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

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(B) Mass $=$ Volume $	imes$ Density. Let the masses be $M_A$ and $M_B$,and radii be $R_A$ and $R_B$. Since $M_A = M_B = M$,we have: $M = \frac{4}{3}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Mass $=$ Volume $	imes$ Density. Let the masses be $M_A$ and $M_B$,and radii be $R_A$ and $R_B$. Since $M_A = M_B = M$,we have: $M = \frac{4}{3} \pi R_A^3 ho_A = \frac{4}{3} \pi R_B^3 ho_B$. From this,$\frac{R_B^3}{R_A^3} = \frac{ho_A}{ho_B}$,which implies $\frac{R_B}{R_A} = (\frac{ho_A}{ho_B})^{1/3}$. The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} MR^2$. Therefore,$\frac{I_B}{I_A} = \frac{\frac{2}{5} M R_B^2}{\frac{2}{5} M R_A^2} = \frac{R_B^2}{R_A^2}$. Substituting the ratio of radii: $\frac{I_B}{I_A} = ((\frac{ho_A}{ho_B})^{1/3})^2 = (\frac{ho_A}{ho_B})^{2/3}$.

Two solid spheres ($A$ and $B$) are made of metals having densities $\rho_A$ and $\rho_B$ respectively. If their masses are equal,then the ratio of their moments of inertia $(\frac{I_B}{I_A})$ about their respective diameters is:

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