Question

The moment of inertia of a thin square plate of mass $1.2kg$ is $0.2kg{m}^2$ when it is made to rotate about an axis perpendicular to plane of plate and passing through a corner of plate. The side length of plate is :

• A. $0.2m$
• B. $0.4m$
• C. $0.5m$
• D. $0.8m$

Answer 1

Answer: (C)

$0.5m$

We know that $I_{cm}=\frac{m{l}^2}{6}$

Distance of center from corner ,$d=\sqrt{\frac{l^2}{4}+\frac{l^2}{4}}=\frac{l}{\sqrt{2}}$

Now using parallel axis theorem $I=I_{cm}+m{d}^2=\frac{m{l}^2}{6}+\frac{m{l}^2}{2}=\frac{2m{l}^2}{3}$

Now $0.2=2m{l}^2/3\Rightarrow l=0.5m$