A metal cylinder of radius 8 cm and height 12 cm is melted and made into two solid spheres such that the radius of the first one is half that of the second. What is the ratio of the surface area of the second sphere to the total surface area of the cylinder?
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1 : 3
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1 : 2
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2 : 3
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4 :5
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3 : 4
D
Correct answer
Explanation
The cylinder volume is pi*r^2*h = pi*64*12 = 768*pi. The two spheres have radii r and 2r, so their volumes are (4/3)pi*r^3 and (4/3)*pi(8r^3) = (32/3)*pi*r^3. Summing these gives (36/3)*pi*r^3 = 12*pi*r^3 = 768*pi, so r^3 = 64, r = 4. The second sphere radius is 8. Surface area of second sphere is 4*pi*8^2 = 256*pi. Total surface area of cylinder is 2*pi*r(r+h) = 2*pi*8(8+12) = 320*pi. Ratio is 256/320 = 4/5.