
${{\rm{V}}_C} = \varepsilon \left[ {1 - {{\rm{e}}^{ - t/{\rm{RC}}}}} \right]$
At $t=100 \mathrm{\,ms}$
${{\rm{V}}_{\rm{R}}} = {{\rm{V}}_{\rm{C}}} \Rightarrow {{\rm{e}}^{ - t/{\rm{RC}}}} = 1 - {{\rm{e}}^{ - t/{\rm{RC}}}} \Rightarrow 2{{\rm{e}}^{ - t/{\rm{RC}}}} = 1$
$ \Rightarrow {{\rm{e}}^{ - t/{\rm{RC}}}} = 1/2 \Rightarrow {{\rm{e}}^{ - t/{\rm{RC}}}} = 2 \Rightarrow \frac{{\rm{t}}}{{{\rm{RC}}}} = \ell {\rm{n}}2$
$ \Rightarrow \frac{{100}}{{{\rm{RC}}}} = \ell {\rm{n}}2 \Rightarrow {\rm{RC}} = \frac{{100}}{{\ell {\rm{n}}(2)}} = 145.45\,{\rm{ms}}$
(as $\ln 2 = 0.693 \simeq 0.7$)


$(A)$ $I =\frac{V_0 t}{\pi \rho} \ln \left(\frac{R_2}{R_1}\right)$
$(B)$ the outer surface is at a higher voltage than the inner surface
$(C)$ the outer surface is at a lower voltage than the inner surface
$(D)$ $\Delta V \propto I ^2$

Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.
