A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is $ρ .$ An electron (charge e, mass m) is released inside the cavity from point P as shown in figure. The center of sphere and center of cavity are separated by a distance a. The time after which the electron again touches the sphere is

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$\sqrt{\frac{6 \sqrt{2} r ε_{0} m}{e p a}}$

$\sqrt{\frac{\sqrt{2} r ε_{0} m}{e p a}}$

$\sqrt{\frac{6 r ε_{0} m}{e p a}}$

$\sqrt{\frac{r ε_{0} m}{e p a}}$

answer is A.

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Detailed Solution

Acceleration of electron $A = e E / m$ (in backward direction).

and $E = \frac{ρ a}{3 ε_{0}} o r A = \frac{e p a}{3 ε_{0} m}$

Distance traveled by electron where it hits the wall of cavity is $S = 2 r \cos θ = \frac{r}{\sqrt{2}}$

Hence using $S = \frac{1}{2} A t^{2}$

$t = \sqrt{\frac{6 \sqrt{2} r ε_{0} m}{e ρ a}}$

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