Class 12th

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \mathrm{n}=100$

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

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

${\mathrm{V}}_{\mathrm{P}}={\mathrm{V}}_{\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}}$

$\begin{array}{rr}& =\left(\frac{\mathrm{K}\mathrm{q}}{5-3}+\frac{\mathrm{K}(-\mathrm{q})}{5+3}\right)*{10}^2\\ & =\left(\frac{\mathrm{K}\mathrm{q}}{2}-\frac{\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\\ & =\left(\frac{3\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\end{array}$

Yes, candidates can get admission in Janki Devi College of Hotel Management with 60% in Class 12. The college has not specified the minimum required aggregate for admission in the courses offered, however since 60% is a good score, candidates can get admission in Janki Devi College of Hotel Management. 

(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = -2 (3a-4b+12c+19)/ (3²+ (-4)²+12²)
(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = (-6a+8b-24c-38)/169
(x, y, z) = (a–6, β, γ)
(a-b)-a)/3 = (β-b)/ (-4) = (γ-c)/12 = (-6a+8b-24c-38)/169
(β-b)/ (-4) = -2
=> β = 8+b
=> 3a – 4b + 12c = 150 . (i)
a + b + c = 5
=> 3a + 3b + 3c = 15 . (ii)
Applying (i) – (ii), we get :
= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137