Question

A solid cylinder of mass 6 kg of length $l$ is inserted tightly into a hollow cylinder of outer radius 0.6 m and length $l$ without any air gap between them. If the moment of inertia of this setup about its own axis is $1.44\text{ kgm}^{2}$, then the mass of the hollow cylinder is

• A. 1 kg
• B. 3 kg
• C. 4 kg
• D. 2 kg
• E. 5 kg

Hint:

Logic Tip: The phrase "inserted tightly... without any air gap" is a classic physics problem indicator that you should treat the multiple components as a single geometric entity rather than calculating complex inner and outer radius dependencies individually.

Answer 1

The Correct Option is D

Concept:
When a solid cylinder is inserted tightly into a hollow cylinder with no air gaps, the two objects effectively merge to form a single composite solid cylinder of uniform radius $R$, assuming uniform density distribution. The moment of inertia for a uniform solid cylinder about its central axis is $I = \frac{1}{2}MR^2$, where $M$ is the total mass of the system.
Step 1: Identify the parameters of the composite system.
Total Moment of Inertia, $I_{total} = 1.44\text{ kgm}^2$ Outer radius of the composite cylinder, $R = 0.6\text{ m}$ Mass of the inner solid cylinder, $M_1 = 6\text{ kg}$ Let the mass of the hollow cylinder be $M_2$. Total mass of the composite system, $M_{total} = M_1 + M_2 = 6 + M_2$
Step 2: Apply the moment of inertia formula.
Treating the entire setup as a single solid cylinder: $$I_{total} = \frac{1}{2} M_{total} R^2$$ Substitute the known values into the equation: $$1.44 = \frac{1}{2} (6 + M_2) (0.6)^2$$
Step 3: Solve for $M_2$.
Calculate the square of the radius: $$1.44 = \frac{1}{2} (6 + M_2) (0.36)$$ $$1.44 = 0.18 (6 + M_2)$$ Divide both sides by $0.18$: $$6 + M_2 = \frac{1.44}{0.18}$$ $$6 + M_2 = 8$$ Subtract 6 from both sides: $$M_2 = 2\text{ kg}$$