JEE Advanced 2020 — Paper 1 — Question 10

Area of the strip = (t.dr)

$\mathrm{dR}=\rho\left(\frac{\pi \mathrm{r}}{\mathrm{tdr}}\right)$

$\sum \frac{1}{\mathrm{dR}}=\int_{\mathrm{R}_{1}}^{\mathrm{R}_{2}} \frac{\mathrm{ddr}}{\pi \rho \mathrm{r}}$

$\frac{1}{\mathrm{R}_{\text {eq }}}=\left(\frac{\mathrm{t}}{\pi \rho}\right) \mathrm{L}_{\mathrm{n}}\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)$

$R_{e q}=\frac{\pi \rho}{t \times \ell n\left(\frac{R_{2}}{R_{1}}\right)}$

So, $I=\frac{V_{0}}{R_{\text {eq }}}=\left(\frac{V_{0} t}{\pi \rho}\right) \ln \left(\frac{R_{2}}{R_{1}}\right)$

So, option (A) is correct.

Electrons are revolving in a circular path so there is a centripetal force on the electron, which provide a

transverse electric field. The direction of transverse electric field is radially out side hence inner surface

is at higher potential.

So, option (C) is correct.

Current density $J(r)=\frac{d l}{(t d r)}$

$J(r)=\frac{V_{0}}{(d r)(t d r)}$

$J(r)=\frac{V_{0}}{\left(\frac{\rho \pi r}{t d r}\right)(t d r)}$

$J(r)=\frac{V_{0}}{(\rho \pi r)}$

$\mathrm{I}=\mathrm{ne}^{\mathrm{A}} \mathrm{V}_{\mathrm{d}}$

$\mathrm{J}(\mathrm{r})=\mathrm{neV}_{\mathrm{d}}$

$\Rightarrow V_{d}=\left(\frac{J(r)}{n e}\right)=\left(\frac{V_{0}}{\rho \pi r n e}\right) …(i)$

Centripetal force

$$\begin{gathered} e E_{T}=\frac{m_{e} v_{d}^{2}}{r} …(ii) \end{gathered}$$

From (i) and (ii) $E_{T} \propto V_{0}^{2}$

$$\begin{gathered} \Delta V=\int E_{T} d r …(iii) \end{gathered}$$ $$\begin{gathered} \Rightarrow \Delta V \propto E_{T} …(iv) \end{gathered}$$

From (iii) and (iv)

$\Delta \mathrm{V} \propto \mathrm{V}_{0}^{2}$

$I \propto V_{0}$

Hence $\Delta V \propto I^{2}$

Hence Option (D) is correct