If the height of a cone is tripled and the radius is halved, how does the volume change?

Correct — B.
Volume of a cone = (1⁄3)πr²h, so the volume changes as r² × h.

Radius halved: r² becomes (1⁄2)² = 1⁄4 of before
Height tripled: h becomes 3 times as much
New volume = 1⁄4 × 3 = 3⁄4 of the original → option (b)

Why the others are wrong

• (a)
  Doubles — Doubles needs a factor of 2, but the factor here is 1⁄4 × 3 = 3⁄4, below 1. Halving the radius takes away more than tripling the height adds.
• (c)
  Becomes one-third of the original — One-third is the constant in V = (1⁄3)πr²h. It sits in both the old and the new volume and cancels, so the change is r² × h alone: 3⁄4.
• (d)
  Remains unchanged — Unchanged would need the height to grow 4 times to cancel the 1⁄4 from halving the radius. Tripling it leaves the volume at 3⁄4.

When dimensions scale, multiply the scale factors through the formula instead of recomputing the volume. A cone's volume depends on r² and h, so a radius factor counts squared and a height factor counts once.

Here: (1⁄2)² × 3 = 3⁄4. Constants such as 1⁄3 and π are the same before and after, so they never affect the ratio.

Key facts

• Volume of a cone = (1⁄3)πr²h.
• Scaling a cone's radius by k scales its volume by k².
• Scaling a cone's height by k scales its volume by k.
• Radius halved and height tripled: 1⁄4 × 3 = 3⁄4 of the original volume.

Common traps

• Applying the radius factor once instead of squaring it: 1⁄2 × 3 = 3⁄2, which reads as an increase.
• Carrying the formula's 1⁄3 into the answer as if it were the change.

No directly related past PYQ was found.