The base of a cone with radius 15, cm and sla | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Radius of the cone $=$ radius of the hemisphere $= r = 15\, cm$ and slant height of the cone $l = 25\, cm$.For a cone,$l^{2} = r^{2} + h^{2}$.$	h

Vedclass · Accepted Answer

$11775$

Vedclass · Answer

(B) Radius of the cone $=$ radius of the hemisphere $= r = 15\, cm$ and slant height of the cone $l = 25\, cm$.For a cone,$l^{2} = r^{2} + h^{2}$.$	herefore h^{2} = l^{2} - r^{2} = 25^{2} - 15^{2} = 625 - 225 = 400 = (20)^{2}$.So,$h = 20\, cm$.Volume of the combined solid $=$ Volume of the cone $+$ Volume of the hemisphere.Volume $= \frac{1}{3} \pi r^{2} h + \frac{2}{3} \pi r^{3} = \frac{1}{3} \pi r^{2} (h + 2r)$.Substituting the values: Volume $= \frac{1}{3} 	imes 3.14 	imes 15 	imes 15 	imes (20 + 2 	imes 15)$.Volume $= 3.14 	imes 5 	imes 15 	imes (20 + 30) = 3.14 	imes 75 	imes 50 = 3.14 	imes 3750 = 11775\, cm^{3}$.Thus,the volume of the given solid is $11775\, cm^{3}$.

The base of a cone with radius $15\, cm$ and slant height $25\, cm$ is hemispherical. Find the volume of this solid. $(\pi=3.14)$ (in $cm^{3}$)

Similar Questions

$A$ cylinder is closed at both the ends by hemispheres. The radius of the cylinder is $20 \, cm$ and the total height of the article is $80 \, cm$. Find the total surface area of this combined article. $(\pi = 3.14)$ (in $cm^2$)

The volume of a cylinder is given by:

How many balls of radius $1.4 \, cm$ can be produced by melting a metallic sphere of radius $42 \, cm$?

The area of the base of a cylinder is $616\,cm^{2}$. Then,the radius of the base of the cylinder is $\ldots \ldots \ldots \ldots cm.$

$A$ top is made by joining a hemisphere to the base of a cone. The radius and the height of the conical part are $2.8 \, cm$ and $6.4 \, cm$ respectively. Find the volume of the top in $cm^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module