For$RC$ charging circuit as shown in figure the switch is closed at $t=0$ It is found that at $t=t_1$ the charge on capacitor is $\frac{\mathrm{CE}}{2}\text{ (at }t=0$ the capacitor is uncharged while $t_2$ is the time taken for the charge on the capacitor to increase from $\frac{CE}{2}\text{ to CE, }$ then mark the correct statements. 

 

For RC charging circuit., charge on capacitor as a function of time is given by $q=CE\left[1-e^{-t/RC}\right]$ It is given that $q=\frac{CE}{2}\text{ at }t=t_1$ , Substjtuting these values in above equation we get 

$\frac{CE}{2}=CE\left[1-e^{\frac{t_1}{RC}}\right]\Rightarrow t_1=RC\ln 2$

It is a known fact that initially charging Is fast and then charging slows down exponentially and complete charging will take infinite time. Hence  $t_1<t_2$.