MCQS Questions
6.1 A wire is stretched by a weight \( w \). If the diameter of the wire is reduced to half of its previous value, the extension will become:
(a) one half
(b) double
(c) one fourth
(d) four times
Correct Option: (d) four times
Reason: Extension is inversely proportional to the square of the diameter, so halving the diameter makes the extension \( 2^2 = 4 \) times larger.
6.2 Four wires of the same material are stretched by the same load. Their dimensions are given below. Which of them will elongate most?
(a) Length \( 1 \, \text{m} \), diameter \( 1 \, \text{mm} \)
(b) Length \( 2 \, \text{m} \), diameter \( 2 \, \text{mm} \)
(c) Length \( 3 \, \text{m} \), diameter \( 3 \, \text{mm} \)
(d) Length \( 4 \, \text{m} \), diameter \( 0.5 \, \text{mm} \)
Correct Option: (d) Length \( 4 \, \text{m} \), diameter \( 0.5 \, \text{mm} \)
Reason: Extension is proportional to length and inversely proportional to the square of the diameter. The wire with the greatest length and smallest diameter elongates the most.
6.3 Two metal plates of area \( 2 \, \text{m}^2 \) and \( 3 \, \text{m}^2 \) are placed in a liquid at the same depth. The ratio of pressures on the two plates is:
(a) 1:1
(b) \( \sqrt{2} : \sqrt{3} \)
(c) 2:3
(d) 4:9
Correct Option: (a) 1:1
Reason: Pressure in a liquid depends only on the depth and not on the area of the plate.
6.4 The pressure at any point in a liquid is proportional to:
(a) density of the liquid
(b) depth of the point below the surface of the liquid
(c) acceleration due to gravity
(d) all of the above
Correct Option: (d) all of the above
Reason: Pressure in a liquid is given by \( P = \rho g h \), which depends on the density (\( \rho \)), gravity (\( g \)), and depth (\( h \)).
6.5 Pressure applied to an enclosed fluid is:
(a) increased and applied to every part of the fluid
(b) diminished and transmitted to the walls of the container
(c) increased in proportional to the mass of the fluid and then transmitted to each part of the fluid
(d) transmitted unchanged to every portion of the fluid and walls of the containing vessel
Correct Option: (d) transmitted unchanged to every portion of the fluid and walls of the containing vessel
Reason: This is Pascal’s law, which states that pressure in an enclosed fluid is transmitted equally in all directions.
6.6 The principle of a hydraulic press is based on:
(a) Hooke’s law
(b) Pascal’s law
(c) Principle of conservation of energy
(d) Principle of conservation of momentum
Correct Option: (b) Pascal’s law
Reason: A hydraulic press uses Pascal’s law, where pressure applied is transmitted uniformly throughout the fluid.
6.7 When a spring is compressed, what form of energy does it possess?
(a) Kinetic
(b) Potential
(c) Internal
(d) Heat
Correct Option: (b) Potential
Reason: Compressed springs store energy in the form of elastic potential energy.
6.8 What is the force exerted by the atmosphere on a rectangular block surface of length \( 50 \, \text{cm} \) and breadth \( 40 \, \text{cm} \)? The atmospheric pressure is \( 100 \, \text{kPa} \).
(a) \( 20 \, \text{kN} \)
(b) \( 100 \, \text{kN} \)
(c) \( 200 \, \text{kN} \)
(d) \( 500 \, \text{kN} \)
Correct Option: (a) \( 20 \, \text{kN} \)
Reason: Force is given by \( F = P \cdot A \). The area \( A = 0.5 \cdot 0.4 = 0.2 \, \text{m}^2 \), so
\[F = 100,000 \cdot 0.2 = 20,000 \, \text{N} = 20 \, \text{kN}\]
Short Questions
6.1 Why heavy animals like an elephant have a large area of the foot?
Elephants have a large foot area to reduce the pressure on the ground. Since pressure is inversely proportional to area, a larger foot area distributes their weight evenly, reducing pressure.
6.2 Why animals like deer who run fast have a small area of the foot?
Deer have small foot areas to increase pressure on the ground. This helps them get better grip and speed while running.
6.3 Why is it painful to walk barefoot on pebbles?
Walking on pebbles is painful because the small surface area of the pebbles increases the pressure exerted on your feet.
