A solid metallic sphere of radius 15 cm is melted and recast into spherical balls of radius 3 cm each. What is the ratio of the surface area of the original sphere and the sum of the surface areas of all balls?
A. 1 : 5
B. 5 : 27
C. 1 : 10
D. 3 : 40
Answer: Option A
Solution (By Examveda Team)
R3 = nr315 × 15 × 15 = n × 3 × 3 × 3
n = 125
(n = number of small spherical balls)
$$\eqalign{ & \frac{{{S_1}}}{{{S_2}}} = \frac{{{R^2}}}{{n{r^2}}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{{15 \times 15}}{{125 \times 3 \times 3}} \cr & {S_1}:{S_2} = 1:5 \cr} $$
Related Questions on Mensuration 3D
A. 1.057 cm3
B. 4.224 cm3
C. 1.056 cm3
D. 42.24 cm3
A sphere and a hemisphere have the same volume. The ratio of their curved surface area is:
A. $${2^{\frac{3}{2}}}:1$$
B. $${2^{\frac{2}{3}}}:1$$
C. $${4^{\frac{2}{3}}}:1$$
D. $${2^{\frac{1}{3}}}:1$$

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