501. A steel wire of length $2 \, m$ and cross-sectional area $1 \, mm^2$ is stretched under a load of $200 \, N$. If Young’s modulus of steel is $2 \times 10^{11} \, Pa$, what is the extension produced?
ⓐ. $2 \, mm$
ⓑ. $1 \, mm$
ⓒ. $0.2 \, mm$
ⓓ. $0.02 \, mm$
Correct Answer: $0.02 \, mm$
Explanation: Stress = $ \frac{F}{A} = \frac{200}{1 \times 10^{-6}} = 2 \times 10^{8} \, Pa$.
Strain = $\sigma / Y = 2 \times 10^{8} / 2 \times 10^{11} = 1 \times 10^{-3}$.
Extension = strain $\times L = 1 \times 10^{-3} \times 2 = 2 \times 10^{-3} \, m = 0.02 \, mm$.
502. A wire of original length $L = 1.5 \, m$ and diameter $1 \, mm$ is stretched by $1.5 \, mm$ under a load of $30 \, N$. Calculate Young’s modulus.
ⓐ. $2.55 \times 10^{10} \, Pa$
ⓑ. $3.82 \times 10^{10} \, Pa$
ⓒ. $5.10 \times 10^{10} \, Pa$
ⓓ. $7.64 \times 10^{10} \, Pa$
Correct Answer: $3.82 \times 10^{10} \, Pa$
Explanation: Stress = $F/A = \frac{30}{\pi (0.5 \times 10^{-3})^2} = 3.82 \times 10^{7} \, Pa$.
Strain = $\Delta L / L = 1.5 \times 10^{-3}/1.5 = 1 \times 10^{-3}$.
$Y = \sigma / \epsilon = 3.82 \times 10^{7} / 1 \times 10^{-3} = 3.82 \times 10^{10} \, Pa$.
503. A steel wire of length $2 \, m$ and radius $0.5 \, mm$ is fixed at one end. A load of $100 \, N$ is applied at the free end. If the elongation is $0.2 \, mm$, calculate Young’s modulus.
ⓐ. $2.55 \times 10^{10} \, Pa$
ⓑ. $2.55 \times 10^{12} \, Pa$
ⓒ. $3.82 \times 10^{11} \, Pa$
ⓓ. $1.27 \times 10^{12} \, Pa$
Correct Answer: $1.27 \times 10^{12} \, Pa$
Explanation: Stress = $F/A = \frac{100}{\pi (0.5 \times 10^{-3})^2} = 1.27 \times 10^{8} \, Pa$.
Strain = $\Delta L/L = 0.2 \times 10^{-3} / 2 = 1 \times 10^{-4}$.
$Y = 1.27 \times 10^{8}/1 \times 10^{-4} = 1.27 \times 10^{12} \, Pa$.
504. A wire is stretched by $1 \, mm$ when a load of $20 \, N$ is applied. If the same wire is stretched by $2 \, mm$, what load is required, assuming Hooke’s law holds?
ⓐ. $20 \, N$
ⓑ. $30 \, N$
ⓒ. $40 \, N$
ⓓ. $80 \, N$
Correct Answer: $40 \, N$
Explanation: Within elastic limit, stress ∝ strain ⇒ Load ∝ extension. So, doubling extension doubles load. Required load = $20 \times 2 = 40 \, N$.
505. A steel rod of cross-sectional area $1 \, cm^2$ is subjected to tensile stress of $2 \times 10^7 \, N/m^2$. If Young’s modulus of steel is $2 \times 10^{11} \, Pa$, find elongation of the rod of length $2 \, m$.
ⓐ. $0.002 \, mm$
ⓑ. $0.02 \, mm$
ⓒ. $0.2 \, mm$
ⓓ. $2 \, mm$
Correct Answer: $0.2 \, mm$
Explanation: Strain = $\sigma / Y = 2 \times 10^7 / 2 \times 10^{11} = 1 \times 10^{-4}$.
Extension = strain × length = $1 \times 10^{-4} \times 2 = 2 \times 10^{-4} \, m = 0.2 \, mm$.
506. Two wires of same material and length but radii in ratio 1:2 are stretched by same load. What is the ratio of their elongations?
ⓐ. 3:4
ⓑ. 4:1
ⓒ. 4:3
ⓓ. 1:4
Correct Answer: 1:4
Explanation: Extension $\Delta L = \frac{FL}{AY}$. Since area ∝ $r^2$, elongation ∝ 1/$r^2$. Ratio = $r_2^2 : r_1^2 = 2^2:1^2 = 4:1$. But elongation is inverse, so = 1:4.
507. A steel wire and a copper wire of same length and cross-sectional area are joined end to end and stretched by 100 N. If Young’s modulus of steel = $2 \times 10^{11} \, Pa$ and copper = $1 \times 10^{11} \, Pa$, ratio of elongations of steel to copper is:
ⓐ. 1:2
ⓑ. 2:1
ⓒ. 1:1
ⓓ. 4:1
Correct Answer: 1:2
Explanation: Elongation $\Delta L \propto 1/Y$. Since $Y_{steel} : Y_{copper} = 2:1$, elongation ratio = 1:2.
508. A rod of length $1 \, m$ and cross-sectional area $1 \, cm^2$ is subjected to tensile force of $1000 \, N$. If Young’s modulus is $2 \times 10^{11} \, Pa$, calculate extension.
ⓐ. $0.0005 \, mm$
ⓑ. $0.005 \, mm$
ⓒ. $0.05 \, mm$
ⓓ. $0.5 \, mm$
Correct Answer: $0.05 \, mm$
Explanation: Stress = $1000 / 1 \times 10^{-4} = 1 \times 10^7 \, Pa$.
Strain = $1 \times 10^7 / 2 \times 10^{11} = 5 \times 10^{-5}$.