6.4 State Pascal’s law. Give an application of Pascal’s law.
Pascal’s law states that pressure applied to an enclosed fluid is transmitted equally in all directions throughout the fluid.
Application: Hydraulic systems like hydraulic brakes in automobiles.
6.5 State what you mean by elasticity of a solid.
Elasticity is the property of a solid to return to its original shape and size after the removal of the deforming force.
6.6 What is Hooke’s law? Does an object remain elastic beyond elastic limit? Give reason.
Hooke’s law states that the force required to stretch a material is proportional to its extension, as long as the elastic limit is not exceeded.
Beyond the elastic limit, an object loses its elasticity and undergoes permanent deformation.
6.7 Distinguish between force and pressure.
| Force | Pressure |
| A push or pull acting on a body. | Force acting per unit area. |
| Measured in newtons (N). | Measured in pascals (Pa). |
| Formula: \(F=ma\) | Formula: \(P=\frac{F}{A}\) |
6.8 What is the relationship between liquid pressure and the depth of the liquid?
Liquid pressure increases with depth. The formula is \( P = \rho g h \), where \( h \) is the depth of the liquid.
6.9 What is the basic principle to measure the atmospheric pressure by a simple mercury barometer?
The atmospheric pressure is balanced by the height of the mercury column in a barometer. The height indicates the pressure.
6.10 State the basic principle used in the hydraulic brake system of the automobiles.
The hydraulic brake system works on Pascal’s law. Pressure applied at one point in the brake fluid is transmitted equally to all parts, amplifying the force.
Constructed Response Questions
6.1 A spring having spring constant \( k \) hangs vertically from a fixed point. A load of weight \( L \), when hung from the spring, causes an extension \( x \), assuming the elastic limit is not exceeded. Complete the table for the following arrangements.
Table Completion
| Arrangement | Total Extension in terms of \(x\) | Spring Constant kkk of the Arrangement |
| Single spring | \(x\) | \(k\) |
| Two springs in series | \(2x\) | \(\frac{k}{2}\) |
| Two springs in parallel | \(\frac{x}{2}\) | \(2k\) |
6.2 Springs are made of steel instead of iron. Why?
Steel is more elastic than iron, meaning it can return to its original shape more efficiently after deformation.
6.3 Which of the following materials is more elastic?
(a) Iron or rubber
(b) Air or water
(a) Iron is more elastic than rubber because it resists deformation more and returns to its original shape.
(b) Water is more elastic than air because it has a greater bulk modulus, making it harder to compress.
6.4 How does water pressure one metre below the surface of a swimming pool compare to water pressure one metre below the surface of a very large and deep lake?
The water pressure is the same in both cases because water pressure depends only on depth (\( h \)), not on the size or volume of the water body. It is calculated as \( P = \rho g h \).
6.5 What will happen to the pressure in all parts of a confined liquid if pressure is increased in one part? Give an example from your daily life where such principle is applied.
When pressure is applied to one part of a confined liquid, it is transmitted equally in all directions (Pascal’s law).
Example: Hydraulic brakes in vehicles, where pressure is applied at one point and transmitted to the brake pads.
6.6 If some air remains trapped within the top of the mercury column of the barometer which is supposed to be vacuum, how would it affect the height of the mercury column?
The trapped air would exert additional pressure, reducing the height of the mercury column, as the pressure inside the barometer is no longer solely due to the atmospheric pressure.
6.7 How does the long neck is not a problem to a giraffe while raising its neck suddenly?
Giraffes have specialized valves in their neck veins that regulate blood flow and prevent blood from rushing to or away from the brain when the neck is raised or lowered suddenly.
6.8 The end of glass tube used in a simple barometer is not properly sealed, some leak is present. What will be its effect?
If the glass tube is not properly sealed, air will enter and disturb the vacuum, causing the mercury column to drop and making the barometer readings inaccurate.
6.9 Comment on the statement, “Density is a property of a material not the property of an object made of that material.”
This statement is correct. Density is a characteristic property of a material (e.g., iron or water) and does not depend on the size or shape of the object made from it.
6.10 How the load of a large structure is estimated by an engineer?
Engineers estimate the load of a structure by calculating the weight of its components, materials used, and the external forces (e.g., wind, earthquakes) it must withstand. They use formulas and simulations to determine structural integrity.
