Question

Four spheres, each of mass $M$ and radius $r$ are situated at the four corners of square of side $R$. The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its centre will be

• A. $\frac { 5 } { 2 } M \left( 4 r ^ { 2 } + 5 R ^ { 2 } \right)$
• B. $\frac { 2 } { 5 } M \left( 4 r ^ { 2 } + 5 R ^ { 2 } \right)$
• C. $\frac { 2 } { 5 } M \left( 4 r ^ { 2 } + 5 r ^ { 2 } \right)$
• D. $\frac { 5 } { 2 } M \left( 4 r ^ { 2 } + 5 r ^ { 2 } \right)$

Answer 1

Answer: (B)

$\frac { 2 } { 5 } M \left( 4 r ^ { 2 } + 5 R ^ { 2 } \right)$

Answer 2

M. I. of sphere A about its diameter $I _ { O ^ { \prime } } = \frac { 2 } { 5 } M r ^ { 2 }$

Now M.I. of sphere A about an axis perpendicular to the plane of square and passing through its centre will be

$I _ { O } = I _ { O ^ { \prime } } + M \left( \frac { R } { \sqrt { 2 } } \right) ^ { 2 }$ $= \frac { 2 } { 5 } M r ^ { 2 } + \frac { M R ^ { 2 } } { 2 }$

[by the theorem of parallel axis]

Moment of inertia of system (i.e. four sphere)= $4 I _ { O }$ $= 4 \left[ \frac { 2 } { 5 } M r ^ { 2 } + \frac { M R ^ { 2 } } { 2 } \right]$

$= \frac { 2 } { 5 } M \left[ 4 r ^ { 2 } + 5 R ^ { 2 } \right]$