Extension = $5 \times 10^{-5} \times 1 = 5 \times 10^{-5} \, m = 0.05 \, mm$.
509. A wire of length $2.5 \, m$ and area $1 \, mm^2$ is subjected to a force of $50 \, N$. If elongation is $0.25 \, mm$, calculate Young’s modulus.
ⓐ. $1 \times 10^{10} \, Pa$
ⓑ. $2 \times 10^{10} \, Pa$
ⓒ. $2 \times 10^{11} \, Pa$
ⓓ. $5 \times 10^{11} \, Pa$
Correct Answer: $5 \times 10^{11} \, Pa$
Explanation: Stress = $50/1 \times 10^{-6} = 5 \times 10^7 \, Pa$.
Strain = $0.25 \times 10^{-3} / 2.5 = 1 \times 10^{-4}$.
$Y = \sigma / \epsilon = 5 \times 10^7 / 1 \times 10^{-4} = 5 \times 10^{11} \, Pa$.
510. A steel rod of length $2 \, m$ and cross-sectional area $1 \, cm^2$ is subjected to a tensile force of $20 \, kN$. If elongation is $1 \, mm$, find Young’s modulus.
ⓐ. $2 \times 10^{10} \, Pa$
ⓑ. $4 \times 10^{13} \, Pa$
ⓒ. $4 \times 10^{11} \, Pa$
ⓓ. $4 \times 10^{9} \, Pa$
Correct Answer: $4 \times 10^{11} \, Pa$
Explanation: Stress = $20000/1 \times 10^{-4} = 2 \times 10^8 \, Pa$.
Strain = $1 \times 10^{-3}/2 = 5 \times 10^{-4}$.
$Y = 2 \times 10^8 / 5 \times 10^{-4} = 4 \times 10^{11} \, Pa$.
511. A steel wire of length $2 \, m$ and radius $1 \, mm$ is subjected to a load of $100 \, N$. If $Y = 2 \times 10^{11} \, Pa$, calculate the extension.
ⓐ. $0.032 \, mm$
ⓑ. $0.16 \, mm$
ⓒ. $0.32 \, mm$
ⓓ. $1.6 \, mm$
Correct Answer: $0.032 \, mm$
Explanation: Area $A = \pi r^2 = 3.14 \times (1 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, m^2$.
Stress = $100 / 3.14 \times 10^{-6} = 3.18 \times 10^7 \, Pa$.
Strain = $3.18 \times 10^7 / 2 \times 10^{11} = 1.59 \times 10^{-4}$.
Extension = strain × length = $1.59 \times 10^{-4} \times 2 = 3.18 \times 10^{-4} \, m = 0.032 \, mm$.
512. A brass rod of length $1.5 \, m$ and cross-sectional area $1 \, cm^2$ is subjected to a tensile force of $3000 \, N$. If $Y = 1 \times 10^{11} \, Pa$, find elongation.
ⓐ. $0.45 \, mm$
ⓑ. $0.30 \, mm$
ⓒ. $0.15 \, mm$
ⓓ. $0.05 \, mm$
Correct Answer: $0.30 \, mm$
Explanation: Stress = $3000 / 1 \times 10^{-4} = 3 \times 10^7 \, Pa$.
Strain = $3 \times 10^7 / 1 \times 10^{11} = 3 \times 10^{-4}$.
Elongation = strain × length = $3 \times 10^{-4} \times 1.5 = 4.5 \times 10^{-4} \, m = 0.30 \, mm$.
513. A wire of length $2 \, m$ is stretched by a load of $100 \, N$. The energy stored per unit volume is $0.5 \, J/m^3$. Find Young’s modulus if cross-sectional area = $1 \, mm^2$.
ⓐ. $2 \times 10^{11} \, Pa$
ⓑ. $1 \times 10^{11} \, Pa$
ⓒ. $5 \times 10^{10} \, Pa$
ⓓ. $1 \times 10^{10} \, Pa$
Correct Answer: $2 \times 10^{11} \, Pa$
Explanation: Energy density = $\frac{1}{2} \sigma \epsilon = 0.5$.
$\sigma = F/A = 100 / 1 \times 10^{-6} = 1 \times 10^8 \, Pa$.
So, $\epsilon = \sigma / Y$.
$\frac{1}{2} \times (1 \times 10^8)(1 \times 10^8 / Y) = 0.5$.
$\Rightarrow Y = 2 \times 10^{11} \, Pa$.
514. A steel rod of length $1 \, m$, area $2 \, cm^2$, and Young’s modulus $2 \times 10^{11} \, Pa$ is subjected to a tensile force of $20 \, kN$. Calculate energy stored per unit volume.
ⓐ. $1 \times 10^5 \, J/m^3$
ⓑ. $2 \times 10^5 \, J/m^3$
ⓒ. $5 \times 10^5 \, J/m^3$
ⓓ. $2 \times 10^5 \, J/m^3$
Correct Answer: $2 \times 10^5 \, J/m^3$
Explanation: Stress = $F/A = 20000 / 2 \times 10^{-4} = 1 \times 10^8 \, Pa$.
Energy density = $\sigma^2 / (2Y) = (1 \times 10^8)^2 / (4 \times 10^{11}) = 2.5 \times 10^4 \, J/m^3$.
515. Two wires of same length and same load but radii in ratio 1:2 are stretched. Find the ratio of energy stored per unit volume.
ⓐ. 1:4
ⓑ. 4:1
ⓒ. 1:16
ⓓ. 16:1
Correct Answer: 16:1
Explanation: Energy density $\propto \sigma^2$. Stress $\sigma \propto 1/r^2$. Thus, energy density $\propto 1/r^4$. Ratio = $2^4 : 1^4 = 16:1$.
516. A steel wire of length $2 \, m$ and radius $1 \, mm$ is stretched by $2 \, mm$ under a load. Calculate strain energy stored if $Y = 2 \times 10^{11} \, Pa$.
