Question

Point size bodies of masses $20,40,60$ and $80g$ are attached at $20,40,60$ and $80cm$ divisions of a thin meter rod of negligible mass. The moment of inertia of all four bodies about an axis perpendicular to the length of rod passing through its centre is:

• A. $5\times {10}^{-3}kg{m}^2$
• B. $4\times {10}^{-3}kg{m}^2$
• C. $2\times {10}^{-3}kg{m}^2$
• D. $10\times {10}^{-3}kg{m}^2$

Answer 1

The correct option is D $10\times {10}^{-3}kg{m}^2$

Center will be at 50 cm ,All unit are to be taken is SI

$I=20\times {10}^{-3}\times (30\times {10}^{-2}{)}^2+40\times {10}^{-3}\times (10\times {10}^{-2}{)}^2+60\times {10}^{-3}\times (10\times {10}^{-2}{)}^2+80\times {10}^{-3}\times (30\times {10}^{-2}{)}^2$

So $I=10\times {10}^{-3}kg{m}^2$