FBISE, NCP & NBF-aligned resources for Classes 9–12
Browse by Class
Browse by Type
Physics Chapter 7 : Topic 3
Stress, Strain and Young Modulus
1 / 40
Stress in a solid is defined as:
Reasoning: Stress is the internal restoring force developed per unit cross-sectional area of a material.
2 / 40
The SI unit of stress is:
Reasoning: Stress has the same unit as pressure, which is Pascal $\mathbf{(}\mathbf{N\ /\ m}^{\mathbf{2}}\mathbf{)}$.
3 / 40
Mathematically, stress is expressed as:
Reasoning: Stress is the applied force divided by the cross-sectional area over which it acts.
4 / 40
If the same force is applied to two wires, the wire with smaller cross-sectional area experiences:
Reasoning: Stress is inversely proportional to area, so smaller area produces larger stress.
5 / 40
A force of $\mathbf{100\ N}$ acts on an area of $\mathbf{2\ m}^{\mathbf{2}}$. The stress produced is:
Reasoning: Stress $\mathbf{=}$ Force $\mathbf{/}$ Area $\mathbf{= 100\ /\ 2 = 50\ Pa}$
6 / 40
The SI unit of strain is:
Reasoning: Strain is a ratio of two lengths, so units cancel.
7 / 40
Strain is defined as:
Reasoning: Strain measures the fractional change in dimension of a material.
8 / 40
A wire of length $\mathbf{2\ m}$ increases its length by $\mathbf{0.01\ m}$. The strain is:
Reasoning: Strain $\mathbf{=}$ Change in length $\mathbf{/}$ Original length $\mathbf{= 0.01\ /\ 2 = 0.005}$
9 / 40
A material with a large Young modulus is:
Reasoning: A large Young modulus means greater stress is required to produce a given strain.
10 / 40
Young modulus is defined as:
Reasoning: Young modulus is the ratio of tensile stress to tensile strain.
11 / 40
The SI unit of Young modulus is:
Reasoning: Since Young modulus is stress divided by strain and strain has no unit, its unit is Pascal.
12 / 40
Young modulus measures the:
Reasoning: Young modulus indicates how strongly a material resists stretching.
13 / 40
Two wires have the same dimensions. Wire $\mathbf{A}$ stretches less than wire $\mathbf{B}$ under the same force. This means:
Reasoning: Less strain for the same stress means a higher Young modulus.
14 / 40
A material obeys Hooke’s law when:
Reasoning: Hooke’s law states that within the elastic limit, stress is directly proportional to strain.
15 / 40
The elastic limit of a material is the point beyond which:
Reasoning: Beyond the elastic limit, permanent deformation may occur.
16 / 40
A wire is stretched by applying a force. The stress developed in the wire depends on:
Reasoning: Stress depends on the applied force and the area over which it acts.
17 / 40
Increasing the length of a wire while keeping the same force and area will:
Reasoning: Stress depends on force and area, not length.
18 / 40
Increasing the length of a wire while keeping force and area constant will:
Reasoning: For the same extension, a longer wire has a smaller strain, but under the same force its extension increases; hence strain increases.
19 / 40
If the stress on a wire is doubled and it remains within elastic limit, the strain will:
Reasoning: According to Hooke’s law, stress is directly proportional to strain.
20 / 40
Which quantity is a measure of deformation?
Reasoning: Strain describes the amount of deformation relative to original size.
21 / 40
A wire with a higher Young modulus compared with another wire will:
Reasoning: Higher Young modulus means greater resistance to deformation.
22 / 40
Which quantity describes the internal resistance of a material against deformation?
Reasoning: Young modulus indicates the stiffness of a material.
23 / 40
A wire has a high Young modulus. To stretch it by a certain amount, the required force will be:
Reasoning: High Young modulus means greater force is needed for the same strain.
24 / 40
The stress produced in a wire is increased by reducing its:
Reasoning: Stress increases when force is applied over a smaller area.
25 / 40
A wire experiences tensile stress when:
Reasoning: Tensile stress occurs when forces stretch a material.
26 / 40
The strain produced in a wire depends on:
Reasoning: Strain is the ratio of extension to original length.
27 / 40
Which of the following has the greatest Young modulus?
Reasoning: High Young modulus represents high stiffness.
28 / 40
A wire of length $\mathbf{L}$ is stretched to length $\mathbf{L + \Delta L}$. The strain is:
Reasoning: Strain is the fractional increase in length.
29 / 40
Young modulus can be written as:
Reasoning: Young modulus $\mathbf{=}$ stress $\mathbf{/}$ strain $\mathbf{= (F\ /\ A)\ /\ (\Delta L\ /\ L) = FL\ /\ A\ \Delta L}$
30 / 40
A wire is twice as long as another wire of the same material and area. Under the same force, it will have:
Reasoning: Extension is directly proportional to original length.
31 / 40
If the cross-sectional area of a wire is doubled while the same force is applied, the stress becomes:
Reasoning: Stress $\mathbf{=}$ Force $/$ Area, so doubling area halves stress.
32 / 40
Two materials experience the same stress. The material with greater strain has:
Reasoning: Young modulus $\mathbf{=}$ stress $/$ strain ; greater strain means smaller Young modulus.
33 / 40
Stress and pressure have the same unit because both are:
Reasoning: Both quantities are measured as force divided by area.
34 / 40
A material returns to its original dimensions after removing force. This indicates:
Reasoning: Elastic materials regain their original shape after removal of applied force.
35 / 40
Which statement about Young modulus is correct?
Reasoning: Young modulus depends on the nature of the material.
36 / 40
If a material has a very small Young modulus, it means:
Reasoning: Small Young modulus indicates low resistance to deformation.
37 / 40
A steel wire and a rubber wire are subjected to the same stress. The rubber wire stretches more because:
Reasoning: Rubber has a much lower Young modulus than steel.
38 / 40
The relationship between stress and strain is linear:
Reasoning: Hooke’s law applies only within the elastic limit.
39 / 40
If the length of a wire is increased but its extension remains unchanged, the strain will:
Reasoning: Strain $\mathbf{=}$ extension $\mathbf{/}$ original lengthSo, increasing length reduces strain.
40 / 40
The purpose of calculating Young modulus is to compare:
Reasoning: Young modulus allows comparison of stiffness and elasticity of different materials.
Your score is
Restart quiz