ⓐ. $0.25 \, J$
ⓑ. $0.50 \, J$
ⓒ. $1.0 \, J$
ⓓ. $2.0 \, J$
Correct Answer: $0.50 \, J$
Explanation: Strain = $\Delta L / L = 2 \times 10^{-3}/2 = 1 \times 10^{-3}$.
Stress = $Y \epsilon = 2 \times 10^{11} \times 1 \times 10^{-3} = 2 \times 10^8 \, Pa$.
Energy density = $\frac{1}{2}\sigma \epsilon = 0.5 \times 2 \times 10^8 \times 1 \times 10^{-3} = 1 \times 10^5 \, J/m^3$.
Volume = $A L = \pi (1 \times 10^{-3})^2 \times 2 = 6.28 \times 10^{-6}$.
Energy = $1 \times 10^5 \times 6.28 \times 10^{-6} = 0.628 \, J$. Closest = 0.50 J.
517. A wire of length $1.5 \, m$ and diameter $2 \, mm$ is subjected to a load of $300 \, N$. If Young’s modulus = $2 \times 10^{11} \, Pa$, find the strain energy per unit volume.
ⓐ. $2.4 \times 10^4 \, J/m^3$
ⓑ. $1.2 \times 10^5 \, J/m^3$
ⓒ. $3.6 \times 10^4 \, J/m^3$
ⓓ. $4.8 \times 10^5 \, J/m^3$
Correct Answer: $2.4 \times 10^4 \, J/m^3$
Explanation: Area = $\pi (1 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, m^2$.
Stress = $300 / 3.14 \times 10^{-6} = 9.55 \times 10^7 \, Pa$.
Energy density = $\sigma^2 / (2Y) = (9.55 \times 10^7)^2 / (4 \times 10^{11}) \approx 2.4 \times 10^4 \, J/m^3$.
518. A steel rod of length $1 \, m$, cross-section $100 \, mm^2$, is subjected to tensile force $10 \, kN$. Calculate the strain energy stored in the rod. $Y = 2 \times 10^{11} \, Pa$.
ⓐ. $0.025 \, J$
ⓑ. $0.25 \, J$
ⓒ. $2.5 \, J$
ⓓ. $25 \, J$
Correct Answer: $2.5 \, J$
Explanation: Stress = $10000/1 \times 10^{-4} = 1 \times 10^8 \, Pa$.
Strain = $\sigma/Y = 1 \times 10^8 / 2 \times 10^{11} = 5 \times 10^{-4}$.
Energy density = $\frac{1}{2}\sigma \epsilon = 0.5 \times 1 \times 10^8 \times 5 \times 10^{-4} = 2.5 \times 10^4 \, J/m^3$.
Volume = $A L = 1 \times 10^{-4} \times 1 = 1 \times 10^{-4} \, m^3$.
Energy = $2.5 \times 10^4 \times 1 \times 10^{-4} = 2.5 \, J$.
519. The length of a steel wire increases by $1.0 \, mm$ when a load of $200 \, N$ is applied. If the same load is applied to a copper wire of same dimensions, elongation is $1.6 \, mm$. What is the ratio of Young’s moduli of steel to copper?
ⓐ. 1:1.6
ⓑ. 1.6:1
ⓒ. 1:2
ⓓ. 2:1
Correct Answer: 1.6:1
Explanation: $Y \propto 1/\Delta L$. Ratio = $Y_{steel}/Y_{copper} = \Delta L_{Cu} / \Delta L_{steel} = 1.6/1 = 1.6$.
520. A steel wire of radius $1 \, mm$ and length $2 \, m$ is clamped at one end and twisted through an angle of $1 \, rad$. Calculate the shear modulus if the torque applied is $0.1 \, Nm$.
ⓐ. $1.6 \times 10^{10} \, Pa$
ⓑ. $2.0 \times 10^{10} \, Pa$
ⓒ. $4.0 \times 10^{10} \, Pa$
ⓓ. $8.0 \times 10^{10} \, Pa$
Correct Answer: $1.6 \times 10^{10} \, Pa$
Explanation: For torsion, $T = \frac{\pi G r^4 \theta}{2L}$.
Here, $0.1 = \frac{\pi G (1 \times 10^{-3})^4 \times 1}{2 \times 2}$.
$\Rightarrow G = \frac{0.1 \times 4}{\pi \times 10^{-12}} \approx 1.27 \times 10^{10} \, Pa$. Closest = $1.6 \times 10^{10}$.
521. A wire of length $2 \, m$ and diameter $1 \, mm$ is stretched under a load of $100 \, N$. If the lateral strain is $1/800$ and longitudinal strain is $1/400$, calculate Poisson’s ratio.
ⓐ. 0.25
ⓑ. 0.50
ⓒ. 0.75
ⓓ. 0.80
Correct Answer: 0.25
Explanation: Poisson’s ratio $\nu = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{1/800}{1/400} = 0.25$.
522. For a material, Young’s modulus $Y = 2.0 \times 10^{11} \, Pa$ and Poisson’s ratio $\nu = 0.25$. Find the bulk modulus.
ⓐ. $6.67 \times 10^{10} \, Pa$
ⓑ. $8.0 \times 10^{10} \, Pa$
ⓒ. $1.0 \times 10^{11} \, Pa$
ⓓ. $1.33 \times 10^{11} \, Pa$
Correct Answer: $1.33 \times 10^{11} \, Pa$
Explanation: $K = \frac{Y}{3(1-2\nu)} = \frac{2 \times 10^{11}}{3(1-0.5)} = \frac{2 \times 10^{11}}{1.5} = 1.33 \times 10^{11} \, Pa$.
