Rotation 1 Question 20

23. Four solid spheres each of diameter $\sqrt{5} cm$ and mass $0.5 kg$ are placed with their centres at the corners of a square of side $4 cm$. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4} kg-m^{2}$, then $N$ is

(2011)

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Answer:

Correct Answer: 23. (9)

Solution:

  1. $r=\frac{d}{2}=\frac{\sqrt{5}}{2} cm=\frac{\sqrt{5}}{2} \times 10^{-2} m \Rightarrow m=0.5 kg$

$I _{X X}=I _1+I _2+I _3+I _4$

$=\frac{2}{5} m r^{2}+m \frac{a}{\sqrt{2}}^{2}+\frac{2}{5} m r^{2}$

$+\frac{2}{5} m r^{2}+m \frac{a}{\sqrt{2}}^{2}+\frac{2}{5} m r^{2}$

Substituting the values, we get

$$ I _{X X}=9 \times 10^{-4} kg-m^{2} \therefore \quad N=9 $$



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