The densities of two solid spheres A and B of the same radii R vary with radial distance r as \rho _A(r) = k( r \over R ) and \rho _B(r) = k( r \over R )^5 , , respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are I_A and I_B , respectively. If, I_B \over I_A = n \over 10 , the value of n is

Rotational Motion PYQ — JEE Advanced 2015 · Shift 2 · Q43 with Solution

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Correct answer: 6

For sphere $A$ : $\rho_A(r)=k\left(\frac{r}{R}\right)$

For sphere $B$ : $\rho_B(r)=k\left(\frac{r}{R}\right)^5$

We need the moment of inertia of each solid sphere about an axis through its centre.

Take a thin shell of radius $r$ and thickness $dr$ .

Its mass is $dm=\rho(r)\cdot 4\pi r^2\,dr$

Moment of inertia of a thin spherical shell about a diameter is $dI=\frac{2}{3}r^2\,dm$

So,

=\frac{8\pi}{3}\rho(r)r^4dr$$ Hence, $$I=\frac{8\pi}{3}\int_0^R \rho(r)r^4dr$$ --- 3. **Compute $I_A$** Substitute $\rho_A(r)=k\left(\frac{r}{R}\right)$: $$I_A=\frac{8\pi}{3}\int_0^R k\left(\frac{r}{R}\right)r^4dr$$ $$I_A=\frac{8\pi k}{3R}\int_0^R r^5dr$$ $$\int_0^R r^5dr=\frac{R^6}{6}$$ Therefore, $$I_A=\frac{8\pi k}{3R}\cdot \frac{R^6}{6} =\frac{8\pi kR^5}{18} =\frac{4\pi kR^5}{9}$$ --- 4. **Compute $I_B$** Substitute $\rho_B(r)=k\left(\frac{r}{R}\right)^5$: $$I_B=\frac{8\pi}{3}\int_0^R k\left(\frac{r}{R}\right)^5 r^4dr$$ $$I_B=\frac{8\pi k}{3R^5}\int_0^R r^9dr$$ $$\int_0^R r^9dr=\frac{R^{10}}{10}$$ Thus, $$I_B=\frac{8\pi k}{3R^5}\cdot \frac{R^{10}}{10} =\frac{8\pi kR^5}{30} =\frac{4\pi kR^5}{15}$$ --- 5. **Find the ratio** $$\frac{I_B}{I_A}=\frac{\frac{4\pi kR^5}{15}}{\frac{4\pi kR^5}{9}}=\frac{9}{15}=\frac{3}{5}$$ Given, $$\frac{I_B}{I_A}=\frac{n}{10}$$ So, $$\frac{n}{10}=\frac{3}{5}$$ $$n=10\cdot \frac{3}{5}=6$$ --- 6. **Final answer** $$\boxed{n=6}$$ The derived answer matches the stored correct answer.