Advertisements
Advertisements
प्रश्न
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. What is the ratio of their volumes?
उत्तर
Let r be the radius of the base.
And h is the height.
Here, h = r.
Now,
The ratio of their volumes will be
Volume of cone: volume of hemisphere: volume of a cylinder
`1/3pir^2 h : 2/3pir^3 : pir^2 h `
`V_1 :V_2 :V_3 = 1/3pir^3 : 2/3 pir^3 : pir^3`
Hence , `V_1 : V_2 :V_3 = 1:2:3`
APPEARS IN
संबंधित प्रश्न
A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends, as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container. at the rate of 21 per litre. [use π=22/7]
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs.20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs.8 per 100 cm2. [Take π = 3.14]
Derive the formula for the volume of the frustum of a cone.
A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out.
If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm respectively. What is the curved surface area of the frustum?
The diameters of the ends of a frustum of a cone are 32 cm and 20 cm. If its slant height is 10 cm, then its lateral surface area is
A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.
By melting a solid sphere of radius 5 cm a solid right circular cone of the same circular base radius is made. The height of cone is ______.
A shuttle cock used for playing badminton has the shape of the combination of ______.
The volume of the frustum of a cone is `1/3 pih[r_1^2 + r_2^2 - r_1r_2]`, where h is vertical height of the frustum and r1, r2 are the radii of the ends.
