# NCERT Exemplar Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 12.1 in Chapter 12 - Surface Areas and Volumes

Question 14 Surface Areas and Volumes - Exercise 12.1

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

14cm

Volume of spherical shell = Volume of cone recast by melting

For Spherical Shell,

Internal diameter, d1 = 4 cm

External diameter, d2 = 8 cm

External radius, r2 = 4 cm

Now,

As volume of spherical shell= 4/3 π (r23 – r13)

where r1 and r2 are internal and external radii respectively.

volume of given shell = 4/3 π (43 – 23)

= 4/3 π (56)

= (224/3) π

We know that,

Volume of cone = 224π /3 cm3

For cone,

Base diameter = 8 cm

Base radius, r = 4 cm

Let Height of cone = ‘h’.

We know,

Volume of cone = (1/3) π r2h,

Where r = Base radius and h = height of cone

Volume of given cone = (1/3) π 42h

⇒ 224π /3 = 16πh /3

⇒ 16h = 224

h = 14 cm

So, Height of cone is 14 cm.

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