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← JEE Advanced 2015 · Paper 2
PhysicsInteger+4Question 3 of 60

JEE Advanced 2015 · Paper 2, Physics question 3

The densities of two solid spheres AA and BB of the same radii RR vary with radial distance rr as ρA(r)=k(rR)\rho_A(r)=k\left(\dfrac rR\right) and ρB(r)=k(rR)5\rho_B(r)=k\left(\dfrac rR\right)^5, respectively, where kk is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IAI_A and IBI_B, respectively. If IBIA=n10\dfrac{I_B}{I_A}=\dfrac n{10}, the value of nn is

Answer and worked solution

Official answer: 6

Moment of inertia of a sphere with radial density:
I=8π3∫0Rρ(r)r4 dr.I=\frac{8\pi}3\int_0^R\rho(r)r^4\,dr.

The two spheres.

  • IA∝∫0Rr5R dr=R56I_A\propto\displaystyle\int_0^R\frac{r^5}R\,dr=\frac{R^5}6.
  • IB∝∫0Rr9R5 dr=R510I_B\propto\displaystyle\int_0^R\frac{r^9}{R^5}\,dr=\frac{R^5}{10}.

Ratio. IBIA=610\dfrac{I_B}{I_A}=\dfrac6{10}, so n=6n=6.

Answer: 6