Class 12th

$F_{\text{convex}}=f$

${\mathrm{F}}_{\text{concave}}=-\mathrm{f}$

$\begin{array}{rr}& \frac{1}{{\mathrm{f}}_{\text{comb}}}=\frac{1}{{\mathrm{F}}_{\text{convex}}}+\frac{1}{{\mathrm{F}}_{\text{concave}}}\\ & =\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{f}}\end{array}$

$\Rightarrow {\mathrm{f}}_{\text{comb}}=\mathrm{\infty }$

$\frac{1}{{\mathrm{f}}_1}=(1.6-1)\left(\frac{1}{\mathrm{\infty }}-\frac{1}{20}\right)=-\frac{3}{100}$

$\frac{1}{{\mathrm{f}}_2}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)$ $=\frac{1}{20}$

$\frac{1}{f_3}=\frac{1}{f_1}=-\frac{3}{100}$

$\frac{1}{{\mathrm{f}}_{\text{net}}}=\frac{1}{{\mathrm{f}}_1}+\frac{1}{{\mathrm{f}}_2}+\frac{1}{{\mathrm{f}}_3}$

$=-\frac{3}{50}+\frac{1}{20}$

$\Rightarrow {\mathrm{f}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=-100\mathrm{c}\mathrm{m}$

Q = ω? L/R ⇒ Δω = R/L = 50/4 = 8rad/sec
Δω
ω? = 1/√LC = 1/√ (5*80*10? ) = 50rad/sec
ω? = ω? - Δω/2 = 46rad/sec
ω? = ω? + Δω/2 = 54rad/sec

$\omega =2\pi \mathrm{f}\Rightarrow \omega =100\pi$

$\mathrm{Z}=\sqrt[]{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$

$=\sqrt[]{{10}^2+{\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)}^2}$

$=\sqrt[]{100+{\left(100\pi *\frac{50}{\pi }*{10}^{-3}-\frac{1}{100\pi *\frac{{10}^3}{\pi }*{10}^{-6}}\right)}^2}$

$=\sqrt[]{100+(5-10{)}^2}$

$=\sqrt[]{100+25}$

$\mathrm{Z}=5\sqrt[]{5}\mathrm{\Omega }$

Kindly go through the solution 

eE = e (dV/dr) = mω²r
ΔV = ∫dV = ∫ (m/e)ω²rdr from R to 2R
ΔV = (3mω²R²)/ (2e)

$\left|\stackrel{?}{\mathrm{F}}\right|=\left|\mathrm{I}\right(\stackrel{?}{\mathrm{L}}*\stackrel{?}{\mathrm{B}}\left)\right|$

$\begin{array}{rr}& =\left|\mathrm{I}\right[\mathrm{L}\stackrel{\mathrm{ˆ}}{\mathrm{i}}*(2\stackrel{\mathrm{ˆ}}{\mathrm{i}}+3\stackrel{\mathrm{ˆ}}{\mathrm{j}}-4\stackrel{\mathrm{ˆ}}{\mathrm{k}})\left]\right|\\ & =5\mathrm{I}\mathrm{L}\end{array}$

${\stackrel{?}{\mathrm{B}}}_{\mathrm{P}}={\stackrel{?}{\mathrm{B}}}_{\text{upper wire}}\otimes +{\stackrel{?}{\mathrm{B}}}_{\text{semi-circle}}?+{\stackrel{?}{\mathrm{B}}}_{\text{lower-wire}}\otimes$

$B_P=-\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4R}-\frac{{\mu }_{0}i}{4\pi R}=\frac{{\mu }_0\mathrm{i}}{4\mathrm{R}}\left(1-\frac{2}{\pi }\right)$ pointing away from the page

$i_{\text{series}}=\frac{E}{R_{\text{series}}}$

$\Rightarrow {\mathrm{i}}_{\text{series}}=\frac{\mathrm{E}}{10\mathrm{R}}$

${\mathrm{i}}_{\text{parallel}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\text{Parallel}}}$

$=\frac{\mathrm{E}}{\mathrm{R}/10}$

${\mathrm{i}}_{\text{parallel}}=\mathrm{n}*{\mathrm{i}}_{\text{series}}$

$\Rightarrow \frac{10\mathrm{E}}{\mathrm{R}}=\frac{\mathrm{n}\mathrm{E}}{10\mathrm{R}}$

$\Rightarrow \mathrm{n}=100$

$\mathrm{R}={\mathrm{R}}_0(1+\alpha \mathrm{\Delta }\mathrm{T})$

$\begin{array}{rr}& \Rightarrow 6.8=2[1+\alpha *(80-0\left)\right]\\ & \Rightarrow \alpha =\frac{3.4-1}{80}\\ & =0.03\\ & =3*{10}^{-2}{}^{?}{\mathrm{C}}^{-1}\end{array}$