A solid cylinder has a radius of 5 cm and a height of 10 cm. A smaller cylindrical hole of radius 3 cm is drilled coaxially through its entire length. What is the approximate volume of the remaining solid?

Correct — C. The drilled solid is a hollow cylinder: the outer cylinder minus the hole.

Outer cylinder = πR²h = π × 5² × 10 = 250π
Hole = πr²h = π × 3² × 10 = 90π
Subtract: 250π − 90π = 160π

Multiply by π ≈ 3.14: 160 × 3.14 = 502.4
≈ 502 cm³ → option (c)

Why the others are wrong

• (a)
  282.6 cm³ — 282.6 cm³ is 90π, the volume of the 3 cm hole itself (3.14 × 9 × 10). That is the material drilled out, not the solid left behind.
• (b)
  785 cm³ — 785 cm³ is 250π, the full 5 cm cylinder before drilling (3.14 × 25 × 10). It never subtracts the hole.
• (d)
  628.3 cm³ — 628.3 cm³ ≈ 200π, which would need a cross-section of 20π cm². The ring's cross-section is π(5² − 3²) = 16π cm², so the volume is 160π.

A coaxial hole leaves a hollow cylinder, a tube. Its cross-section is a ring between two circles, so its volume is the ring's area times the length:

V = π(R² − r²)h

Here R² − r² = 25 − 9 = 16. Factorising it as (R + r)(R − r) = 8 × 2 gives the same 16, which is quicker when the radii are large.

Key facts

• Volume of a hollow cylinder = π(R² − r²)h, where R and r are the outer and inner radii.
• R² − r² = (R + r)(R − r): here 8 × 2 = 16.
• 160π ≈ 502.65 cm³.

Common traps

• Picking 282.6 cm³, the volume of the hole, instead of the solid that remains
• Writing π(R − r)²h = π × 2² × 10 = 40π instead of π(R² − r²)h

No directly related past PYQ was found.