A top is made by joining a hemisphere to the | vedclass

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

Question

(C) The top consists of a cone and a hemisphere joined at their bases. The radius of both the cone and the hemisphere is $r = 2.8 \, cm$. The height o

Vedclass · Accepted Answer

$98.56$

Vedclass · Answer

(C) The top consists of a cone and a hemisphere joined at their bases. The radius of both the cone and the hemisphere is $r = 2.8 \, cm$. The height of the cone is $h = 6.4 \, cm$.Volume of the cone $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} 	imes \frac{22}{7} 	imes (2.8)^2 	imes 6.4 = \frac{1}{3} 	imes \frac{22}{7} 	imes 7.84 	imes 6.4 \approx 52.53 \, cm^3$.Volume of the hemisphere $V_{hemi} = \frac{2}{3} \pi r^3 = \frac{2}{3} 	imes \frac{22}{7} 	imes (2.8)^3 = \frac{2}{3} 	imes \frac{22}{7} 	imes 21.952 \approx 46.03 \, cm^3$.Total volume $V = V_{cone} + V_{hemi} = 52.53 + 46.03 = 98.56 \, cm^3$.

$A$ 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 $cm^3$.

Similar Questions

The base of a cone is hemispherical. The radius of the cone is $15 \, cm$ and the total height of the solid is $55 \, cm$. Find the volume of this solid (in $cm^3$).

Write 'True' or 'False' and justify your answer:
$A$ spherical steel ball is melted to make eight new identical balls. Then,the radius of each new ball is $\frac{1}{8}$th the radius of the original ball.

Surface area of a sphere $= \dots \dots \dots \dots .$

Water flows through a cylindrical pipe,whose inner radius is $1 \,cm$,at the rate of $80 \,cm/sec$ into an empty cylindrical tank,the radius of whose base is $40 \,cm$. What is the rise of water level in the tank in half an hour? (in $cm$)

The volume of a frustum of a cone is equal to $\ldots \ldots \ldots . .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module