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QJEEAdv15Ph8ee9: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph8ee9 Solution: The densities of two solid spheres A and B of the same radii R…

The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r)=k(rR){\rho _A}(r) = k\left( {{r \over R}} \right) and ρB(r)=k(rR)5{\rho _B}(r) = k{\left( {{r \over R}} \right)^5} , , respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IA{I_A} and IB{I_B} , respectively. If, IBIA=n10{{{I_B}} \over {{I_A}}} = {n \over {10}} , the value of n is

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Consider a spherical shell of radius r and small thickness dr.

JEE Advanced 2015 Paper 2 Offline Physics - Rotational Motion Question 49 English Explanation

The volume of the shell is dV=4πr2drdV = 4\pi {r^2}dr and its mass is

dm=ρdV=4πρr2drdm = \rho dV = 4\pi \rho {r^2}dr .

The moment of inertia of the spherical shell of mass dm and radius r about an axis passing through its centre O is given by dI=23dm r2dI = {2 \over 3}dm\,{r^2} . Substitute the expressions for dm and ρ\rho and then integrate to get the moment of inertia of the two spheres.

IA=∫0R23(4πρAr4)dr=8πk3R∫0Rr5dr=8πkR518{I_A} = \int_0^R {{2 \over 3}(4\pi {\rho _A}{r^4})dr = {{8\pi k} \over {3R}}\int_0^R {{r^5}dr = {{8\pi k{R^5}} \over {18}}} } ,

IB=∫0R23(4πρBr4)dr=8πk3R5∫0Rr9dr=8πkR530{I_B} = \int_0^R {{2 \over 3}(4\pi {\rho _B}{r^4})dr = {{8\pi k} \over {3{R^5}}}\int_0^R {{r^9}dr = {{8\pi k{R^5}} \over {30}}} } .

Divide to get IB/IA=6/10{I_B}/{I_A} = 6/10 .

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