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dk_ch Share
Feb 10, 2018

drawn

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

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