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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?
Answer & Solution
Correct Answer:
Option
A
R3 = nr3
15 × 15 × 15 = n × 3 × 3 × 3
n = 125
(n = number of small spherical balls)
$$\eqalign{ & \frac{{{S_1}}}{{{S_2}}} = \frac{{{R^2}}}{{n{r^2}}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{{15 \times 15}}{{125 \times 3 \times 3}} \cr & {S_1}:{S_2} = 1:5 \cr} $$
15 × 15 × 15 = n × 3 × 3 × 3
n = 125
(n = number of small spherical balls)
$$\eqalign{ & \frac{{{S_1}}}{{{S_2}}} = \frac{{{R^2}}}{{n{r^2}}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{{15 \times 15}}{{125 \times 3 \times 3}} \cr & {S_1}:{S_2} = 1:5 \cr} $$
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