$(a)\;5.76\times10^{-14}\;m\qquad(b)\;57.6\times10^{-14}\;m\qquad(c)\;0.576\times10^{-14}\;m\qquad(d)\;None$

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Answer : (a) $\;5.76\times10^{-14}\;m$

Explanation :

Since there is no external force

$\bigtriangleup T\;.E=0$

$\bigtriangleup K\;.E + \bigtriangleup P\;.E=0$

$ \bigtriangleup K\;.E=10\;MeV$

$\large\frac{9\times10^{9}\times100\times1.6\times10^{-19}\times4\times1.6\times10^{-19}}{r}=10^{7}\times1.6 \times10^{-19}$

$r=9\times10^2\times100\times1.6\times10^{-19}\times4$

$r=57.6\times10^{-15}\;m\;.$

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