523. A rod of length $2 \, m$, area $2 \, cm^2$, and Young’s modulus $2 \times 10^{11} \, Pa$ is stretched by $2 \, mm$. Calculate the applied force.
ⓐ. $2 \, kN$
ⓑ. $4 \, kN$
ⓒ. $8 \, kN$
ⓓ. $10 \, kN$
Correct Answer: $8 \, kN$
Explanation: Strain = $\Delta L/L = 2 \times 10^{-3}/2 = 1 \times 10^{-3}$.
Stress = $Y \epsilon = 2 \times 10^{11} \times 1 \times 10^{-3} = 2 \times 10^{8} \, Pa$.
Force = Stress × Area = $2 \times 10^{8} \times 2 \times 10^{-4} = 4 \times 10^4 \, N = 8 \, kN$.
524. The length of a rod increases by $0.5 \, mm$ when subjected to tensile stress of $1.0 \times 10^8 \, Pa$. If Young’s modulus = $2.0 \times 10^{11} \, Pa$, find original length of rod.
ⓐ. $0.5 \, m$
ⓑ. $1.0 \, m$
ⓒ. $2.0 \, m$
ⓓ. $5.0 \, m$
Correct Answer: $1.0 \, m$
Explanation: Strain = $\sigma/Y = 1 \times 10^8 / 2 \times 10^{11} = 5 \times 10^{-4}$.
So, $\Delta L / L = 5 \times 10^{-4}$.
Thus, $L = \Delta L / strain = 0.5 \times 10^{-3}/5 \times 10^{-4} = 1 \, m$.
525. For a certain material, $Y = 2.1 \times 10^{11} \, Pa$, bulk modulus $K = 1.6 \times 10^{11} \, Pa$. Find Poisson’s ratio.
ⓐ. 0.20
ⓑ. 0.25
ⓒ. 0.30
ⓓ. 0.35
Correct Answer: 0.30
Explanation: Relation $Y = 3K(1-2\nu)$.
So, $\nu = \frac{1}{2}\left[1 – \frac{Y}{3K}\right] = \frac{1}{2}\left(1 – \frac{2.1}{4.8}\right) = 0.30$.
526. A cylindrical rod of radius $1 \, cm$, length $1 \, m$, is subjected to shear stress of $2 \times 10^6 \, Pa$. If shear modulus $G = 8 \times 10^{10} \, Pa$, find the lateral displacement of top face.
ⓐ. $0.025 \, mm$
ⓑ. $0.025 \, cm$
ⓒ. $0.25 \, mm$
ⓓ. $0.25 \, cm$
Correct Answer: $0.025 \, mm$
Explanation: Shear strain = $\tau/G = 2 \times 10^6 / 8 \times 10^{10} = 2.5 \times 10^{-5}$.
Displacement = strain × length = $2.5 \times 10^{-5} \times 1 = 2.5 \times 10^{-5} \, m = 0.025 \, mm$.
527. A steel wire of radius $1.0 \, mm$ is clamped at one end and twisted by torque of $0.2 \, Nm$. If length = $2.0 \, m$, calculate angle of twist. (Shear modulus $G = 8 \times 10^{10} \, Pa$)
ⓐ. 0.1 rad
ⓑ. 0.2 rad
ⓒ. 0.4 rad
ⓓ. 0.8 rad
Correct Answer: 0.1 rad
Explanation: $T = \frac{\pi G r^4 \theta}{2L}$.
So, $\theta = \frac{2TL}{\pi G r^4} = \frac{2 \times 0.2 \times 2}{3.14 \times 8 \times 10^{10} \times (1 \times 10^{-3})^4} \approx 0.1 \, rad$.
528. A beam of rectangular cross-section breadth $2 \, cm$, depth $4 \, cm$, length $1 \, m$, is loaded centrally with $100 \, N$. Find the central deflection if $Y = 2 \times 10^{11} \, Pa$.
ⓐ. $0.001 \, mm$
ⓑ. $0.01 \, mm$
ⓒ. $0.1 \, mm$
ⓓ. $1.0 \, mm$
Correct Answer: $0.01 \, mm$
Explanation: Deflection $\delta = \frac{WL^3}{48 YI}$.
Moment of inertia $I = bd^3/12 = (0.02 \times (0.04)^3)/12 = 1.07 \times 10^{-7}$.
$\delta = \frac{100 \times 1}{48 \times 2 \times 10^{11} \times 1.07 \times 10^{-7}} \approx 1 \times 10^{-5} \, m = 0.01 \, mm$.
529. A steel rod of length $L$, area $A$, Young’s modulus $Y$, is subjected to stress $\sigma$. What is the strain energy stored per unit volume?
ⓐ. $\frac{\sigma}{Y}$
ⓑ. $\frac{\sigma^2}{2Y}$
ⓒ. $\frac{\sigma^2}{Y}$
ⓓ. $\frac{\sigma}{2Y}$
Correct Answer: $\frac{\sigma^2}{2Y}$
Explanation: Energy density $= \frac{1}{2} \sigma \epsilon = \frac{1}{2}\sigma \cdot \frac{\sigma}{Y} = \frac{\sigma^2}{2Y}$.
530. A metal rod has $Y = 2 \times 10^{11} \, Pa$, $K = 1.6 \times 10^{11} \, Pa$. Find shear modulus $G$.
ⓐ. $7.7 \times 10^{10} \, Pa$
ⓑ. $8.0 \times 10^{10} \, Pa$
ⓒ. $8.5 \times 10^{10} \, Pa$
ⓓ. $9.0 \times 10^{10} \, Pa$
Correct Answer: $7.7 \times 10^{10} \, Pa$
Explanation: Relation: $\frac{1}{G} = \frac{3}{Y} + \frac{1}{3K}$.
Or, using $Y = 9KG / (3K+G)$.
Solve: $2 \times 10^{11} = \frac{9(1.6 \times 10^{11})G}{3(1.6 \times 10^{11})+G}$.
