Question

What is the reduction in the surface area of a new sphere by reducing the radius of the
sphere by 10%?
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Answer 1

Answer:

Let radius of sphere be r

Volume of sphere=(4/3)πr^3

Radius reduced by 10%

Now volume=(4/3)π(90r/100)^3

=(4/3)π(729r^3/1000)

Difference=(4/3)πr^3(1–729/1000)

=(4/3)πr^3(271/1000)

271πr^3/750

For (4/3)πr^3 the difference is 271πr^3/750

For 100

100(271πr^3/750)/(4/3)πr^3

Upon simplification,27.1Let radius of sphere be r

Volume of sphere=(4/3)πr^3

Radius reduced by 10%

Now volume=(4/3)π(90r/100)^3

=(4/3)π(729r^3/1000)

Difference=(4/3)πr^3(1–729/1000)

=(4/3)πr^3(271/1000)

271πr^3/750

For (4/3)πr^3 the difference is 271πr^3/750

For 100.

100(271πr^3/750)/(4/3)πr^3

Upon simplification,27.1%

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Answer 2

Answer:

Hope it's helpful to you.