Question

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 moment of inertia $I_B:I_A$ about their respective diameter is :

• A. ${\left(\frac{{\rho }_B}{{\rho }_A}\right)}^{\frac{2}{3}}$
• B. ${\left(\frac{{\rho }_A}{{\rho }_B}\right)}^{\frac{2}{3}}$
• C. $\frac{{\rho }_A}{{\rho }_B}$
• D. $\frac{{\rho }_B}{{\rho }_A}$

Answer 1

The correct option is B ${\left(\frac{{\rho }_A}{{\rho }_B}\right)}^{\frac{2}{3}}$

The Let the radii of $A$ and $B$ be $R_A$ and $R_B$.

So $M_A=\frac{4}{3}\pi R_A^3{\rho }_A$

and, $M_B=\frac{4}{3}\pi R_B^3{\rho }_B$

As $M_A=M_B$

$\Rightarrow \frac{4}{3}\pi R_A^3{\rho }_A=\frac{4}{3}\pi R_B^3{\rho }_B$

$\Rightarrow R_B/R_A=({\rho }_A/{\rho }_B{)}^{1/3}$

Now $I_A=\frac{2}{5}{M}_A{R}_A^2$

and, $I_B=\frac{2}{5}{M}_B{R}_B^2$

$\Rightarrow I_B/I_A=R_B^2/R_A^2$

$\Rightarrow I_B/I_A=({\rho }_A/{\rho }_B{)}^{2/3}$