Cross-solving → $G \approx 7.7 \times 10^{10} \, Pa$.
531. A steel rod of length $2 \, m$ and cross-sectional area $2 \, cm^2$ is subjected to a tensile force of $10 \, kN$. If $Y = 2 \times 10^{11} \, Pa$, calculate elongation.
ⓐ. $0.25 \, mm$
ⓑ. $0.50 \, mm$
ⓒ. $1.00 \, mm$
ⓓ. $2.00 \, mm$
Correct Answer: $0.50 \, mm$
Explanation: Stress = $F/A = 10000 / 2 \times 10^{-4} = 5 \times 10^7 \, Pa$.
Strain = $\sigma / Y = 5 \times 10^7 / 2 \times 10^{11} = 2.5 \times 10^{-4}$.
Elongation = strain × length = $2.5 \times 10^{-4} \times 2 = 5.0 \times 10^{-4} \, m = 0.50 \, mm$.
532. A cantilever beam of length $1 \, m$, breadth $2 \, cm$, depth $4 \, cm$ is loaded at the free end with $50 \, N$. If $Y = 2 \times 10^{11} \, Pa$, calculate deflection at the free end.
ⓐ. $0.002 \, mm$
ⓑ. $0.02 \, mm$
ⓒ. $0.20 \, mm$
ⓓ. $2.00 \, mm$
Correct Answer: $0.20 \, mm$
Explanation: Deflection of cantilever under end load: $\delta = \frac{WL^3}{3 Y I}$.
$I = \frac{bd^3}{12} = \frac{0.02 (0.04)^3}{12} = 1.07 \times 10^{-7} \, m^4$.
$\delta = \frac{50 \times 1}{3 \times 2 \times 10^{11} \times 1.07 \times 10^{-7}} = 2.0 \times 10^{-4} \, m = 0.20 \, mm$.
533. A steel wire of length $1.5 \, m$ and radius $1 \, mm$ is clamped at one end and twisted by torque $0.1 \, Nm$. If $G = 8 \times 10^{10} \, Pa$, find angle of twist.
ⓐ. $0.05 \, rad$
ⓑ. $0.10 \, rad$
ⓒ. $0.20 \, rad$
ⓓ. $0.40 \, rad$
Correct Answer: $0.10 \, rad$
Explanation: $T = \frac{\pi G r^4 \theta}{2L}$.
$\theta = \frac{2 T L}{\pi G r^4} = \frac{2 \times 0.1 \times 1.5}{3.14 \times 8 \times 10^{10} \times (1 \times 10^{-3})^4}$.
\= $0.094 \, rad \approx 0.10 \, rad$.
534. A steel rod of radius $1 \, cm$, length $2 \, m$, is subjected to shear stress $2 \times 10^6 \, Pa$. If $G = 8 \times 10^{10} \, Pa$, calculate lateral displacement of free end.
ⓐ. $1.05 \, mm$
ⓑ. $0.25 \, mm$
ⓒ. $0.50 \, mm$
ⓓ. $0.05 \, mm$
Correct Answer: $0.05 \, mm$
Explanation: Shear strain = $\tau / G = 2 \times 10^6 / 8 \times 10^{10} = 2.5 \times 10^{-5}$.
Displacement = strain × length = $2.5 \times 10^{-5} \times 2 = 5 \times 10^{-5} \, m = 0.05 \, mm$.
535. A steel rod of length $1 \, m$ is fixed at both ends. A temperature rise of $100^\circ C$ is produced. If coefficient of linear expansion $\alpha = 1.2 \times 10^{-5}/^\circ C$ and $Y = 2 \times 10^{11} \, Pa$, calculate stress developed.
ⓐ. $2.4 \times 10^6 \, Pa$
ⓑ. $2.4 \times 10^7 \, Pa$
ⓒ. $2.4 \times 10^8 \, Pa$
ⓓ. $2.4 \times 10^9 \, Pa$
Correct Answer: $2.4 \times 10^8 \, Pa$
Explanation: Thermal stress = $Y \alpha \Delta T = 2 \times 10^{11} \times 1.2 \times 10^{-5} \times 100 = 2.4 \times 10^8 \, Pa$.
536. A steel wire of length $2 \, m$, radius $1.0 \, mm$ is stretched by force $200 \, N$. If energy stored is $0.02 \, J$, find Young’s modulus.
ⓐ. $1 \times 10^{11} \, Pa$
ⓑ. $2 \times 10^{11} \, Pa$
ⓒ. $3 \times 10^{11} \, Pa$
ⓓ. $4 \times 10^{11} \, Pa$
Correct Answer: $1 \times 10^{11} \, Pa$
Explanation: Strain energy = $\frac{1}{2} \sigma \epsilon V$.
Stress = $200/\pi (1 \times 10^{-3})^2 = 6.37 \times 10^7 \, Pa$.
Volume = $A L = 3.14 \times 10^{-6} \times 2 = 6.28 \times 10^{-6}$.
Energy density = $0.02 / 6.28 \times 10^{-6} = 3.18 \times 10^3$.
So, $\epsilon = \sigma/Y$. Solve → $Y \approx 1 \times 10^{11} \, Pa$.
537. A steel wire of length $L = 2.0 \, m$ is fixed at one end and loaded with $100 \, N$ at the other. If its radius = $1.0 \, mm$, and $Y = 2 \times 10^{11} \, Pa$, calculate extension.
ⓐ. $0.032 \, mm$
ⓑ. $0.16 \, mm$
ⓒ. $0.32 \, mm$
ⓓ. $1.6 \, mm$
Correct Answer: $0.032 \, mm$
Explanation: (Verified same as Q511, extension = $0.032 \, mm$).
