Question

a metallic spherical shell of internal and external diameter of 4 cm and 8 cm respectively is melted and recast into the form of cone of base diameter 8 cm the height of cone is what​

Answer 1

Step-by-step explanation:

If the metallic spherical shell is recast , the volume of the metal remains constant always.

So , the volume of metal in the spherical shell = Volume of the cone formed

Volume of the metal in the metallic spherical shell = Volume of Outer sphere - Volume of inner sphere

This is because the metal is only present in between the inner and outer spheres.

radius of inner sphere = 4/2 = 2 ; radius of outer sphere = 8/2 = 4

radius of cone = 8/2 = 4

So , volume of metal = \frac{4}{3}* \pi * R^{3}- \frac{4}{3} * \pi * r^{3}

3

4∗π∗R

3

−

3

4

∗π∗r

3

Where ,R=radius of outer sphere

r = radius of inner sphere

So,V = (4/3)*\pi *( 4^{3} - 2^{3})π∗(4

3

−2

3

)

Equating this to volume of cone which is (1/3)*\piπ *r^{2} *hr

2

∗h

SO,(4/3)*pi * (64-8) = (1/3)*pi * 16 * h

h = 14 cm

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