The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm² . The length of the 400 g rod is nearly :

For a thin uniform rod rotating about an axis through its centre and perpendicular to the rod, the moment of inertia is given by

$I=\frac{1}{12}m{L}^2$

where $I$ is the moment of inertia, $m$ the mass of the rod and $L$ its length.

We are given $I=\; 2400 \mathrm{g·cm}{2}^2$ and $m=\; 400 g$. Solve the formula for $L$:

$L=\sqrt{\frac{12\times I}{m}}$

Insert the numbers:

$12\times 2400=28800$

$\frac{28800}{400}=72$ cm22

Therefore

$L=\sqrt{72}$ cm

Factor $72=36\times 2$, so

$L=\sqrt{36}\times \sqrt{2}=6\times \sqrt{2} \mathrm{cm}$

Numeric evaluation:

$6\times \sqrt{2}\approx 6\times 1.414=8.485$ cm

Rounded to the appropriate precision, the length of the 400 g rod is about 8.5 cm.