Question

The percentage change in the flat surface before and after-when a solid metal cylinder of height 10 units and radius 7 units is melted and recast into two cones in the volume proportion of $3:4$, keeping the height same would be_____

Answer 1

Answer: 50

Let the radius of cylinder $=R$
And that of cone 1 and cone $2=r_1$ and $r_2$
Let volume be $V,V_1$ and $V_2$ and $V_2$, respectively.
Cones are recasted from molten cylinder
So $V=V_1+V_2$
Ratio of volume of the two cones $V_1:V_2:$ :
$3:4$
$\Rightarrow V_1=3x$ and $V_2=4x$
Thus, $V=V_1+V_2=3x+4x=7x$
Volume of cylinder $=\pi R^{2}h$
Volume of cone $1=\frac{1}{3}\pi r_1^{2}h$
Volume of cone $2=\frac{1}{3}\pi r_2^{2}h$
We know ratio of volume is $7:3:4$
$\Rightarrow \pi R^{2}h:\frac{1}{3}\pi r_1^{2}h:\frac{1}{3}\pi r_2^{2}h$
$R^2:\frac{1}{3}{r}_1^2:\frac{1}{3}{r}_2^2::7:3:4$
$R^2:r_1^2:r_2^2=7:9:12=k$
$R^2=7k$
$r_1^2=9k$
$r_2^2=12k$
Flat surface area of cylinder is sum of the area of two circles at the top and bottom.
So, flat surface area of cylinder $=2\pi R^2$
Flat surface area of cone 1 , and cone $2=\pi r_1^2$ and $\pi r_2^2$
Ratio of flat surface area of cylinder so that of two cone 1 and cone 2
$2\pi R^2:\pi r_1^2+\pi r_2^2$
$2R^2:r_1^2+r_2^2$
$2(7k):9k+12k$
$14k:21k$
Change in flat surface area $=21k-14k=7k$ $\%$ change $=\frac{7k}{14k}\times 100=50\%$