JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 16)
A circular hole of radius $$\left( {{a \over 2}} \right)$$ is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :
_24th_February_Evening_Shift_en_16_1.png)
$${1 \over 6}a$$
$${2 \over 3}a$$
$${5 \over 6}a$$
$${10 \over 11}a$$
Explanation
_24th_February_Evening_Shift_en_16_2.png)
Let $$\sigma$$ is the surface mass density of disc.$${X_{com}} = {{(\sigma \times \pi {a^2} \times a) - (\sigma {{\pi {a^2}} \over 4} \times {{3a} \over 2})} \over {\sigma \pi {a^2} - {{\sigma \pi {a^2}} \over 4}}}$$
$$ \Rightarrow $$ $${X_{com}} = {{a - 3{a \over 8}} \over {1 - {1 \over 4}}}$$
$$ \Rightarrow $$ $${X_{com}} = {{{{5a} \over 8}} \over {{3 \over 4}}}$$
$$ \Rightarrow $$ $${X_{com}} = {{5a} \over 6}$$
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