A solid cylinder of diameter ‘D’ and height equal to ‘D’, and a s
| A solid cylinder of diameter ‘D’ and height equal to ‘D’, and a solid cube of side ‘2D’ are being sand-cast using the same material. Assuming there is no superheat in both the cases, the ratio of solidification time of the cylinder to the solidification time of the cube is ______.
A. 0.50
B. 0.8
C. 0.4
D. 0.25
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
Solidification Time: \(t = c{\left( {\frac{V}{A}} \right)^2}\)
Calculation:
For solid cylinder
\(V = \frac{\pi }{4}{D^2} \times D = \frac{{\pi {D^3}}}{4}\)
\(A = 2 \times \frac{\pi }{4}{D^2} + \pi D.D = \frac{{3\pi {D^2}}}{2}\)
\({\left( {\frac{V}{A}} \right)_{cylinder}} = \frac{{{D^3}}}{4} \times \frac{2}{{3{D^2}}} = \frac{D}{6}\)
For cube
V = (2D)3 = 8D3
A = 6 × (2D)2 = 24 D2
\({\left( {\frac{V}{A}} \right)_{cube}} = \frac{{8{D^3}}}{{24{D^2}}} = \frac{D}{3}\)
Solidification Time: \(t = c{\left( {\frac{V}{A}} \right)^2}\)
\(\frac{{{t_{cylinder}}}}{{{t_{cube}}}} = \frac{{{{\left( {\frac{D}{6}} \right)}^2}}}{{{{\left( {\frac{D}{3}} \right)}^2}}} = {\left( {\frac{1}{2}} \right)^2} = 0.25\)