Energy corresponding to wavelength 4000 $\stackrel{o}{A}$

$E=\frac{hc}{\pi }$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}\times 1.6\times {10}^{-19}}$ eV

$=\frac{12400}{4000}$

$=3.1$ eV

$\approx$3 eV

The Correct Answer is Option (B)

$\begin{array}{rl}& {\mathrm{R}}_{\mathrm{eq}}=\frac{150\times 150}{300}+25=30\mathrm{\Omega }\\ \\ & {\mathrm{I}}_1=\frac{5}{10}=0.05\mathrm{A}\\ \\ & {\mathrm{I}}_2={\mathrm{I}}_4=\frac{0.05}{2}=0.025\mathrm{A}\\ \\ & \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=2\end{array}$

The Correct Answer is Option (D)

Till $|V|\le 0.6V$

$|V_0|=|V|$

So correct graph will be D.

$R=\frac{20\times 20}{40}+15=25\mathrm{\Omega }$

D₁ : conducting

D₂ : open circuit

$i=\frac{1-0.6}{60+40}A$

$=\frac{0.4}{100}A$

$\Rightarrow i=4$ mA