Two solid spheres (A and B) are made of metal | vedclass

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(B) Given that the masses of the two solid spheres are equal, $m_A = m_B$.Since mass $m = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3} \pi R^3

Vedclass · Accepted Answer

$(\rho_A / \rho_B)^{2/3}$

Vedclass · Answer

(B) Given that the masses of the two solid spheres are equal, $m_A = m_B$.Since mass $m = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3} \pi R^3 ho$, we have:$\frac{4}{3} \pi R_A^3 ho_A = \frac{4}{3} \pi R_B^3 ho_B$$\Rightarrow \frac{R_A^3}{R_B^3} = \frac{ho_B}{ho_A} \Rightarrow \frac{R_A}{R_B} = \left(\frac{ho_B}{ho_A}ight)^{1/3}$.The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} m R^2$.Since $m_A = m_B$, the ratio of moments of inertia is:$\frac{I_B}{I_A} = \frac{\frac{2}{5} m_B R_B^2}{\frac{2}{5} m_A R_A^2} = \left(\frac{R_B}{R_A}ight)^2$.Substituting the ratio of radii:$\frac{I_B}{I_A} = \left[ \left( \frac{ho_A}{ho_B} ight)^{1/3} ight]^2 = \left( \frac{ho_A}{ho_B} ight)^{2/3}$.

Two solid spheres ($A$ and $B$) are made of metals of different densities $\rho_A$ and $\rho_B$ respectively. If their masses are equal, the ratio of their moments of inertia $(I_B / I_A)$ about their respective diameter is

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