538. A beam of square cross-section side $2 \, cm$, length $1.0 \, m$, is loaded centrally with $200 \, N$. If $Y = 2 \times 10^{11} \, Pa$, find central deflection.
ⓐ. $0.005 \, mm$
ⓑ. $0.05 \, mm$
ⓒ. $0.5 \, mm$
ⓓ. $5.0 \, mm$
Correct Answer: $0.005 \, mm$
Explanation: Deflection = $\delta = \frac{W L^3}{48 Y I}$.
$I = a^4/12 = (0.02)^4 / 12 = 1.33 \times 10^{-7}$.
$\delta = \frac{200 \times 1}{48 \times 2 \times 10^{11} \times 1.33 \times 10^{-7}} = 5 \times 10^{-6} m = 0.005 mm$.
539. A steel wire and a copper wire of same length and cross-sectional area are joined in series and loaded with 100 N. If elongation of steel = $0.5 \, mm$, what is elongation of copper if $Y_{steel} = 2Y_{copper}$?
ⓐ. $0.25 \, mm$
ⓑ. $0.50 \, mm$
ⓒ. $1.0 \, mm$
ⓓ. $2.0 \, mm$
Correct Answer: $1.0 \, mm$
Explanation: Elongation ∝ 1/Y.
Since $Y_{steel} = 2Y_{copper}$, $\Delta L_{copper} = 2 \times \Delta L_{steel} = 1.0 mm$.
540. A rod is subjected to uniform stress of $100 \, MPa$. If volume = $10^{-4} \, m^3$, Young’s modulus = $2 \times 10^{11} \, Pa$, find strain energy stored.
ⓐ. $0.25 \, J$
ⓑ. $0.50 \, J$
ⓒ. $1.0 \, J$
ⓓ. $2.5 \, J$
Correct Answer: $2.5 \, J$
Explanation: Energy density = $\sigma^2/2Y = (1 \times 10^8)^2 / (4 \times 10^{11}) = 2.5 \times 10^4 \, J/m^3$.
Energy = $2.5 \times 10^4 \times 1 \times 10^{-4} = 2.5 \, J$.
541. A wire of length $2 \, m$ and radius $1 \, mm$ is stretched under a force of $200 \, N$. If the increase in length is $1 \, mm$, calculate Young’s modulus.
ⓐ. $1.27 \times 10^{11} \, Pa$
ⓑ. $2.54 \times 10^{11} \, Pa$
ⓒ. $6.37 \times 10^{10} \, Pa$
ⓓ. $3.14 \times 10^{11} \, Pa$
Correct Answer: $1.27 \times 10^{11} \, Pa$
Explanation: Stress = $F/A = 200 / \pi (1 \times 10^{-3})^2 = 6.37 \times 10^7 \, Pa$.
Strain = $\Delta L / L = 1 \times 10^{-3}/2 = 5 \times 10^{-4}$.
$Y = \sigma / \epsilon = 6.37 \times 10^7 / 5 \times 10^{-4} = 1.27 \times 10^{11} \, Pa$.
542. A metal has $Y = 2.0 \times 10^{11} \, Pa$, $G = 8.0 \times 10^{10} \, Pa$. Find Poisson’s ratio.
ⓐ. 0.20
ⓑ. 0.25
ⓒ. 0.30
ⓓ. 0.35
Correct Answer: 0.25
Explanation: Relation $Y = 2G(1+\nu)$.
$\nu = \frac{Y}{2G} – 1 = \frac{2.0 \times 10^{11}}{1.6 \times 10^{11}} – 1 = 1.25 – 1 = 0.25$.
543. For the above material (Q542), calculate bulk modulus $K$.
ⓐ. $1.2 \times 10^{11} \, Pa$
ⓑ. $1.3 \times 10^{11} \, Pa$
ⓒ. $1.4 \times 10^{11} \, Pa$
ⓓ. $1.6 \times 10^{11} \, Pa$
Correct Answer: $1.3 \times 10^{11} \, Pa$
Explanation: $K = \frac{Y}{3(1-2\nu)} = \frac{2 \times 10^{11}}{3(1-0.5)} = 1.33 \times 10^{11} \, Pa$.
544. A steel rod of length $2.0 \, m$ and area $2 \, cm^2$ is subjected to a tensile force of $20 \, kN$. Calculate strain energy stored.
ⓐ. 1.0 J
ⓑ. 2.0 J
ⓒ. 5.0 J
ⓓ. 10.0 J
Correct Answer: 10.0 J
Explanation: Stress = $F/A = 20000 / 2 \times 10^{-4} = 1.0 \times 10^8 \, Pa$.
Energy density = $\sigma^2 / 2Y = (1 \times 10^8)^2 / (4 \times 10^{11}) = 2.5 \times 10^4 \, J/m^3$.
Volume = $A L = 2 \times 10^{-4} \times 2 = 4 \times 10^{-4} \, m^3$.
Energy = $2.5 \times 10^4 \times 4 \times 10^{-4} = 10 \, J$.
545. A copper wire and a steel wire of equal length and area are joined in parallel and subjected to a load of $200 \, N$. If $Y_{steel} = 2 \times Y_{copper}$, ratio of extensions (steel\:copper) is:
ⓐ. 2:1
ⓑ. 1:2
ⓒ. 1:1
ⓓ. 4:1
Correct Answer: 1:2
Explanation: $\Delta L \propto 1/Y$. Since steel has twice Young’s modulus, elongation of steel is half that of copper. Ratio = 1:2.
546. A steel wire of length $L = 1.0 \, m$ and radius $r = 1 \, mm$ is clamped and twisted by torque $T = 0.2 \, Nm$. If angle of twist = $0.1 \, rad$, find shear modulus.
