Question

The surface area of a solid sphere is increased by 12% without changing its shape. Find the percentage increase in its:

1. radius
2. volume

Answer 1

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%