Physics - Electrostatics Question with Solution | TestHub

$\left(\mathrm{A}\right)$$\Rightarrow h>2Rr>R$

$\varphi =\frac{Q}{{\epsilon }_0}$
Clearly from Gauss' Law$\Rightarrow$Option$\left(\mathrm{A}\right)$is correct.
$\Rightarrow$for$h=2rR=\frac{4R}{5}$

$\mathrm{s}\mathrm{h}\mathrm{a}\mathrm{d}\mathrm{e}\mathrm{d}\text{\hspace{0.17em}charge}=2\pi (1-\mathrm{c}\mathrm{o}\mathrm{s}{53}^{\mathrm{o}})\times \frac{Q}{4\pi }=\frac{Q}{5}$
Where,$2\pi (1-\mathrm{c}\mathrm{o}\mathrm{s}53°)=\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{d}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}$
$\therefore q_{\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}}=\frac{2Q}{5}\left(\frac{Q}{5}\mathrm{a}\mathrm{b}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}\frac{Q}{5}\mathrm{b}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\right)$
$\therefore \varphi =\frac{2Q}{5{\epsilon }_0}$
$\therefore forh>2R,r=\frac{4R}{5}$
$\therefore \varphi =\frac{2Q}{5{\epsilon }_0}\Rightarrow$Option$\left(\mathrm{D}\right)$is incorrect.
$\left(\mathrm{C}\right)\Rightarrow$suppose$h=\frac{8R}{5}\text{\hspace{0.17em}���\hspace{0.17em}}r=\frac{3R}{5}$

$\varphi =0$(as no charge is enclosed inside the cylinder, as shown in diagram)
So for$h<\frac{8R}{5},\varphi =0\Rightarrow$Option$\left(\mathrm{C}\right)$is correct.
$\left(B\right)\Rightarrow$like option$\left(\mathrm{D}\right)$for$h=2R,r=\frac{3R}{5}$
$q_{\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}}=2\times 2\pi \left(1-\mathrm{c}\mathrm{o}\mathrm{s}{37}^{\mathrm{o}}\right)\frac{Q}{4\pi }=\frac{Q}{5}$
$\therefore \varphi =\frac{Q}{5{\epsilon }_0}\Rightarrow$Option$\left(\mathrm{B}\right)$is correct.