JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 9)
The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2.
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14.9 kg/m3
7.5 $$\times$$ 101 kg/m3
7.5 $$\times$$ 102 kg/m3
1.49 $$\times$$ 102 kg/m3
Explanation
Parallel axis theorem
I = ICM + Md2
$$I = {{M{r^2}} \over 2} + M{\left( {{L \over 2}} \right)^2}$$
$$2.7 = M{{{{(0.2)}^2}} \over 2} + M{\left( {{{0.8} \over 2}} \right)^2}$$
$$2.7 = M\left[ {{2 \over {100}} + {{16} \over {100}}} \right]$$
M = 15 kg
$$ \Rightarrow \rho = {M \over {\pi {r^2}L}} = {{15} \over {\pi {{(0.2)}^2} \times 0.8}}$$
= 0.1492 $$\times$$ 103
I = ICM + Md2
$$I = {{M{r^2}} \over 2} + M{\left( {{L \over 2}} \right)^2}$$
$$2.7 = M{{{{(0.2)}^2}} \over 2} + M{\left( {{{0.8} \over 2}} \right)^2}$$
$$2.7 = M\left[ {{2 \over {100}} + {{16} \over {100}}} \right]$$
M = 15 kg
$$ \Rightarrow \rho = {M \over {\pi {r^2}L}} = {{15} \over {\pi {{(0.2)}^2} \times 0.8}}$$
= 0.1492 $$\times$$ 103
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