Farm Machinery Questions and Answers – Strength of Materials

This set of Farm Machinery Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

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
View Answer

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
View Answer

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
View Answer

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
View Answer

Answer: a
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
View Answer

Answer: d
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
View Answer

Answer: a
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
View Answer

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) \(\frac{1}{9} \)
View Answer

Answer: d
Explanation: Elongation is inversely proportional to the cross-sectional area
∆l = 1 * \((\frac{1}{3})^2=\frac{1}{9}\)mm

Sanfoundry Global Education & Learning Series – Farm Machinery.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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