If the radius of the spherical base is decreased by $20 %$ , what will be decrease in volume in percent?

Open in App

Solution

Decrease in volume (in percent)
$= [{(1 - \frac{20}{100})}^{3} \sim 1] \times 100$
$= [1 - {(\frac{4}{5})}^{3}] \times 100$
$= \frac{61}{125} \times 100$
$= 48.8$ %

Suggest Corrections

Similar questions

Q. A right circular cone has a base radius of

r

and a height of

h .

If the radius decrease by

20

percent and the height increase by

10

percent, which of the following is the resulting percent change in the volume of the cone?

Q. If the water level of River – A in July is decreased by 20 percent, then what will be the water level of River – A in July?

Q. If the radius of a right circular cylinder is decreased by

50 %

and its height is increased by

60 %

, its volume will be decreased by:

If the radius and height of a cylinder are increased by $10 %$ and decreased by $5 %$ respectively, then what will be the percentage change in volume? [4 MARKS]

Q. If the diameter of wire is decreased by

10 %

, by how much percent approximately will the length be increased to keep the volume constant?

Join BYJU'S Learning Program

If the radius of the spherical base is decreased by 20%, what will be decrease in volume in percent?

Decrease in volume (in percent)=[(1−20100)3∼1]×100=[1−(45)3]×100=61125×100=48.8%

If the radius of the spherical base is decreased by $20 %$ , what will be decrease in volume in percent?

Decrease in volume (in percent)
$= [{(1 - \frac{20}{100})}^{3} \sim 1] \times 100$
$= [1 - {(\frac{4}{5})}^{3}] \times 100$
$= \frac{61}{125} \times 100$
$= 48.8$ %