JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 21)
A square shaped hole of side l = $${a \over 2}$$ is carved
out at a distance d = $${a \over 2}$$ from the centre ‘O’ of
a uniform circular disk of radius a. If the
distance of the centre of mass of the
remaining portion form O is
$$ - {a \over X}$$ , value of X (to
the nearest integer) is :
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Answer
23
Explanation
Xcom = $${{\sigma \pi {a^2} \times 0 - \sigma .{{{a^2}} \over 4}.d} \over {\sigma \pi {a^2} - \sigma .{{{a^2}} \over 4}}}$$
= $${{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}$$
= $${{ - d} \over {4\pi - 1}}$$ = $${{ - a} \over {2\left( {4\pi - 1} \right)}}$$
$$ \therefore $$ $$ - {a \over X}$$ = $${{ - a} \over {2\left( {4\pi - 1} \right)}}$$
$$ \Rightarrow $$ X = $${2\left( {4\pi - 1} \right)}$$ = $${\left( {8\pi - 2} \right)}$$ = 23.12
So, nearest integer value of X = 23
= $${{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}$$
= $${{ - d} \over {4\pi - 1}}$$ = $${{ - a} \over {2\left( {4\pi - 1} \right)}}$$
$$ \therefore $$ $$ - {a \over X}$$ = $${{ - a} \over {2\left( {4\pi - 1} \right)}}$$
$$ \Rightarrow $$ X = $${2\left( {4\pi - 1} \right)}$$ = $${\left( {8\pi - 2} \right)}$$ = 23.12
So, nearest integer value of X = 23
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