The given metal cylinder is as follows
Here, radius(r) =
=7cm
height (h) =10 cm
∴ Volume of cylinder =
πr2h=×72×10 =1540cm
3 This cylinder is melted and recast into two cones in the proportion of 3: 4 (volume), keeping the height 10 cm. Let radius of first cone be
r1 cm and radius of another cone be r
2 cm and volume of first cone be V
1 cm
3 and volume of second cone be V
2 cm
3Then,
=⇒=⇒=⇒r12=r22 ...(i)
where, h = Height
r1 = Radius of 1st cone,
r2 = Radius of 2nd cone
l1 = Slant height of 1st cone
l2 = Slant height of 2nd cone
NOW,V1+V2=1540cm3⇒
πr12h+πr22h=1540⇒
πh(r12+r22)=1540⇒
××10(r12+r22)=1540⇒
r12+r22=⇒
r12+r22=147...(ii)
⇒
r22+r22=147....[from Eq. (i)]
⇒
7r22=147×4⇒
r22=147×47⇒
r22=84 ∴r2=2√21cmFrom Eq. (i),
r12=×84=63∴
r1=√63cmNow, flat surface area of the cylinder
=2πr2=2π×(7)2=98πcm2and flat surface area of the cones
=πr12+πr22=63π+84π=147πcm2∴ Required percentage change
=147π−98π98π×100==50%