What is the volume of a cone (in cubic units) with radius 3 units and slant height 3√2 units?

Correct — B. The stem is an image asking for the volume of a cone with radius 3 units and slant height 3√2 units, and the four options are images of expressions.

Radius, height and slant height form a right triangle: l² = r² + h²
l² = (3√2)² = 18 and r² = 9
h² = 18 − 9 = 9, so h = 3

Volume = ⅓ π r² h
= ⅓ × π × 9 × 3 = 9π cubic units → option (b), the picture reading 9π

Why the others are wrong

• (a)
  Option (a) reads 10π⁄3, which would need r²h = 10. The figures given force r²h = 9 × 3 = 27, and a third of 27π is 9π.
• (c)
  Option (c) reads 18π, exactly twice the answer, and needs a height of 6. The 3 and 3√2 right triangle gives h = 3, so 18π overshoots by a factor of two.
• (d)
  Option (d) reads 9π⁄2, which requires r²h = 13.5, that is a height of 1.5. Nothing in the question produces 1.5 — Pythagoras pins the height at 3.

A cone's three lengths are locked together by l² = r² + h², because the radius, the axis and the slant line form a right-angled triangle.

SSC gives the slant height rather than the height precisely to force that step. Volume needs the perpendicular height, so the slant height can never go straight into ⅓ π r² h.

Here r = 3 and h = 3 come out equal, so the cone's axial cross-section is a right isosceles triangle and the slant height is r√2. Recognising that shape gives the height instantly.

Key facts

• For a right circular cone, l² = r² + h².
• Volume of a cone = ⅓ π r² h.
• With r = h = 3 the axial section is a right isosceles triangle, so l = r√2 = 3√2.
• Curved surface area is π r l, which for this cone is 9√2 π — a quantity often confused with the volume.

Common traps

• Substituting the slant height for the height in the volume formula.
• Dropping the ⅓ and reporting the cylinder volume 27π.
• Squaring 3√2 as 3 × 2 = 6 instead of 9 × 2 = 18.

No directly related past PYQ was found.