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# Six identical particles each of mass $'m'$ are arranged at the corners of a regular hexagon of side length . If the mass of one of the particle is doubled, the shift in the centre of mass is:A) $L$B) $\dfrac{{6L}}{7}$C) $\dfrac{L}{7}$D) $\dfrac{L}{{\sqrt 3 }}$

Last updated date: 13th Jun 2024
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Hint: We can find the shift within the center of mass because the product of mass removed and therefore the distance between original center of mass and away from each other when divided by the mass of the remaining portion is also known as centre of mass.

The center of mass of the primary arrangement is six equal masses on the six corners of a regular hexagon side length and is at the center of the hexagon. Then the new arrangement if the mass of one of the particles is doubled, and find the value of the shift in the center of mass.
The total masses will be $6m + m = 7m$.
When, the six identical particles are present at six different positions of the hexagon. So, the moment will be $2m \times L$ for that shifted particle. So, the total will be
$\dfrac{2mL+\dfrac{-mL}{2}+\dfrac{-mL}{2} +mL-mL}{7m}$
$\Rightarrow \dfrac{{2mL - mL}}{{7m}}$
Which one is solving we will get the value of position of the center is,
$\Rightarrow \dfrac{L}{7}$
Since, the masses are the same for all positions except the shifted mass of $2m$ hence the moment can be added.

Hence, the correct answer is option (C).