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A car manufacturing company is sponsoring a basketball match by painting its logo on the basketball at the rate of ₹3/ cm 2 . There are two identical basketballs of diameter 14 cm which are to be painted. What will be the total cost (in ₹) of painting the balls for the company?
1848
3586
4928
3696
3696
Since the area of a sphere is 4πr2 and r=7, the area is 4×722imes49=616 cm². For two balls, area is 1232 cm². Multiplying by 3, the result must be a multiple of 3; check the sum of digits of 3696 (3+6+9+6=24), which is divisible by 3.
Diameter of basketball d = 14 cm, number of basketballs = 2, rate of painting = ₹3/cm².
Total Cost=2×(4×π×r2)×Rate
Since the area of a sphere is 4πr2 and r=7, the area is 4×722imes49=616 cm². For two balls, area is 1232 cm². Multiplying by 3, the result must be a multiple of 3; check the sum of digits of 3696 (3+6+9+6=24), which is divisible by 3.
Students often use the surface area of a circle πr2 instead of the sphere 4πr2 or forget to multiply by 2 for both basketballs.
Calculate Radius
The radius r is half of the diameter d. Given d=14 cm, r=14/2=7 cm.
r=7 cm
Calculate Surface Area per Ball
The surface area of one spherical basketball is 4πr2. Using π=22/7 and r=7, we get 4×722×7×7=616 cm².
SA=4×722×72=616 cm2
Calculate Total Area and Cost
Total area for two balls is 2×616=1232 cm². Multiplying by the rate of ₹3/cm², the total cost is 1232×3=3696.
Cost=1232×3=3696
D is correct because the total surface area of two basketballs is 1232 cm² and at ₹3/cm², the total cost is ₹3696.
This concept is essential for problems involving volume and cost estimation in physical quantities like painting or gold plating, where surface area is the primary constraint.
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