A solid cylinder of length L = 80 , text{cm} | vedclass

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(D) Given: Length $L = 0.8 \, 	ext{m}$,Radius $r = 0.2 \, 	ext{m}$,Moment of Inertia $I = 2.7 \, 	ext{kg m}^2$.Using the parallel axis theorem,the

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$149$

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(D) Given: Length $L = 0.8 \, 	ext{m}$,Radius $r = 0.2 \, 	ext{m}$,Moment of Inertia $I = 2.7 \, 	ext{kg m}^2$.Using the parallel axis theorem,the moment of inertia about axis $CD$ is given by:$I = I_{CM} + Md^2$Here,$I_{CM}$ is the moment of inertia about the central axis $AB$,which is $\frac{Mr^2}{2}$,and the distance $d$ between the axes is $\frac{L}{2}$.$I = \frac{Mr^2}{2} + M\left(\frac{L}{2}ight)^2$$2.7 = M \left[ \frac{(0.2)^2}{2} + \left(\frac{0.8}{2}ight)^2 ight]$$2.7 = M [0.02 + 0.16] = M(0.18)$$M = \frac{2.7}{0.18} = 15 \, 	ext{kg}$The density $ho$ is given by $ho = \frac{M}{V} = \frac{M}{\pi r^2 L}$.$ho = \frac{15}{\pi 	imes (0.2)^2 	imes 0.8} = \frac{15}{\pi 	imes 0.04 	imes 0.8} = \frac{15}{0.032 \pi} \approx 149.2 \, 	ext{kg/m}^3$.Rounding to the nearest option,the density is $149 \, 	ext{kg/m}^3$.

$A$ solid cylinder of length $L = 80 \, \text{cm}$ and mass $M$ has a radius $r = 20 \, \text{cm}$. Calculate the density of the material used if the moment of inertia of the cylinder about an axis $CD$ parallel to the central axis $AB$ (as shown in the figure) is $2.7 \, \text{kg m}^2$.

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