the radius of a hemisphere is increased by 50%.find the % increase in its volume
Answers
Answer:
237.5%.
Step-by-step explanation:
Let the radius be x
Volume of sphere = \frac{4}{3} \pi r ^3
3
4
πr
3
Where r is the radius
So, radius of given sphere = \frac{4}{3} \pi x^3
3
4
πx
3
Now radius of a sphere is increased 50%
So, New radius = \frac{50}{100}x+x =\frac{150x}{100}
100
50
x+x=
100
150x
So, New Volume = \frac{4}{3} \pi (\frac{150x}{100})^3
3
4
π(
100
150x
)
3
= \frac{4}{3} \pi (\frac{3x}{2})^3
3
4
π(
2
3x
)
3
= \frac{9}{2} \pi x^3
2
9
πx
3
Change in volume = New volume - Original volume
= \frac{9}{2} \pi x^3-\frac{4}{3} \pi x ^3
2
9
πx
3
−
3
4
πx
3
= \frac{19}{6} \pi x^3
6
19
πx
3
So, Increase in volume in percent =\frac{\text{Change in volume}}{\text{original volume}} \times 100
original volume
Change in volume
×100
=\frac{frac{19}{6} \pi x^3}{\frac{4}{3} \pi x ^3} \times 100
3
4
πx
3
frac196πx
3
×100
=237.5\%237.5%
Hence increase volume in percent is 237.5%.