Question
A solid cylinder of radius R and height H, is melted and recast into three identical cylinders, each having the same height as the original cylinder but half the radius of the original cylinder. What fraction of the original cylinder's volume remains unused?
This question was previously asked in
RRB NTPC 2025 Under Graduate CBT 1 Question Paper PDF (20-Jun-2026) (Shift 3)
The correct answer is
\(\dfrac14\)
Original volume: \(\pi R^2 H\).
Volume of each new cylinder (radius R/2, same height H): \(\pi\left(\dfrac{R}{2}\right)^2H = \dfrac{\pi R^2H}{4}\).
Volume of three such cylinders: \(3\times\dfrac{\pi R^2H}{4} = \dfrac{3\pi R^2H}{4}\), which is \(\tfrac34\) of the original volume.
Unused fraction: \(1-\tfrac34 = \tfrac14\).
Hence, one-fourth of the original cylinder's volume remains unused.
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