Comprehensive Questions
6.1 What is Hook’s law? Give three applications of this law.
Hooke’s Law: The extension of a spring is directly proportional to the force applied to it, as long as the elastic limit is not exceeded.
\[F = kx\]
where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the extension.
Applications:
Designing springs in mechanical systems like suspension systems in vehicles.
Measuring forces using a spring balance.
Manufacturing mattresses and trampolines with elastic properties.
6.2 Describe the working and applications of a simple mercury barometer and a manometer.
Mercury Barometer:
Working: Measures atmospheric pressure. A glass tube filled with mercury is inverted into a mercury reservoir. The height of the mercury column represents atmospheric pressure.
Applications: Used in weather forecasting and determining altitude.
Manometer:
Working: Measures the pressure of gases or liquids. The pressure difference is indicated by the liquid height in the U-shaped tube.
Applications: Used in laboratories, HVAC systems, and pressure testing in pipelines.
6.3 Describe Pascal’s Law. State its applications with examples.
Pascal’s Law: Pressure applied to a confined fluid is transmitted equally in all directions.
\[P = \frac{F}{A}\]
where \( P \) is pressure, \( F \) is force, and \( A \) is area.
Applications:
Hydraulic Brakes: Pressure is applied through brake fluid to stop vehicles.
Hydraulic Lifts: Used to lift heavy loads in industries and garages.
Syringes: Force applied on the plunger transmits pressure to expel liquid.
6.4 On what factors does the pressure of a liquid in a container depend? How is it determined?
Factors:
Depth (\( h \)): Pressure increases with depth.
Density (\( \rho \)) of the Liquid: Greater density results in higher pressure.
Acceleration due to Gravity (\( g \)): Pressure increases with gravity.
Formula:
\[P = \rho g h\]
6.5 Explain that atmosphere exerts pressure. What are its applications? Give at least three examples.
The atmosphere exerts pressure due to the weight of air. This pressure is measured using a barometer.
Applications:
Suction Pumps: Function based on atmospheric pressure.
Drinking with a Straw: Reduced pressure in the straw causes liquid to rise.
Vacuum Packaging: Removing air extends the shelf life of food.
Numerical Problems
6.1 A spring is stretched 20 mm by a load of 40 N. Calculate the value of the spring constant. If an object causes an extension of 16 mm, what will be its weight?
Given Data:
Extension (\( x = 20 \, \text{mm} = 0.02 \, \text{m} \))
Force (\( F = 40 \, \text{N} \))
To Find:
1. Spring constant (\( k \))
2. Weight for \( x = 16 \, \text{mm} = 0.016 \, \text{m} \)
Formula:
\[k = \frac{F}{x}\]
\[F = k \cdot x\]
Solution:
1. Calculate spring constant:
\[k = \frac{40}{0.02} = 2000 \, \text{N/m} \, (2 \, \text{kN/m})\]
2. Calculate force for \( x = 0.016 \, \text{m} \):
\[F = 2000 \cdot 0.016 = 32 \, \text{N}\]
Answer:
Spring constant: \( 2 \, \text{kN/m} \)
Weight: \( 32 \, \text{N} \)
6.2 The mass of 5 litres of milk is 4.5 kg. Find its density in SI units.
Given Data:
Mass (\( m = 4.5 \, \text{kg} \))
Volume (\( V = 5 \, \text{litres} = 0.005 \, \text{m}^3 \))
To Find:
Density (\( \rho \))
Formula:
\[\rho = \frac{m}{V}\]
Solution:
\[\rho = \frac{4.5}{0.005} = 900 \, \text{kg/m}^3\]
Answer:
\( \rho = 0.9 \times 10^3 \, \text{kg/m}^3 \)
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 Data:
Mass (\( m = 60 \, \text{g} = 0.06 \, \text{kg} \))
Volume displaced (\( V = 44 – 40 = 4 \, \text{cm}^3 = 4 \times 10^{-6} \, \text{m}^3 \))
To Find:
Density (\( \rho \))
Formula:
\[\rho = \frac{m}{V}\]
Solution:
\[\rho = \frac{0.06}{4 \times 10^{-6}} = 15 \times 10^3 \, \text{kg/m}^3\]
Answer:
\( \rho = 15 \times 10^3 \, \text{kg/m}^3 \)
