Question

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?

• A. 9%
• B. 16%
• C. 25%
• D. 50%
• E. None of the above

Hint:

When a solid is re-cast, always conserve the \textbf{volume} first, then relate to the required dimensions (like radius/height). Finally, compare the new and old surface areas.

Answer 1

The Correct Option is D

Step 1: Volume of the cylinder.
The cylinder has radius \(r = \tfrac{14}{2} = 7 \, \text{cm}\), height \(h = 10 \, \text{cm}\).
\[ V_{\text{cylinder}} = \pi r^2 h = \pi (7^2)(10) = 490\pi \, \text{cm}^3 \] Step 2: Volumes of cones (in ratio 3:4).
Let the two cones have volumes in ratio 3:4.
\[ V_1 = \frac{3}{7} \times 490\pi = 210\pi, \quad V_2 = \frac{4}{7} \times 490\pi = 280\pi \] Step 3: Relation between volume and radius of cones.
For a cone: \[ V = \tfrac{1}{3}\pi r^2 h \] Here, \(h = 10\). For cone 1: \[ 210\pi = \tfrac{1}{3}\pi r_1^2 (10) \;\Rightarrow\; r_1^2 = 63 \;\Rightarrow\; r_1 = \sqrt{63} \] For cone 2: \[ 280\pi = \tfrac{1}{3}\pi r_2^2 (10) \;\Rightarrow\; r_2^2 = 84 \;\Rightarrow\; r_2 = \sqrt{84} \] Step 4: Flat surface area before and after.
- For cylinder: flat surface area = area of top + bottom = \(2 \pi r^2\).
\[ A_{\text{cylinder}} = 2 \pi (7^2) = 98\pi \] - For cones: flat surface area = area of base circles = \(\pi r_1^2 + \pi r_2^2\).
\[ A_{\text{cones}} = \pi(63) + \pi(84) = 147\pi \] Step 5: Percentage change.
\[ %\;\text{Change} = \frac{147\pi - 98\pi}{98\pi} \times 100 = \frac{49\pi}{98\pi} \times 100 = 50% \] \[ \boxed{\text{Percentage increase = 50%}} \]