Question

If radius of the sphere is 14cm. If the radius is increased by 50%find how much % is the volume increased?????

Plezzzzz answer fast I have exam
I want correct ans

Answer 1

of the sphere :-
original radius = 14 cm
volume = 4/3 pi (r)^3
= 4/3 x 22/7 x ( 14 )^3
= 11498.667

now
New radius = 14(1+50/100)
= 14(150/100)
= 21 cm

New volume = 4/3 pi (r)^3
= 4/3 x 22/7 x (21)^3
= 38808

increase in volume = 38808 - 11498.667
= 27309.333

percentage increase
=increase volume/ original volume x 100
= 27309.333/11498.667 x 100
= 237.49999 %

Answer 2

Answer:

237.5%

Step-by-step explanation:

Radius = 14 cm

Volume of sphere = $\frac{4}{3} \pi r^3$

                             = $\frac{4}{3} \pi (14)^3$

The radius is increased by 50%

So, new radius = $14+50\% \times 14 =14+\frac{50}{100} \times 14 = 21 cm$

New Volume = $\frac{4}{3} \pi (21)^3$

Change in volume = $\frac{\text{new volume - original volume }}{\text{original volume }} \times 100$

Change in volume = $\frac{\frac{4}{3} \pi (21)^3 - \frac{4}{3} \pi (14)^3}{\frac{4}{3} \pi (14)^3} \times 100$

Change in volume = $\frac{\frac{4}{3} \pi ((21)^3 - (14)^3)}{\frac{4}{3} \pi (14)^3} \times 100$

Change in volume = $\frac{((21)^3- (14)^3)}{(14)^3} \times 100$

Change in volume = $237.5$

So, Change in volume is 237.5%