1030)(10)(10)
P = 10300 Pa
Result
Pressure P = 1.03 × 105 Nm-2
6.9.The area of cross-section of the small and large pistons of a hydraulic press is respectively 10 cm2 and 100 cm2. What force should be exerted on the small piston in order to lift a car of weight 4000 N?
Given
Area of small piston a = 10 cm2 = 10 × 10-4 m2
Area of large piston A = 100 cm2 = 100 × 10-4 m2
Force on large piston (car weight) F2 = 4000 N
Force on small piston F1 = ?
Formula
Force relationship F1 = aA × F2
Working
F1 = frac{10 × 10-4{100 × 10-4 × 4000
F1 = (0.1)(4000)
Result
Force on small piston F1 = 400 N
6.10.In a hot air balloon, the following data was recorded. Draw a graph between the altitude and pressure and find out: (a) What would the air pressure have been at sea level? (b) At what height the air pressure would have been 90 kPa?
Given
Data points (altitude in m, pressure in kPa) 150:99.5, 500:95.7, 800:92.4, 1140:88.9, 1300:87.2, 1500:85.3
Sea level pressure and height at 90 kPa P(h=0) = ? and h(P=90) = ?
(a) Rate of pressure change ΔPΔh = 95.7 - 99.5500 - 150 = -3.8350 = -0.01086 kPa/m
(a) Linear extrapolation to sea level P = 99.5 + (-0.01086)(0 - 150)
P = 99.5 + 1.629
Result
(a) Sea level pressure P = 101.1 kPa
(b) Rate using 800m and 1140m ΔPΔh = 88.9 - 92.41140 - 800 = -3.5340 = -0.01029 kPa/m
(b) Solving for height at 90 kPa 90 = 92.4 + (-0.01029)(h - 800)
-2.40.01029 = h - 800
233.3 = h - 800
(b) Height at 90 kPa h = 1033.3 m ≈ 1.02 km
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?
Given
Pressure increase ΔP = 10 N cm-2
Area of output platform A = 50 cm2 = 50 × 10-4 m2
Extra load/force F = ?
Formula
Force F = ΔP × A
Working
F = (10)(50)
Result
Extra load F = 500 N
6.12.The force exerted normally on the hydraulic brake system of a car, with its piston of cross sectional area 5 cm2 is 500 N. What will be the: (a) pressure transferred to the brake oil? (b) force on the brake piston of area of cross section 20 cm2?
Given
Force F = 500 N
Piston area A = 5 cm2 = 5 × 10-4 m2
Result
(a) Pressure = 1.0 × 106 N m-2
(b) Force on brake piston = 2000 N