Rotational MotionMedium
Question
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400\text{ }g{\text{ }cm}^{2}$. The length of the 400 g rod is nearly:
Options
A.20.7 cm
B.72.0 cm
C.8.5 cm
D.17.5 cm
Solution
Sol. [Key Concept: Moment of inertia of a thin rod about its centre]
Moment of inertia of rod $= I = \frac{m\mathcal{l}^{2}}{12}$
$$\begin{matrix} & \ \Rightarrow \ 2400 = 400\frac{\mathcal{l}^{2}}{12} \\ & \ \Rightarrow \ 72 = \mathcal{l}^{2} \\ & \ \Rightarrow \ \mathcal{l} = \sqrt{72} = 8.48\text{ }cm \simeq 8.5\text{ }cm \end{matrix}$$
Create a free account to view solution
View Solution FreeMore Rotational Motion Questions
A uniform disc is rolling on a horizontal surface. At a certain instant B is the point of contact and A is at height 2R ...A stone is tied to a string and is rotated round the finger and is then released. The string wrapps around the finger, t...A hoop of radius r and mass m rotating with an angular velocity ω0 is placed on a rough horizontal surface. The ini...For the same total mass, which of the following will have the largest moment of inertia about an axis passing through th...A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. T...