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A solid metallic sphere of radius 4 cm is melted and recast into 64 identical spheres. What is the ratio of the surface area of the original sphere to the total surface area of 5 smaller spheres so formed?
32 : 3
14 : 5
16 : 15
16 : 5
16 : 5
When a solid metallic sphere is melted and recast, the total volume remains constant, while the surface area increases as the material is divided into smaller units. For a sphere of radius R converted into n smaller spheres of radius r, the conservation of volume dictates R3=n├Чr3.
When a solid metallic sphere is melted and recast, the total volume remains constant, while the surface area increases as the material is divided into smaller units. For a sphere of radius R converted into n smaller spheres of radius r, the conservation of volume dictates R3=n├Чr3.
Think of it like play-dough: you have one large ball, and if you divide it into many smaller ones, the total amount of dough is the same, but the total exposed surface area increases significantly.
V=34тАЛ╧Аr3 тАФ Volume of a sphere
A=4╧Аr2 тАФ Surface area of a sphere
The principle of conservation of matter (volume in this case) ensures that the initial volume V=34тАЛ╧АR3 equals the sum of the volumes of the n resulting spheres. By finding the new radius r=R├Ч(n)тИТ1/3, we calculate the surface area S=4╧Аr2. The total surface area of k smaller spheres is k├Ч4╧Аr2.
Volume remains invariant during the melting and recasting process.
The total surface area of the resulting spheres is inversely proportional to the cube root of the number of spheres.
Smaller fragments always possess a higher surface-area-to-volume ratio than the larger parent object.
High surface area increases reaction rates in chemical processes
Enhanced cooling efficiency in metallic casting
Increased surface area can lead to faster oxidation or corrosion of metallic particles
Powder metallurgy for creating complex components
Catalytic surfaces in chemical reactors
Radius of large sphere R=4 cm. Number of small spheres n=64. Radius of small sphere r=4/(64)1/3=4/4=1 cm.
Surface area of original sphere A1тАЛ=4╧А(4)2=64╧А. Total surface area of 5 small spheres A2тАЛ=5├Ч4╧А(1)2=20╧А. Ratio 64╧А/20╧А=16:5.
Option A is incorrect as it assumes a different geometric scaling.
Option B is incorrect due to calculation error in identifying the ratio.
D is correct тАФ The ratio of the original surface area (64╧А) to the combined surface area of five smaller spheres (20╧А) is 16:5.
Always equate the total volumes first to find the new radius (r) before calculating the surface area ratios; remember that volume scales by r3 while surface area scales by r2.
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