Rotational Motion

JEE Main level, single correct, medium

50 of 280

Question 50 of 280

The solid cylinder of length 80cm80cm and mass MM has a radius of 20cm.20cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CDCD parallel to ABAB as shown in figure is 2.7kgm22.7kg{m}^{2}.

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Answer: (A) 1.49×102kgm−31.49\times {10}^{2}kg{m}^{-3}

I=Mr22+ML24=(ρπr2L)r22+(ρπr2L)L24I=\dfrac{M{r}^{2}}{2}+\dfrac{M{L}^{2}}{4}=\dfrac{\left(\rho \pi {r}^{2}L\right){r}^{2}}{2}+\left(\rho \pi {r}^{2}L\right)\dfrac{{L}^{2}}{4}
=ρπr2L(r22+L24)=\rho \pi {r}^{2}L\left(\dfrac{{r}^{2}}{2}+\dfrac{{L}^{2}}{4}\right)
⇒ρ=Iπr2L(r22+L24)\Rightarrow \rho =\dfrac{I}{\pi {r}^{2}L\left(\dfrac{{r}^{2}}{2}+\dfrac{{L}^{2}}{4}\right)}
On putting the values, we get ρ=1.49×102kgm−3\rho =1.49\times {10}^{2}kg{m}^{-3}.

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