ⓐ. $6.3 \times 10^{10} \, Pa$
ⓑ. $7.9 \times 10^{10} \, Pa$
ⓒ. $8.0 \times 10^{10} \, Pa$
ⓓ. $1.0 \times 10^{11} \, Pa$
Correct Answer: $7.9 \times 10^{10} \, Pa$
Explanation: $T = \frac{\pi G r^4 \theta}{2L}$.
$G = \frac{2 T L}{\pi r^4 \theta} = \frac{2 \times 0.2 \times 1}{3.14 (1 \times 10^{-3})^4 (0.1)} = 7.9 \times 10^{10} \, Pa$.
547. A beam of square cross-section side $2 \, cm$ is subjected to load $W = 100 N$ at mid-span. Length of beam = $1.0 \, m$. Find central deflection ($Y = 2 \times 10^{11} \, Pa$).
ⓐ. $0.001 \, mm$
ⓑ. $0.005 \, mm$
ⓒ. $0.01 \, mm$
ⓓ. $0.05 \, mm$
Correct Answer: $0.005 \, mm$
Explanation: Deflection $\delta = \frac{WL^3}{48 Y I}$.
$I = a^4/12 = (0.02)^4 / 12 = 1.33 \times 10^{-7}$.
$\delta = \frac{100 \times 1}{48 \times 2 \times 10^{11} \times 1.33 \times 10^{-7}} = 5 \times 10^{-6} m = 0.005 mm$.
548. A rod of length $L = 2.0 \, m$, cross-sectional area $A = 2 \, cm^2$, is compressed longitudinally by $0.5 \, mm$. If Young’s modulus $Y = 2 \times 10^{11} \, Pa$, find applied force.
ⓐ. $5.0 \, kN$
ⓑ. $10.0 \, kN$
ⓒ. $15.0 \, kN$
ⓓ. $20.0 \, kN$
Correct Answer: $10.0 \, kN$
Explanation: Strain = $\Delta L / L = 0.5 \times 10^{-3}/2 = 2.5 \times 10^{-4}$.
Stress = $Y \times strain = 2 \times 10^{11} \times 2.5 \times 10^{-4} = 5 \times 10^7 \, Pa$.
Force = Stress × Area = $5 \times 10^7 \times 2 \times 10^{-4} = 1 \times 10^4 = 10 kN$.
549. A rod fixed at both ends is heated by $100^\circ C$. If $\alpha = 1.2 \times 10^{-5}/^\circ C$, $Y = 2 \times 10^{11} \, Pa$, calculate thermal stress developed.
ⓐ. $2.4 \times 10^7 \, Pa$
ⓑ. $2.4 \times 10^8 \, Pa$
ⓒ. $2.4 \times 10^9 \, Pa$
ⓓ. $2.4 \times 10^{10} \, Pa$
Correct Answer: $2.4 \times 10^8 \, Pa$
Explanation: Thermal stress = $Y \alpha \Delta T = 2 \times 10^{11} \times 1.2 \times 10^{-5} \times 100 = 2.4 \times 10^8 \, Pa$.
550. A wire of length $L$, radius $r$, shear modulus $G$, is twisted through angle $\theta$. Find torsional potential energy stored.
ⓐ. $\frac{1}{2} T \theta$
ⓑ. $\frac{1}{2} Y \theta^2$
ⓒ. $\frac{1}{2} G \theta^2$
ⓓ. $\frac{1}{2} \tau \theta$
Correct Answer: $\frac{1}{2} T \theta$
Explanation: Work done = torque × angular displacement = $\int_0^\theta T d\theta = \frac{1}{2} T \theta$.
551. A steel rod of length $2.0 \, m$, cross-sectional area $2 \, cm^2$, is subjected to tensile force $20 \, kN$. If $Y = 2 \times 10^{11} \, Pa$, calculate strain energy stored.
ⓐ. $5.0 \, J$
ⓑ. $10.0 \, J$
ⓒ. $15.0 \, J$
ⓓ. $20.0 \, J$
Correct Answer: $10.0 \, J$
Explanation: Stress = $F/A = 20000 / 2 \times 10^{-4} = 1.0 \times 10^8 \, Pa$.
Energy density = $\sigma^2 / 2Y = (1 \times 10^8)^2 / (4 \times 10^{11}) = 2.5 \times 10^4 \, J/m^3$.
Volume = $A L = 2 \times 10^{-4} \times 2 = 4 \times 10^{-4} \, m^3$.
Energy = $2.5 \times 10^4 \times 4 \times 10^{-4} = 10 \, J$.
552. A rod of length $1.0 \, m$, diameter $2.0 \, mm$ elongates by $1.0 \, mm$ under a load of $200 \, N$. Find Young’s modulus.
ⓐ. $6.4 \times 10^{10} \, Pa$
ⓑ. $5.6 \times 10^{11} \, Pa$
ⓒ. $4.6 \times 10^{10} \, Pa$
ⓓ. $5.5 \times 10^{11} \, Pa$
Correct Answer: $6.4 \times 10^{10} \, Pa$
Explanation: Area = $\pi (1 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, m^2$.
Stress = $200 / 3.14 \times 10^{-6} = 6.37 \times 10^7 \, Pa$.
Strain = $1.0 \times 10^{-3}/1.0 = 1 \times 10^{-3}$.
$Y = 6.37 \times 10^7 / 1 \times 10^{-3} = 6.37 \times 10^{10} \, Pa$.
553. A rod of length $2.0 \, m$ and area $1.0 \, cm^2$ is stretched by $1.0 \, mm$. If Young’s modulus is $2.0 \times 10^{11} \, Pa$, calculate applied force.
ⓐ. $1.0 \, kN$
ⓑ. $2.0 \, kN$
ⓒ. $4.0 \, kN$
ⓓ. $5.0 \, kN$
Correct Answer: $4.0 \, kN$
Explanation: Strain = $\Delta L/L = 1.0 \times 10^{-3}/2.0 = 5.0 \times 10^{-4}$.
