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Question
The surface area of a solid sphere is increased by 12% without changing its shape. Find the percentage increase in its:
- radius
- volume
Solution
Let the radius of the sphere be 'r'.
Total surface area the sphere, S = 4πr2
New surface area of the sphere, S’
= `4pir^2 + 21/100 xx 4pir^2`
= `121/100 4pir^2`
i. Let the new radius be r1
`S^I = 4pir_1^2`
`S^I = 121/100 4pir^2`
`=> 4pir_1^2 = 121/100 4pir^2`
`=> r_1^2 = 121/100r^2`
`=> r_1 = 11/10r`
`=> r_1 = r + r/10`
`=> r_1 - r = r/10`
`=>` Change in radius = `r/10`
Percentage change in radius = `"Change in radius"/"Original in radius" xx 100`
= `(r/10)/r xx 100`
= 10
Percentage change in radius = 10%
ii. Let the volume of the sphere be V
Let the new volume of the sphere be V'.
`V = 4/3pir^3`
`V^I = 4/3pir_1^3`
`=> V^I = 4/3pi((11r)/10)^3`
`=> V^I = 4/3pi1331/1000 r^3`
`=> V^I = 4/3pir^3 1331/1000 `
`=> V^I = 1331/1000 V`
`=> V^I = V + 1331/1000 V`
`=> V^I - V = 331/1000 V`
∴ Change in volume = `331/1000 V `
Percentage change in volume = `"Change in volume"/"Original volume" xx 100`
= `(331/1000V)/V xx 100`
= `331/10`
= 33.1
Percentage change in volume = 33.1%
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