6.4 A block of density \( 8 \times 10^3 \, \text{kg/m}^3 \) has a volume \( 60 \, \text{cm}^3 \). Find its mass.
Given Data:
Density (\( \rho = 8 \times 10^3 \, \text{kg/m}^3 \))
Volume (\( V = 60 \, \text{cm}^3 = 60 \times 10^{-6} \, \text{m}^3 \))
To Find:
Mass (\( m \))
Formula:
\[m = \rho \cdot V\]
Solution:
\[m = 8 \times 10^3 \cdot 60 \times 10^{-6} = 0.48 \, \text{kg}\]
Answer:
\( m = 0.48 \, \text{kg} \)
6.5 A brick measures \( 5 \, \text{cm} \times 10 \, \text{cm} \times 20 \, \text{cm} \). If its mass is 5 kg, calculate the maximum and minimum pressure which the brick can exert on a horizontal surface.
Given Data:
Mass (\( m = 5 \, \text{kg} \))
Gravitational force (\( F = m \cdot g = 5 \cdot 9.8 = 49 \, \text{N} \))
Areas of faces:
Maximum pressure: \( A_{\text{min}} = 10 \times 20 = 200 \, \text{cm}^2 = 0.02 \, \text{m}^2 \)
Minimum pressure: \( A_{\text{max}} = 5 \times 10 = 50 \, \text{cm}^2 = 0.005 \, \text{m}^2 \)
To Find:
Maximum and minimum pressure (\( P \))
Formula:
\[P = \frac{F}{A}\]
Solution:
1. Maximum pressure:
\[P_{\text{max}} = \frac{49}{0.005} = 9800 \, \text{Pa} \, (1 \times 10^4 \, \text{Pa})\]
2. Minimum pressure:
\[P_{\text{min}} = \frac{49}{0.02} = 2450 \, \text{Pa} \, (2.5 \times 10^3 \, \text{Pa})\]
Answer:
\( P_{\text{max}} = 1 \times 10^4 \, \text{Pa}, \, P_{\text{min}} = 2.5 \times 10^3 \, \text{Pa} \)
6.6 What will be the height of the column in a barometer at sea level if mercury is replaced by water of density \( 1000 \, \text{kg/m}^3 \), where density of mercury is \( 13.6 \times 10^3 \, \text{kg/m}^3 \)?
Given Data:
Density of water (\( \rho_w = 1000 \, \text{kg/m}^3 \))
Density of mercury (\( \rho_m = 13.6 \times 10^3 \, \text{kg/m}^3 \))
Height of mercury column (\( h_m = 0.76 \, \text{m} \))
To Find:
Height of water column (\( h_w \))
Formula:
\[\rho_w \cdot h_w = \rho_m \cdot h_m\]
Solution:
\[h_w = \frac{\rho_m \cdot h_m}{\rho_w}\]
\[h_w = \frac{13.6 \cdot 0.76}{1} = 10.3 \, \text{m}\]
Answer:
\( h_w = 10.3 \, \text{m} \)
6.7 Suppose in the hydraulic brake system of a car, the force exerted normally on its piston of cross-sectional area \( 5 \, \text{cm}^2 \) is \( 500 \, \text{N} \). What will be the pressure transferred to the brake oil? What will be the force on the second piston of area \( 20 \, \text{cm}^2 \)?
Given Data:
Force (\( F_1 = 500 \, \text{N} \))
Area (\( A_1 = 5 \, \text{cm}^2 = 0.0005 \, \text{m}^2 \))
Area (\( A_2 = 20 \, \text{cm}^2 = 0.002 \, \text{m}^2 \))
To Find:
1. Pressure (\( P \))
2. Force on second piston (\( F_2 \))
Formula:
\[P = \frac{F_1}{A_1}\]
\[F_2 = P \cdot A_2\]
Solution:
1. Pressure:
\[P = \frac{500}{0.0005} = 1 \times 10^6 \, \text{Pa}\]
2. Force on second piston:
\[F_2 = 1 \times 10^6 \cdot 0.002 = 2000 \, \text{N}\]
Answer:
Pressure: \( 1 \times 10^6 \, \text{Pa} \)
Force: \( 2000 \, \text{N} \)
6.8 Find the water pressure on a deep-sea diver at a depth of 10 m, where the density of sea water is \( 1030 \, \text{kg/m}^3 \).
