NCERT Exemplar Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 12.1 in Chapter 12 - Surface Areas and Volumes
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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
Internal radius, r1 = 2 cm[ as radius = 1/2 diameter]
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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