Four sphere of diameter 2a and mass M are placed with their centres of the four comers of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{8}{5} {Ma}^{2} + 2 {Mb}^{2}$

$\frac{8}{5} {Ma}^{2}$

$\frac{4}{5} {Ma}^{2} + 2 {Mb}^{2}$

$\frac{4}{5} {Ma}^{2} + 4 {Mb}^{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The situation is shown in fig

The moment of inertia of each of spheres A and D about AD
$= \frac{2}{5} {Ma}^{2}$
The moment of inertia of each of spheres I and C about AD
$= \frac{2}{5} {Ma}^{2} + {Mb}^{2}$
$∴$ Moment of inertia of whole arrangement about the side AD is given by
$\begin{array}{l} = 2 [\frac{2}{5} {Ma}^{2} + \{\frac{2}{5} {Ma}^{2} + {Mb}^{2}\}] \\ = \frac{8}{5} {Ma}^{2} + 2 {Mb}^{2} \end{array}$