The gravitational potential energy of interactions of a system of six identical particles,each of mass $m$ placed at the vertices of a regular hexagon of side $4 a^{'}$ is [PE=0 at infinite separation]

- \frac{G m^{2}}{a} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})

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\frac{- 3 G m^{2}}{a} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})

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\frac{- 6 G m^{2}}{a} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})

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\frac{- 2 G m^{2}}{a} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})

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Solution

The correct option is C $\frac{- 6 G m^{2}}{a} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})$
$Explanation:-$

Total potential energy is

$V = \sum U_{i j}$

$= \frac{- G m^{2}}{a} [6 \times (\frac{1}{a})] + g (\frac{1}{\sqrt{3 a}}) + 3 (\frac{1}{2 a})$

$= \frac{- 6 G m^{2}}{a} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})$

$Hence the correct option is C$

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The gravitational potential energy of interactions of a system of six identical particles,each of mass m placed at the vertices of a regular hexagon of side 4a′ is [PE=0 at infinite separation]

The correct option is C −6Gm2a(1+1√3+14)Explanation:- Total potential energy is V=∑Uij=−Gm2a[6×(1a)]+g(1√3a)+3(12a)=−6Gm2a(1+1√3+14)Hence the correct option is C

The gravitational potential energy of interactions of a system of six identical particles,each of mass $m$ placed at the vertices of a regular hexagon of side $4 a^{'}$ is [PE=0 at infinite separation]