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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

PhysicsEasynumerical

The moment of inertia of a hollow cubical box of mass $M$ and side length $a$ , about an axis passing through centres of two opposite faces, is equal to $\frac{x M a^{2}}{18}$ . The value of $x$ is

Answer:

5.00

Solution:

Let $m$ be the mass of each face of the cubical box.

$I = 2 \times (\frac{m a^{2}}{6}) + 4 \times [\frac{m a^{2}}{12} + m {(\frac{a}{2})}^{2}]$

$= \frac{5 m a^{2}}{3}$

But, $m = \frac{M}{6}$

$∴ I = \frac{5 M a^{2}}{18}$

$\Rightarrow x = 5$

Subject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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