Six identical particles each of mass 'm' are arranged at the corners of a regular hexagon of side length "L”. If the mass of one of the particles is doubled, the shift in the centre of mass?

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6L/7

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L/7

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Solution

The correct option is C
L/7

For regular hexagon having identical masses at the six vertices, the centre of mass of the system will be at its centre.if the mass of one of particle is doubled.

New position of centre of mass from 'C' is

$x_{c m} = \frac{60 (0) + m (L)}{6 m + m}$

$x_{c m} = \frac{m L}{7 m}$

$∴ S h i f t i n t h e c e n t r e o f m a s s = \frac{L}{7}$

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