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

From a solid cube of side $7 \, cm$,a conical cavity of height $7 \, cm$ and radius $3 \, cm$ is hollowed out. Find the volume of the remaining solid (in $cm^3$).

$1\,m^3 = \dots \dots \dots cm^3$.

The volume of a cylinder is given by:

The ratio of radii of two cylinders is $3:4$ and the ratio of their heights is $4:5$. Then,the ratio of their volumes is ..........

How many silver coins,$1.75 \,cm$ in diameter and of thickness $2 \,mm$,must be melted to form a cuboid of dimensions $5.5 \,cm \times 10 \,cm \times 3.5 \,cm$? [Take $\pi = \frac{22}{7}$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module