JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 22)
Moment of inertia of a cylinder of mass M,
length L and radius R about an axis passing
through its centre and perpendicular to the
axis of the cylinder is
I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$. If such a cylinder is to be made for a given mass of a material, the ratio $${L \over R}$$ for it to have minimum possible I is
I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$. If such a cylinder is to be made for a given mass of a material, the ratio $${L \over R}$$ for it to have minimum possible I is
$${3 \over 2}$$
$$\sqrt {{3 \over 2}} $$
$$\sqrt {{2 \over 3}} $$
$${{2 \over 3}}$$
Explanation
_3rd_September_Morning_Slot_en_22_1.png)
Given I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$
M = $$\rho $$.V = $$\rho $$$$\pi $$R2L
$$ \Rightarrow $$ R2 = $${M \over {\rho \pi L}}$$
$$ \therefore $$ I = $$M\left( {{M \over {4\rho \pi L}} + {{{L^2}} \over {12}}} \right)$$
$$ \Rightarrow $$ $${{dI} \over {dL}}$$ = $$M\left( {{M \over {4\rho \pi }}\left( { - {1 \over {{L^2}}}} \right) - {{2L} \over {12}}} \right)$$
For minimum I, $${{dI} \over {dL}} = 0$$
$$ \therefore $$ $$M\left( {{M \over {4\rho \pi }}\left( { - {1 \over {{L^2}}}} \right) - {{2L} \over {12}}} \right)$$ = 0
$$ \Rightarrow $$ $${{{M^2}} \over {4\rho \pi {L^2}}} = {{2LM} \over {12}}$$
$$ \Rightarrow $$ $${M \over {4\rho \pi {L^2}}} = {L \over 6}$$
$$ \Rightarrow $$ $${{\rho \pi {R^2}L} \over {4\rho \pi {L^2}}} = {L \over 6}$$
$$ \Rightarrow $$ $${{{R^2}} \over {{L^2}}} = {2 \over 3}$$
$$ \Rightarrow $$ $${R \over L} = \sqrt {{2 \over 3}} $$
$$ \Rightarrow $$ $${L \over R} = \sqrt {{3 \over 2}} $$
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