Question

M.I of a solid sphere about its diameter is
64 kg m². If that sphere is recast into 8
identical small spheres, then M.I. of such
small sphere about its diameter is​

Answer 1

Answer:

M.I. of such  small sphere about its diameter is​ 2 kg m^2

Explanation:

Let the mass of given sphere is M and its radius is R

so the moment of inertia of the sphere is given as

$I = \frac{2}{5}MR^2$

now we know that one sphere is divided into 8 equal spheres

so we will have

$8(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$

$r = \frac{R}{2}$

now moment of inertia of each small sphere is given as

$I' = \frac{2}{5}(\frac{M}{8})(\frac{R}{2})^2$

$I' = \frac{64}{32}$

$I' = 2 kg m^2$

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Topic : Moment of inertia

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Answer 2

Answer:

2kg/m square is correct answer