Question

A solid metal cylinder of 10cm height and 14cm diameter is melted and recast into two cones in the proportion of 3:4 (volume), keeping the height 10cm. what would be the percentage change in the flat surface area before and after?

How come the volume of come is pi*r*r*height?
Volume of cone is 1/3* (radius) ^2 * height. Right?
Correct me if I am wrong.

Thank you.

Answer 1

Answer:

change in flat surface area is = 14k(diameter of cylinder) -7k (3:4. i.e r1+r2=3+4=7))

percentage change is =. 7k/14k×100=. 50%

so ans is 50%

Answer 2

Given:

The diameter of cylinder = 14cm

Height of cylinder = 10cm

Ratio of volumes of cones = 3:4

Height of cones = 10cm

To find:

Percentage change in the flat surface area before and after.

Solution:

The diameter of cylinder = 14cm

The radius of cylinder = 7cm

Height of cylinder = 10cm

Volume of cylinder = π$r^2h$ = 3.14 x 7 x 7 x 10 = 1538.6 $cm^3$

Now

volume  of cone 1 =   $\frac{\pi r1^2h}{3}$

volume  of cone 2 =   $\frac{\pi r2^2h}{3}$

Ratio of volumes = 3:4

If V1 = 3x then V2 = 4x

Now,

V1 + V2 = 1538.6  $cm^3$

3x + 4x = 1538.6  $cm^3$\

7x = 1538.6  $cm^3$

x = 219.8 $cm^3$

Therefore V1 = 3x = 3 x 219.8 = 659.4 $cm^3$

Therefore V2 = 4x = 4 x 219.8 = 879.2 $cm^3$

Now,

V1 =  $\frac{\pi r1^2h}{3}$ = 659.4

$r1^2$ = $\frac{659.4 (3) }{3.14 (10)}$

$r1^2$ = 63

r1 = 8cm (approx.)

Similarly,

V2 =  $\frac{\pi r2^2h}{3}$  = 879.2

$r2^2$ = $\frac{879.2(3)}{3.14(10)}$

$r2^2$ = 84

r2 = 9cm (approx.)

Now flat surface area of cylinder = 2($\pi r^{2}$) = 307.72 $cm^2$

Flat surface area of cone 1 = $\pi r1^{2}$ = 3.14 x 8 x 8 = 197.82 $cm^2$

Flat surface area of cone 2 = $\pi r2^{2}$ = 3.14 x 9 x 9 = 263.76 $cm^2$

Total surface area of cone 1 + cone 2 = 197.82 + 263.76 = 461.58   $cm^2$Change in area = 461.58 - 307.72 = 153.86

% change = 50%

Therefore, there will be a 50% change in flat surface area.