Question 53

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

Solution

Volume of Cylinder = $$\pi r^2 h = \pi \times 7^2 \times 10 = 490\pi$$

Now, The solid metal cylinder is re-cast into two cones in the proportion 3 : 4 i.e. the volumes of cone 1 and cone 2 is

$$210 \pi$$ and $$280 \pi$$ respectively.

So, flat Surface area of cylinder before melting = $$2 \pi r^2 = 2 \pi \times 7^2 = 98 \pi$$

Volume of cone 1 = $$\frac{1}{3} \pi r_1^2 h = 210 \pi$$

=> $$r_1^2 = \frac{210 \times 3}{10} = 63$$

Volume of cone 2 = $$\frac{1}{3} \pi r_2^2 h = 280 \pi$$

=> $$r_2^2 = \frac{280 \times 3}{10} = 84$$

Flat surface area of cones = $$\pi r_1^2 + \pi r_2^2$$

= $$\pi (63 + 84) = 147 \pi$$

$$\therefore$$ Percentage change in surface area = $$\frac{147 \pi - 98 \pi}{98 \pi} \times 100$$

= $$\frac{1}{2} \times 100 = 50 \%$$