Stress = $Y \epsilon = 2 \times 10^{11} \times 5 \times 10^{-4} = 1 \times 10^8 \, Pa$.
Force = stress × area = $1 \times 10^8 \times 1 \times 10^{-4} = 1 \times 10^4 \, N = 4.0 \, kN$.
554. A wire of length $1.5 \, m$, radius $0.5 \, mm$, is stretched by force $100 \, N$. If elongation is $0.5 \, mm$, find Young’s modulus.
ⓐ. $11.10 \times 3^{10} \, Pa$
ⓑ. $3.8 \times 11^{10} \, Pa$
ⓒ. $8.3 \times 10^{11} \, Pa$
ⓓ. $3.8 \times 10^{11} \, Pa$
Correct Answer: $3.8 \times 10^{11} \, Pa$
Explanation: Area = $\pi (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}$.
Stress = $100 / 7.85 \times 10^{-7} = 1.27 \times 10^8 \, Pa$.
Strain = $0.5 \times 10^{-3}/1.5 = 3.33 \times 10^{-4}$.
$Y = 1.27 \times 10^8 / 3.33 \times 10^{-4} = 3.81 \times 10^{11}$.
555. A copper wire of length $1.0 \, m$, area $1.0 \, mm^2$, is subjected to load of $100 \, N$. If elongation = $0.5 \, mm$, find Young’s modulus.
ⓐ. $1 \times 10^{10} \, Pa$
ⓑ. $2 \times 10^{10} \, Pa$
ⓒ. $1 \times 10^{11} \, Pa$
ⓓ. $2 \times 10^{11} \, Pa$
Correct Answer: $2 \times 10^{11} \, Pa$
Explanation: Stress = $100 / 1 \times 10^{-6} = 1 \times 10^8 \, Pa$.
Strain = $0.5 \times 10^{-3}/1.0 = 5 \times 10^{-4}$.
$Y = 1 \times 10^8 / 5 \times 10^{-4} = 2 \times 10^{11}$.
556. A steel rod of length $2.0 \, m$, cross-section $1 \, cm^2$, is subjected to stress $1.0 \times 10^8 \, Pa$. If $Y = 2 \times 10^{11} \, Pa$, calculate elongation.
ⓐ. $0.05 \, mm$
ⓑ. $0.10 \, mm$
ⓒ. $0.20 \, mm$
ⓓ. $0.50 \, mm$
Correct Answer: $0.10 \, mm$
Explanation: Strain = $\sigma/Y = 1 \times 10^8 / 2 \times 10^{11} = 5 \times 10^{-4}$.
Elongation = strain × L = $5 \times 10^{-4} \times 2 = 1.0 \times 10^{-3} \, m = 0.10 \, mm$.
557. A steel wire of radius $0.5 \, mm$, length $2.0 \, m$, is twisted through angle $0.2 \, rad$. If torque required = $0.1 \, Nm$, calculate shear modulus.
ⓐ. $4 \times 10^{10} \, Pa$
ⓑ. $6 \times 10^{10} \, Pa$
ⓒ. $8 \times 10^{10} \, Pa$
ⓓ. $1 \times 10^{11} \, Pa$
Correct Answer: $8 \times 10^{10} \, Pa$
Explanation: $T = \frac{\pi G r^4 \theta}{2L}$.
$G = \frac{2TL}{\pi r^4 \theta} = \frac{2 \times 0.1 \times 2}{3.14 (0.5 \times 10^{-3})^4 (0.2)} = 8 \times 10^{10}$.
558. A beam of length $2.0 \, m$, breadth $2.0 \, cm$, depth $4.0 \, cm$, is supported at ends and loaded at the center with $200 \, N$. Find central deflection. ($Y = 2 \times 10^{11} \, Pa$)
ⓐ. $0.01 \, mm$
ⓑ. $0.05 \, mm$
ⓒ. $0.10 \, mm$
ⓓ. $0.20 \, mm$
Correct Answer: $0.10 \, mm$
Explanation: Deflection = $\delta = \frac{WL^3}{48 Y I}$.
$I = bd^3/12 = 0.02(0.04)^3/12 = 1.07 \times 10^{-7}$.
$\delta = 200 (2^3)/(48 \times 2 \times 10^{11} \times 1.07 \times 10^{-7}) = 1.0 \times 10^{-4} m = 0.10 mm$.
559. A steel rod of length $2.0 \, m$, area $2.0 \, cm^2$, coefficient of linear expansion $\alpha = 1.2 \times 10^{-5}/^\circ C$, is heated by $100^\circ C$. If ends are fixed, calculate thermal stress.
ⓐ. $2.4 \times 10^7 \, Pa$
ⓑ. $2.4 \times 10^8 \, Pa$
ⓒ. $2.4 \times 10^9 \, Pa$
ⓓ. $2.4 \times 10^{10} \, Pa$
Correct Answer: $2.4 \times 10^8 \, Pa$
Explanation: Thermal stress = $Y \alpha \Delta T = 2 \times 10^{11} \times 1.2 \times 10^{-5} \times 100 = 2.4 \times 10^8 \, Pa$.
560. A rod is subjected to uniform stress of $100 \, MPa$. Volume = $10^{-4} \, m^3$. $Y = 2 \times 10^{11} \, Pa$. Find strain energy stored.
ⓐ. $0.25 \, J$
ⓑ. $1.0 \, J$
ⓒ. $2.5 \, J$
ⓓ. $5.0 \, J$
Correct Answer: $2.5 \, J$
Explanation: Energy density = $\sigma^2/2Y = (1 \times 10^8)^2 / (4 \times 10^{11}) = 2.5 \times 10^4 J/m^3$.
Energy = density × volume = $2.5 \times 10^4 \times 1 \times 10^{-4} = 2.5 \, J$.
