Question

if the radius of a sphere is doubled.then what is the increase in percentage of the surface of the sphere​

Answer 1

Answer:

Let

initial surface area = 4↓0

final radius = 2r↓0

final surface area = 4π(2r

0

)

2

=4.4πr

0

2

=16πr

0

2

increase in surface area = 16πr

0

2

−4πr

0

2

=12πr

0

2

percent increase in surface area =

4πr

0

2

12πr

0

2

×100=300%