JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 16)
The radius of gyration of a uniform rod of length $$l$$, about an axis passing through a
point $${l \over 4}$$ away from the centre of the rod,
and perpendicular to it, is :
$${1 \over 8}l$$
$${1 \over 4}l$$
$$\sqrt {{7 \over {48}}} l$$
$$\sqrt {{3 \over 8}} l$$
Explanation
_7th_January_Morning_Slot_en_16_1.png)
I = $${{M{l^2}} \over {12}} + M{\left( {{l \over 4}} \right)^2}$$
$$ \Rightarrow $$ I = $${{7M{l^2}} \over {48}}$$
$$ \Rightarrow $$ MK2 = $${{7M{l^2}} \over {48}}$$
$$ \Rightarrow $$ K = $$\sqrt {{7 \over {48}}} l$$
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