Q1: The sphericity of a cylinder of 1 mm diameter and length 3 mm is

ANS:A - 0.9

Sphericity (Ψ\PsiΨ) is a measure of how closely the shape of an object approaches that of a perfect sphere. It is defined as the ratio of the surface area of a sphere (having the same volume as the object) to the surface area of the object. For a cylinder with diameter ddd and length LLL, the sphericity Ψ\PsiΨ can be calculated using the formula: Ψ=Surface area of sphere with same volume as the cylinderSurface area of the cylinder\Psi = \frac{\text{Surface area of sphere with same volume as the cylinder}}{\text{Surface area of the cylinder}}Ψ=Surface area of the cylinderSurface area of sphere with same volume as the cylinder​ First, calculate the volume of the cylinder: Vcylinder=π(d2)2L=π(1 mm2)2×3 mm=π×0.25×3 mm3=0.75π mm3V_{\text{cylinder}} = \pi \left(\frac{d}{2}\right)^2 L = \pi \left(\frac{1 \text{ mm}}{2}\right)^2 \times 3 \text{ mm} = \pi \times 0.25 \times 3 \text{ mm}^3 = 0.75\pi \text{ mm}^3Vcylinder​=π(2d​)2L=π(21 mm​)2×3 mm=π×0.25×3 mm3=0.75π mm3 The radius rrr of a sphere with the same volume is found from: Vsphere=43πr3=0.75π mm3  ⟹  r3=0.75×34=0.5625  ⟹  r=0.56253≈0.825 mmV_{\text{sphere}} = \frac{4}{3}\pi r^3 = 0.75\pi \text{ mm}^3 \implies r^3 = 0.75 \times \frac{3}{4} = 0.5625 \implies r = \sqrt[3]{0.5625} \approx 0.825 \text{ mm}Vsphere​=34​πr3=0.75π mm3⟹r3=0.75×43​=0.5625⟹r=30.5625​≈0.825 mm Next, calculate the surface areas:

1. Surface area of the sphere:

Asphere=4πr2=4π(0.825 mm)2≈4π×0.6806=2.722π mm2A_{\text{sphere}} = 4\pi r^2 = 4\pi (0.825 \text{ mm})^2 \approx 4\pi \times 0.6806 = 2.722\pi \text{ mm}^2Asphere​=4πr2=4π(0.825 mm)2≈4π×0.6806=2.722π mm2

2. Surface area of the cylinder:

Acylinder=2π(d2)2+πdL=2π(1 mm2)×3 mm+2π(1 mm2)2A_{\text{cylinder}} = 2\pi \left(\frac{d}{2}\right)^2 + \pi d L = 2\pi \left(\frac{1 \text{ mm}}{2}\right) \times 3 \text{ mm} + 2\pi \left(\frac{1 \text{ mm}}{2}\right)^2Acylinder​=2π(2d​)2+πdL=2π(21 mm​)×3 mm+2π(21 mm​)2 Acylinder=2π×0.5×3+2π×0.25A_{\text{cylinder}} = 2\pi \times 0.5 \times 3 + 2\pi \times 0.25Acylinder​=2π×0.5×3+2π×0.25 Acylinder=3π+0.5π=3.5π mm2A_{\text{cylinder}} = 3\pi + 0.5\pi = 3.5\pi \text{ mm}^2Acylinder​=3π+0.5π=3.5π mm2 Finally, calculate the sphericity: Ψ=AsphereAcylinder=2.722π3.5π=2.7223.5≈0.777\Psi = \frac{A_{\text{sphere}}}{A_{\text{cylinder}}} = \frac{2.722\pi}{3.5\pi} = \frac{2.722}{3.5} \approx 0.777Ψ=Acylinder​Asphere​​=3.5π2.722π​=3.52.722​≈0.777 So, the sphericity of the cylinder is approximately 0.78.