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Four particle each of mass 1kg are at the four corners of a square of side 1m. The work done to move one of the particles to infinity

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a

$2 \sqrt{2} G$

b

3G

c

$(\frac{2 \sqrt{2} + 1}{\sqrt{2}}) G$

d

$(2 \sqrt{2} + 1) G$

answer is B.

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Detailed Solution

$U_{i} = (\frac{- G (1) (1)}{1}) 4 + (\frac{- G (1) (1)}{\sqrt{2}}) 2$
Question Image
$\begin{array}{l} U_{f} = (\frac{- G (1) (1)}{1}) 2 + (\frac{- G (1) (1)}{\sqrt{2}}) (1) \\ w = U_{f} - U_{i} \\ = (- 2 G - \frac{G}{\sqrt{2}}) - (- 4 G - \sqrt{2} G) \\ = G (\frac{2 \sqrt{2} + 1}{\sqrt{2}}) \end{array}$

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