Question

Question #01460

Answer 1

The surface area of the cone=area of the given metal lamina in the shape of a sector of radius r is given by

`=(360-144)/360xxpir^2=216/360xxpir=3/5pir^2`

The slant height of the cone will be radius of the sector `r`

If the radius of the cone be `r_c` then
its surface area will be `=pir_cxxr=3/5pir^2`
`=>r_c=3/5r`

Height of the cone `h=sqrt("slant height"^2-"radius"^2)`

`=sqrt(r^2-r_c^2)=sqrt(r^2-9/25r^2)=4/5r`

So volume of the cone `V=1/3pir_c^2h=1/3pi(3/5r)^2xx4/5r`

Volume of n pieces of spheres each of radius `a` will be

`=4/3pia^3n`

By the condition of the problem we have

`4/3pia^3n=1/3pi(3/5r)^2xx4/5r=(3xx4)/125pir^3`

`=>125na^3=9r^3`

Proved