Question

Q2. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8cm. The height of the cone is (A) 12cm (B) 14cm (C) 15cm (D) 18cm

Answer 1

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 = $\frac{4}{3}* \pi * R^{3}- \frac{4}{3} * \pi * r^{3}$
Where ,R=radius of outer sphere
             r = radius of inner sphere

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

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

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

h = 14 cm

Answer 2

Answer:

B 14

Step-by-step explanation: