250+ TOP MCQs on Strength of Materials and Answers

This set of Farm Machinery Multiple Choice Questions on “Strength of Materials”.

1. A load of 2 kg resulted an elongation of 1mm in a wire of 3 m length with 1 mm diameter. What will be the Young’s modulus of wire?
a) 7.484*10¹⁰ Pa
b) 9.484*10¹⁰ Pa
c) 6.317*10¹⁰ Pa
d) 7.484*10²⁰ Pa
Answer: a
Explanation: Cross-sectional area of wire = πr² = π*(0.5 x 10^-3)² m²
Y = (frac{F.L.}{A.l}) = (2*9.8*3)/[π*(0.5*10^-3})²*10^-3
Y = 7.484 x 10¹⁰ Pa

2. What would be the percent increase in length of wire of diameter 2.5 x I0^-3m stretched by a force of 980 N. Young’s modulus of elasticity of wire is 8 x 10¹⁰ N/m².
a) 0.129
b) 0.364
c) 0.249
d) 0.098
Answer: c
Explanation: Cross-sectional area of wire = πr² = π * (1.25 x 10^-3)² m²
From the relation
Y = (frac{F.L.}{A.l} )
Percent increase in length = (frac{l}{L} = frac{F}{A.Y} )
(frac{l}{L})*100=(frac{F}{A.Y})*100=(frac{980*7*100}{[22(1.25*10^{-3})^2*8*10^{10}]} )
= 0.249

3. A force of 10⁶ N/m² is required for breaking a material of density 3 x 10³ kg/m³. What would be the length of wire, which can break by its own weight?
a) 43 m
b) 98 m
c) 85m
d) 34 m
Answer: d
Explanation: L * ρ * g = 10⁶ N/m²
L = (frac{(10^6)}{ρ.g} = frac{10^6}{3*10^3*9.8} )
= 34m

4. What will be the elastic potential energy per unit volume when a metal wire is suspended along with a suspending weight on it?
a) Y²α
b) (frac{1}{2}) Yα²
c) 4Yα
d) 8Y²α²
Answer: b
Explanation: The elastic potential energy per unit volume, u = (frac{1}{2}) *stress*strain
Y = (frac{Stress}{Strain} )
Stress = Y * Strain = Y*α
U = (frac{1}{2})*Y*α*α = (frac{1}{2})*Y*α²

5. The length of wire is increased by 8 mm in length, when a weight of 5 kg is hung through it. If all the conditions are kept same, but radius of wire is doubled, what would be the new length of wire?
a) 2mm
b) 4mm
c) 6mm
d) 8mm
Answer: a
Explanation:
Y = (frac{M.g.l}{pi.r^2 l} )
From this relation, it is clear that l α (frac{1}{r^2}), accordingly putting the radius of wire as double, the elongation will be
l = (frac{1}{4})*8 = 2 mm

6. Steel wire of length 5.7 m and cross-sectional area 3*10^-5 m² is stretched by the same amount as a Copper wire of 3.5 m length and 4*10^-5m² cross-sectional area, under a given load. What would be the ratio of their Young’s modulus?
a) 2.3:1.9
b) 1.5:1.5
c) 1.6:1.8
d) 2.2:1.3
Answer: d
Explanation:
Y₁ = (frac{F1.L1}{A1.l1} = frac{F1*4.7}{3*10^{-5} l1} )
Y₂ = (frac{F2.L2}{A2.l2} = frac{F2*3.5}{4*10^{-5} l1} )
(frac{Y1}{Y2} = frac{5.7*4*10^{-5}}{4.5*3*10^{-5}}) = 2.2:1.3

7. The upper face of a 4cm cube is displaced by 0.1 cm due to a tangential force of 0.8 dyne. What would be the shear modulus of the cube?
a) 2 dyne/cm²
b) 1.3 dyne/cm²
c) 2.5 dyne/cm²
d) 5 dyne/cm²
Answer: a
Explanation:
Area of the face over which the force was applied = A = L² = 16cm²
η = (frac{F.L}{A.x}=frac{0.8*4}{16*0.1}) = 2 dyne/cm²

8. A material has Poisson’s ratio as 0.5. If a uniform rod of it is subjected by a longitudinal strain of 2 x 10^-3, then what would be the percent increase in volume?
a) 2
b) 0
c) 9
d) 10
Answer: b
Explanation: (frac{∆l}{l}=2*10^{-3} )
σ = 0.5
σ = (frac{[(frac{∆r}{r})]}{[(frac{∆l}{l})]} )
(frac{∆r}{r}) = -0.5*10^-3*2=1*10^-3
Volume of rod = πr²l
Fractional increase in volume = (frac{∆v}{v}=frac{[∆(πr^2 l)]}{[πr^2 l]} = frac{∆l}{l}+frac{2∆r}{r} )
= 2*10^-3+2(-1*10^-3)
= 0

9. A wire is stretched 1 mm by applying a force of 1 KN. How far a wire of same material and the length but of 3 times diameter can be stretched?
a) (frac{1}{2} )
b) (frac{1}{32} )
c) (frac{1}{16} )
d) (frac{1}{9} )
Answer: d
Explanation: Elongation is inversely proportional to the cross-sectional area
∆l = 1 * ((frac{1}{3})^2=frac{1}{9})mm