If a sphere of radius r is divided into 4 ide | vedclass

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

Question

(C) When a sphere of radius $r$ is divided into $4$ identical parts (by cutting it along two perpendicular planes passing through the center),each par

Vedclass · Accepted Answer

$8 \pi r^{2}$ square units

Vedclass · Answer

(C) When a sphere of radius $r$ is divided into $4$ identical parts (by cutting it along two perpendicular planes passing through the center),each part is a spherical wedge (specifically,a spherical quadrant).$1$. The curved surface area of the original sphere is $4 \pi r^{2}$. Since the sphere is divided into $4$ equal parts,the curved surface area of each part is $\frac{1}{4} 	imes 4 \pi r^{2} = \pi r^{2}$.$2$. The total curved surface area of all $4$ parts is $4 	imes \pi r^{2} = 4 \pi r^{2}$.$3$. Each of the $4$ parts has two flat semi-circular faces created by the cuts. Since there are $4$ parts,there are $4 	imes 2 = 8$ semi-circular faces in total.$4$. The area of one semi-circular face is $\frac{1}{2} \pi r^{2}$.$5$. The total area of these $8$ flat faces is $8 	imes \frac{1}{2} \pi r^{2} = 4 \pi r^{2}$.$6$. The total surface area of the $4$ parts is the sum of the total curved surface area and the total flat surface area:Total Surface Area $= 4 \pi r^{2} + 4 \pi r^{2} = 8 \pi r^{2}$ square units.

If a sphere of radius $r$ is divided into $4$ identical parts,then the total surface area of the $4$ parts is

Similar Questions

The area of the iron sheet required to prepare a cone $24 \text{ cm}$ high with base radius $7 \text{ cm}$ is .......... $\text{cm}^2$ (Take $\pi = \frac{22}{7}$)

$A$ cylindrical iron rod is $70\, cm$ long. The diameter of its end portion is $2\, cm$. What is its weight,reckoning a $cm^3$ of iron to weigh $10\, grams$? (in $kg$)

$3$ spherical balls of radii $1\, cm$,$2\, cm$,and $3\, cm$ are melted to form a single spherical ball. In the process,the loss of material is $25\%$. The radius of the new ball is.........$cm$.

$A$ ball of lead,$4\, cm$ in diameter,is covered with gold. If the volume of the gold and lead are equal,then the thickness of gold is approximately (in $cm$).
$[$ Given $\sqrt[3]{2} = 1.259 ]$

$A$ solid consists of a circular cylinder with an exact fitting right circular cone placed on the top. The height of the cone is $h$. If the total volume of the solid is three times the volume of the cone,then the height of the cylinder is

Vedclass Test Series

Exam Paper Generator

Online Exam Module