The base of a cone with radius 0.6 m and hei | vedclass

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

Question

(A) The solid consists of a cone and a hemisphere joined at their bases.Radius of the cone $(r)$ = $0.6 \ m$.Height of the cone $(h)$ = $1.6 \ m$.Radi

Vedclass · Accepted Answer

$1.056$

Vedclass · Answer

(A) The solid consists of a cone and a hemisphere joined at their bases.Radius of the cone $(r)$ = $0.6 \ m$.Height of the cone $(h)$ = $1.6 \ m$.Radius of the hemisphere $(r)$ = $0.6 \ m$.Volume of the cone = $\frac{1}{3} \pi r^{2} h = \frac{1}{3} 	imes \pi 	imes (0.6)^{2} 	imes 1.6 = 0.192 \pi \ m^{3}$.Volume of the hemisphere = $\frac{2}{3} \pi r^{3} = \frac{2}{3} 	imes \pi 	imes (0.6)^{3} = 0.144 \pi \ m^{3}$.Total volume = $0.192 \pi + 0.144 \pi = 0.336 \pi \ m^{3}$.Using $\pi \approx 3.14159$,Total volume $\approx 0.336 	imes 3.14159 \approx 1.05557 \ m^{3}$,which rounds to $1.056 \ m^{3}$.

The base of a cone with radius $0.6 \ m$ and height $1.6 \ m$ is hemispherical. Find the volume of this combined solid in $m^{3}$.

Similar Questions

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

Total surface area of a cylinder $= \dots$

Total surface area of a cube open at one end $= \dots$

If a marble of radius $2.1 \, cm$ is put into a cylindrical cup full of water of radius $5 \, cm$ and height $6 \, cm$,then how much water flows out of the cylindrical cup? (in $cm^3$)

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 $cm^3$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module