JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 18)
A ring is hung on a nail. It can oscillate, without
slipping or sliding
(i) in its plane with a time period T1 and,
(ii) back and forth in a direction perpendicular to its plane,
with a period T2. The ratio $${{{T_1}} \over {{T_2}}}$$ will be :
(i) in its plane with a time period T1 and,
(ii) back and forth in a direction perpendicular to its plane,
with a period T2. The ratio $${{{T_1}} \over {{T_2}}}$$ will be :
$${{\sqrt 2 } \over 3}$$
$${2 \over {\sqrt 3 }}$$
$${2 \over 3}$$
$${3 \over {\sqrt 2 }}$$
Explanation
_5th_September_Evening_Slot_en_18_1.png)
Moment of inertia in case (i) is I1
Moment of inertia in case (ii) is I2
I1 = 2MR2
I2 = $${1 \over 2}M{R^2}$$ + MR2 = $${3 \over 2}M{R^2}$$
T1 = $$2\pi \sqrt {{{{I_1}} \over {Mgd}}} $$
and T2 = $$2\pi \sqrt {{{{I_2}} \over {Mgd}}} $$
$$ \Rightarrow $$ $${{{T_1}} \over {{T_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}} $$
= $$\sqrt {{{2M{R^2}} \over {{3 \over 2}M{R^2}}}} $$
= $${2 \over {\sqrt 3 }}$$
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