Rotational Motion

JEE Main level, numerical answer, medium

38 of 280

Question 38 of 280

Two discs of same mass and different radii are made of different materials such that their thicknesses are 1cm1cm and 0.5cm0.5cm respectively. The densities of materials are in the ratio 3:53:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of x6\dfrac{x}{6}. The value of xx is ______.

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Answer: 5

Mass of the disc in terms of density can be written as, m=ρ(πR2t)m=\rho \left(\pi {R}^{2}t\right).
⇒R2=mρπt\Rightarrow {R}^{2}=\dfrac{m}{\rho \pi t}
Now moment of inertia of the disc about the diameter is, mR24\dfrac{m{R}^{2}}{4}.
Therefore,
I=mR24=m24ρπt⇒I∝1ρtI=\dfrac{m{R}^{2}}{4}=\dfrac{{m}^{2}}{4\rho \pi t} \Rightarrow I\propto \dfrac{1}{\rho t}
Required ratio, I1I2=ρ2t2ρ1t1=53×0.51=56\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{{\rho }_{2}{t}_{2}}{{\rho }_{1}{t}_{1}}=\dfrac{5}{3}\times \dfrac{0.5}{1}=\dfrac{5}{6}
Hence, x=5x=5.

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