Given Data:
Depth (\( h = 10 \, \text{m} \))
Density (\( \rho = 1030 \, \text{kg/m}^3 \))
Acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \))
To Find:
Pressure (\( P \))
Formula:
\[P = \rho \cdot g \cdot h\]
Solution:
\[P = 1030 \cdot 9.8 \cdot 10 = 1.03 \times 10^5 \, \text{Pa}\]
Answer:
\( P = 1.03 \times 10^5 \, \text{Pa} \)
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 Data:
Area of small piston (\( A_s \)) = \( 10 \, \text{cm}^2 = 0.001 \, \text{m}^2 \)
Area of large piston (\( A_l \)) = \( 100 \, \text{cm}^2 = 0.01 \, \text{m}^2 \)
Force on large piston (\( F_l \)) = \( 4000 \, \text{N} \)
To Find:
Force on small piston (\( F_s \))
Formula:
\[\frac{F_s}{A_s} = \frac{F_l}{A_l}\]
Solution:
\[F_s = \frac{F_l \cdot A_s}{A_l}\]
\[F_s = \frac{4000 \cdot 0.001}{0.01} = 400 \, \text{N}\]
Answer:
\( F_s = 400 \, \text{N} \)
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:
| Altitude | Pressure |
| 150 | 99.5 |
| 500 | 95.7 |
| 800 | 92.4 |
| 1140 | 88.9 |
| 1300 | 87.2 |
| 1500 | 85.3 |
To Find:
(a) Air pressure at sea level (\( h = 0 \, \text{m} \))
(b) Altitude at \( P = 90 \, \text{kPa} \)
Solution:
(a) Extrapolating the pressure data, air pressure at sea level can be estimated as \( 101 \, \text{kPa} \).
(b) By interpolation between \( P = 92.4 \, \text{kPa} \) at \( h = 800 \, \text{m} \) and \( P = 88.9 \, \text{kPa} \) at \( h = 1140 \, \text{m} \):
\[h = 800 + \frac{(90 – 92.4)}{(88.9 – 92.4)} \cdot (1140 – 800)\]
\[h = 800 + \frac{(-2.4)}{-3.5} \cdot 340 = 800 + 233 = 1020 \, \text{m}\]
Answer:
(a) Air pressure at sea level: \( 101 \, \text{kPa} \)
(b) Altitude for \( P = 90 \, \text{kPa} \): \( 1020 \, \text{m} \)
6.11 If the pressure in a hydraulic press is increased by an additional \( 10 \, \text{N/cm}^2 \), how much extra load will the output platform support if its cross-sectional area is \( 50 \, \text{cm}^2 \)?
Given Data:
Additional pressure (\( P = 10 \, \text{N/cm}^2 = 10^5 \, \text{Pa} \))
Cross-sectional area (\( A = 50 \, \text{cm}^2 = 0.005 \, \text{m}^2 \))
To Find:
Extra load (\( F \))
Formula:
\[F = P \cdot A\]
Solution:
\[F = 10^5 \cdot 0.005 = 500 \, \text{N}\]
Answer:
\( F = 500 \, \text{N} \)
6.12 The force exerted normally on the hydraulic brake system of a car, with its piston of cross-sectional area \( 5 \, \text{cm}^2 \), is \( 500 \, \text{N} \). What will be:
(a) Pressure transferred to the brake oil?
(b) Force on the brake piston of area \( 20 \, \text{cm}^2 \)?
Given Data:
Force on small piston (\( F_s = 500 \, \text{N} \))
Area of small piston (\( A_s = 5 \, \text{cm}^2 = 0.0005 \, \text{m}^2 \))
Area of large piston (\( A_l = 20 \, \text{cm}^2 = 0.002 \, \text{m}^2 \))
To Find:
(a) Pressure transferred (\( P \))
(b) Force on large piston (\( F_l \))
Formula:
(a) \( P = \frac{F_s}{A_s} \)
(b) \( F_l = P \cdot A_l \)
Solution:
(a) Calculate pressure:
\[P = \frac{500}{0.0005} = 1 \times 10^6 \, \text{Pa}\]
(b) Calculate force:
\[F_l = 1 \times 10^6 \cdot 0.002 = 2000 \, \text{N}\]
Answer:
(a) Pressure transferred: \( 1.0 \times 10^6 \, \text{Pa} \)
(b) Force on large piston: \( 2000 \, \text{N} \)
