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Question

(B) Given,internal diameter of the spherical shell $= 4 \, cm$ and external diameter $= 8 \, cm$.Internal radius of the spherical shell,$r_1 = \frac{4

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Given,internal diameter of the spherical shell $= 4 \, cm$ and external diameter $= 8 \, cm$.Internal radius of the spherical shell,$r_1 = \frac{4}{2} = 2 \, cm$.External radius of the spherical shell,$r_2 = \frac{8}{2} = 4 \, cm$.Volume of the spherical shell $= \frac{4}{3} \pi (r_2^3 - r_1^3) = \frac{4}{3} \pi (4^3 - 2^3) = \frac{4}{3} \pi (64 - 8) = \frac{4}{3} \pi (56) = \frac{224}{3} \pi \, cm^3$.Let the height of the cone be $h \, cm$ and the radius of the base be $R = \frac{8}{2} = 4 \, cm$.Since the shell is recast into a cone,their volumes are equal:Volume of cone $= \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi (4)^2 h = \frac{16}{3} \pi h$.Equating the volumes: $\frac{16}{3} \pi h = \frac{224}{3} \pi$.$16h = 224$.$h = \frac{224}{16} = 14 \, cm$.Thus,the height of the cone is $14 \, cm$.

$A$ metallic spherical shell of internal and external diameters $4 \, cm$ and $8 \, cm$,respectively,is melted and recast into the form of a cone of base diameter $8 \, cm$. The height of the cone is (in $cm$):

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