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

# Six identical particles each of mass m are arranged at the corners of a regular hexagon of side length a. If the mass of one of the particle is doubled, the shift in the centre of mass is

1. A

a

2. B

6a/7

3. C

a/7

4. D

a/$\sqrt{3}$

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### Solution:

m1 = 6m m2 = m
Shift in CM
$=\frac{{\mathrm{m}}_{2}\mathrm{d}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}=\frac{\mathrm{m}\left(\mathrm{a}\right)}{6\mathrm{m}+\mathrm{m}}=\frac{\mathrm{ma}}{7\mathrm{m}}=\frac{\mathrm{a}}{7}$

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