Question

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

Accepted Answer

D. Radius of cylinder $r = 7	ext{ cm}$, height $h = 10	ext{ cm}$. Flat surface area of cylinder $= 2 \pi r^2 = 2 \pi (7^2) = 98\pi$. Volume of cylinder $= \pi r^2 h = 490\pi$. Volumes of two cones are $210\pi$ and $280\pi$. For cone 1: $\frac{1}{3} \pi r_1^2 (10) = 210\pi \implies r_1^2 = 63$. For cone 2: $\frac{1}{3} \pi r_2^2 (10) = 280\pi \implies r_2^2 = 84$. Flat surface area of the two cones $= \pi r_1^2 + \pi r_2^2 = \pi(63 + 84) = 147\pi$. Percentage change $= \frac{147\pi - 98\pi}{98\pi} 	imes 100\% = \frac{49}{98} 	imes 100\% = 50\%$.

Cylinder Melted into Cones Surface Area Change

Explanation

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