Question

The diagram shows a uniform rod of mass M and length $2l$. The moment of inertia of the rod about XY is l. The moment of inertia of the rod about the axis X'Y' ( which is parallel to XY ) is

• A. $\frac{m{L}^2}{3}{\sin }^2\alpha$
• B. $\frac{m{L}^2}{12}{\sin }^2\alpha$
• C. $\frac{m{L}^2}{6}co{s}^2\alpha$
• D. $\frac{m{L}^2}{12}co{s}^2\alpha$

Answer 1

The correct option is B $\frac{m{L}^2}{12}{\sin }^2\alpha$

$If\alpha ={90}^0$

$1st$ Option =Motion of inertia $=\frac{M{L}^2}{3}$

$2nd$ Option=Motion of inertia $=\frac{M{L}^2}{12}$

$3rd$ Option =Moment of inertia $=0$........$\left[{\cos }^{2}90=0\right]$

$4th$ Option =Motion of inertia=0 ........$\left[{\cos }^{2}90=0\right]$

$If\alpha ={90}^0$

Moment of inertia of rod about an axis perpendicular to row through centre is $\frac{m{L}^2}{12}$

$\therefore 2nd$ Option is correct.