Question

The Surface area of a solid metallic sphere is 616 cm2. It is melted and recast into a cone
of height 28cm. Find the diameter of the base of cone so formed. [Use π = 22/7]

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Answer 1

Answer:solution surface area of the sphere -4πr square616=4*22/7R square ={616*7}/88R square=4312/88R square =49r= √49 that is 7 radius of the sphere is 7 cmas sphere is melted and recast into a cone then, volume of the sphere is = volume of the cone4/3πr cube =1/33π33πr square h 4/3*22/7*7*7=1/3*22/7*r square*284/3*7*7*7=1/3*r square*28r square=49r=√ 49 that is 7 cmSo,radius of the cone is 7 cm and diameter= 2*radius=7*2=14cm Answer

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Answer 2

Given :-

Height of cone = 28 cm

Surface area of Sphere = 616 cm²

To Find :-

Diameter of the base of formed cone

SoluTion :-

$\sf {\ \because \ Surface \ area \ of \ Sphere = 4 \pi r^{2}}\\\\\\\sf {\ \therefore \ 616 \ cm^{2} = 4\pi r^{2}}\\\\\\\sf {\Rightarrow r^{2}=\frac{616 \times 7}{4 \times 22}}\\\\\\\sf {\Rightarrow r^{2}=\frac{4312}{88}}\\\\\\\sf {\Rightarrow r^{2}=49}\\\\\\\sf {\Rightarrow r = \sqrt{49} }\\\\\\\sf {\Rightarrow r = 7 \ cm}$

Let the radius of cone be 1r

According to the Question,

Volume of Cone = Volume of Sphere

$\rm {Volume \ of \ cone = \frac{1}{3} \pi (r^{2}) h}\\\\\\\sf {V_{1}=\frac{1}{3} \pi (r_{1})^{2} \times 28 \ cm^{3} \ ....(1)}\\\\\\\\\rm {Volume \ of \ sphere \ V_{2}=\frac{4}{3} \pi r^{3}}\\\\\\\sf {V_{2}=\frac{4}{3} \pi \ 7^{3} \ cm^{3} \ ....(2)}\\\\\\\\\\\bf {(1)=(2)}\\\\\bf {V_{1}=V_{2}}\\\\\\\sf {\implies \frac{1}{3} \pi (r_{1})^{2} \times 28 = \frac{4}{3} \pi \ 7^{3}}\\\\\\\sf {(r_{1})^{2}=49}\\\\\\\sf {r_{1}=7 \ cm}$

Radius of Cone = 7 cm

Diameter of Cone = 7 × 2 = 14 cm