Correct Answer: 162
Moment of inertia of a solid cylinder about its own axis is given by:
I = ½ M R²
Let density of the material be ρ.
Mass of original cylinder:
M₁ = ρ × πR² × L
Moment of inertia of original cylinder:
I₁ = ½ (ρπR²L) R² I₁ = ½ ρπ L R⁴
Now consider the smaller co-centric cylinder.
Its radius = R/3 Its length = L/2
Mass of smaller cylinder:
M₂ = ρ × π (R/3)² × (L/2)
M₂ = ρπR²L / 18
Moment of inertia of smaller cylinder:
I₂ = ½ M₂ (R/3)²
I₂ = ½ × (ρπR²L / 18) × (R² / 9)
I₂ = ρπ L R⁴ / 324
Now take the ratio:
I₁ / I₂ = (½ ρπ L R⁴) ÷ (ρπ L R⁴ / 324)
I₁ / I₂ = ½ × 324
I₁ / I₂ = 162
Hence, the required ratio is 162.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.