# Physics Questions and Answers – Solids Mechanical Properties – Stress and Strain

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Solids Mechanical Properties – Stress and Strain”.

1. Stress in a solid body is defined as ___________ per unit area.
a) external force applied
b) strain
c) pressure
d) internal forces developed due to externally applied forces

Explanation: When a force is applied on a solid body, internal forces develop inside it. Pressure is the external force per unit area while stress is the internal force per unit area. Note that pressure and stress aren’t the same.

2. Stress, like pressure, is always perpendicular to a plane. True or False?
a) True
b) False

Explanation: Stress can act along any direction to a plane. If it is parallel to the plane it is called shearing stress & if it is perpendicular to the plane it’s called normal stress. Pressure is always perpendicular to a unit area.

3. A wire with a radius of 5mm is hung freely from the ceiling. A load of 5N is applied to its free end. Find the elongation in the wire if its volume is 7.85*10-5m3 & young’s modulus is 1011N/m2.
a) 6.21*10-7m
b) 7.00*10-7m
c) 6.36*10-7m
d) 8.00*10-9m

Explanation: The initial length of wire is Vol / πr2 = 7.85*10-5/ π*0.0052 = 1m.
Stress = Y*strain. F/A = Y*Δl / l.
Δl = F/A * l/Y = (5/πr2)*(1/1011)
= 6.37*10-7m.

4. A wire has a young’s modulus of 105N/m2, length 1m & radius 3mm. Assuming a uniform cross sectional area, find the radius of wire after it is under a force of 1N from both ends.
a) 2.58m
b) 2.30m
c) 3.54m
d) 2.24m

Explanation: Force = 1N. Initial area = πr2 = 2.82*10-5m2.
Stress = Y*Strain
Δl = F/A * l/Y = (1/2.82*10-5)*(1/105) = 0.35m
As volume will remain same (we can also say that product of l & r2 will be constant as other terms in expression of volume are constants).
1*32 = 1.35*R2
⇒R = 2.58m.

5. Strain can be negative. True or False?
a) True
b) False

Explanation: Strain is defined as change in length or volume divided by initial length or volume respectively. There can be cases when length or volume decreases like when rod is compressed or body is inserted in fluid (it volume decreases due to pressure from all sides). Hence, strain can be negative.
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6. In the given system, masses are released from rest. The young’s modulus of wire is 1011N/m2, length = 1m & radius = 2mm. Find elongation in wire when masses are moving. Assume pulley to be frictionless. a) 1.05*10-5m
b) 2*10-5m
c) 3*10-5m
d) 0.5*10-5m

Explanation: Let the tension in rope be ‘T’ & acceleration of masses be ‘a’.
2g-T=2a & T-1g=1a.
On solving these equations we get, T = 4g/3 = 1.33g.
For rope, Stress = Y*Strain.
∴ T/A = Y*Δl / l (where A is area of rope & l is initial length)
∴ Δl = (1.33g/πr2)*(1/1011) = 1.05*10-5 m.

Sanfoundry Global Education & Learning Series – Physics – Class 11. 