The densities of two solid spheres $A$ and $B$ of the same radii $R$ vary with radial distance $r$ as $\rho_{A}(r)=k\left(\frac{r}{R}\right)$ and $\rho_{B}(r)=k\left(\frac{r}{R}\right)^{5}$, 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 $\frac{I_{B}}{I_{A}}=\frac{n}{10}$, the value of $n$ is

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Rotational Mechanics

The densities of two solid spheres

A

and

B

of the same radii

R

vary with radial distance

r

ρ_{A} (r) = k (\frac{r}{R})

and

ρ_{B} (r) = k (\frac{r}{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

\frac{I _{B}}{I _{A}} = \frac{n}{10}

, the value of

n

Answer: 6

Consider a spherical shell of radius

r

and thickness

d r

. The moment of inertia of this shell about the central axis is:

d I = \frac{2}{3} (ρ \cdot 4 π r^{2} d r) r^{2}

For sphere A:

I_{A} = \int_{0}^{R} \frac{2}{3} k \frac{r}{R} 4 π r^{2} d r \cdot r^{2} = \frac{8 π k}{3 R} \int_{0}^{R} r^{5} d r = \frac{8 π k R ^{5}}{18}

For sphere B:

I_{B} = \int_{0}^{R} \frac{2}{3} k (\frac{r}{R})^{5} 4 π r^{2} d r \cdot r^{2} = \frac{8 π k}{3 R ^{5}} \int_{0}^{R} r^{9} d r = \frac{8 π k R ^{5}}{30}

Therefore:

\frac{I _{B}}{I _{A}} = \frac{18}{30} = \frac{6}{10}

Hence

n = 6

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer