Question

a sphere is made by reshaping a solid right circular cone of the height 14 CM and diameter 10 CM find the radius of the sphere thus formed


it's very very very urgent​

Answer 1

$\huge \frak \red{Question : }$

→ A sphere is made by reshaping a solid right circular cone of the height 14 CM and diameter 10 CM find the radius of the sphere thus formed.

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$\leadsto\huge\frak\red{Answer} :$

GIVEN:

. Cone → h = 14 cm ; r = 5 cm

When an object is reshaped, the volume remains the same ,

. Volume of cone = Volume of sphere

$\implies \large {\frac{1}{3} \pi {r}^{2} h = \frac{4}{3} \pi {R}^{3} }$

$\implies \large {\frac{1}{3} \pi( {r}^{2} h) = \frac{1}{3}\pi (4 {R}^{3} )}$

. ❌ Cancelling out 1/3π , ❌

$\implies \large{ {5}^{2} \times 14 = 4 {R}^{3} }$

$\implies \large {R}^{3} = \frac{7}{2} \times 25$

$\implies \large \: {R}^{3} = 875$

$\huge \boxed {\leadsto \: {R}^{3} = \sqrt{875} \: cm}$

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. hope it helps....!

Answer 2

Step-by-step explanation:

⇝Answer:

GIVEN:

. Cone → h = 14 cm ; r = 5 cm

When an object is reshaped, the volume remains the same ,

. Volume of cone = Volume of sphere

\implies \large {\frac{1}{3} \pi {r}^{2} h = \frac{4}{3} \pi {R}^{3} }⟹31πr2h=34πR3

\implies \large {\frac{1}{3} \pi( {r}^{2} h) = \frac{1}{3}\pi (4 {R}^{3} )}⟹31π(r2h)=31π(4R3)

. ❌ Cancelling out 1/3π , ❌

\implies \large{ {5}^{2} \times 14 = 4 {R}^{3} }⟹52×14=4R3

\implies \large {R}^{3} = \frac{7}{2} \times 25⟹R3=27×25

\implies \large \: {R}^{3} = 175/2⟹R3=87.5

\huge \boxed {\leadsto \: {R}^{3} = \sqrt{875} \: cm}⇝R3=875cm