QJEEAdv08Ph141d: JEE Advanced 2008 Physics - Class 12 JEEAdv08Ph141d Solution: For a = 0, the value of d (maximum value of as shown in the…

For a = 0, the graph is as shown, the equation for the graph line is

The charge in the dotted element shown in figure is

$dq = \sigma \times 4\pi {r^2}dr$

$\therefore$ $dq = \left( {d - {d \over R}r} \right)4\pi {r^2}dr$

$\Rightarrow Ze = \int_0^R {4\pi d{r^2}dr - \int_0^R {{{4\pi d} \over R}{r^3}dr} }$

$Ze = 4\pi d{{{R^3}} \over 3} - {{4\pi d} \over R}{{{R^4}} \over 4}$

$\therefore$ ${{Ze} \over {4\pi d{R^3}}} = {1 \over 3} - {1 \over 4} = {1 \over {12}}$

$\therefore$ $d = {{3Ze} \over {\pi {R^3}}}$