Test Index

UPSC CDS 2 2021 Math Paper

Directions (Q.Nos. 46-48) Consider the following for the next three questions that follow.

A solid consisting of a right circular cone of radius

x

and height

2x

standing on a hemisphere of radius

x

(take

\pi = \frac{22}{7}

)

Question : 46 of 100

Marks: +1, -0

The volume of the solid is equal to that of a

Solution: 👈: Video Solution

Given dimensions :

Cone : radius

(r_1)=x

height

(h)=2x

Hemisphere : radius

(r_2)=x

\therefore

Volume of the solid = Volume of cone + Volume of hemisphere

\;\; =\;\frac{1}{3} \pi r_1^2 h + \frac{2}{3} \pi r_2^3

\;\; =\;\frac{1}{3} \pi \cdot x^2 \cdot (2x) + \frac{2}{3} \pi \cdot (x)^3

\;\; =\;\frac{2}{3} \pi x^3 + \frac{2}{3} \pi x^3 = \frac{4}{3} \pi x^3

Hence, the volume of the solid is equal to that of a sphere of radius

x

Go to Question:

Prev Question

Next Question