Question

If v1, v2 and v3 be the volumes of a right circular cone, a sphere and a right circular cylinder having the same radius and same height, then

Answer 1

v1 = volume of right circular cone

v2 = volume of sphere

v3 = volume of right circular cylinder

a/c to question, right circular cone, sphere and right circular cylinder having same radius and same height.
Let radius = R and height = H

so, v1 = volume of cone = πR²H/3

v2 = volume of sphere = 4/3 πR³ = 4/3πR² × R

[ as height of all shapes are same. so, diameter of sphere = height
so, R = H/2 ]
so, v2 = 4/3 πR² × H/2 = 2/3 πR²H

v3 = volume of cylinder = πR²H


hence, it is clear that,
2v1 = v2 = 2v3/3

or, 6v1 = 3v2 = 2v3

hence, v1/1 = v2/2 = v3/3

Answer 2

Answer:

1:2:3

Step-by-step explanation:

$V_{1}=volume\:of\:right\:circular\;cone \\V_{1}=\frac{1}{3}\pi\:r^2h \\\\V_{2}=volume\:of\:the\:sphere\\V_{2}=\frac{4}{3}\pi\:r^3\\\\\\V_{3}=volume\:of\:the\:right\:circular\;cylinder\\V_{3}=\pi\:r^2h$

since the height of the cone and cylinder are equal to height of the sphere,

we get h=2r

Now,

$V_{1}:V_{2}:V_{3}=\frac{1}{3}\pi\:r^2\:h:\frac{4}{3}\pi\:r^3:\pi\:r^2\:h\\\\V_{1}:V_{2}:V_{3}=\frac{1}{3}\pi\:r^2(2r):\frac{4}{3}\pi\:r^3:\pi\:r^2(2r)\\\\simplifying\:we\:get, \\\\V_{1}:V_{2}:V_{3}=\frac{2}{3}\pi\:r^3:\frac{4}{3}\pi\:r^3:2\pi\:r^3 \\\\\\V_{1}:V_{2}:V_{3}=2\pi\:r^3:4\pi\:r^3:6\pi\:r^3\\\\V_{1}:V_{2}:V_{3}=2:4:6 \\\\V_{1}:V_{2}:V_{3}=1:2:3$