Class 12th

$k_{\alpha }=2eV\Rightarrow P_{\alpha }=\sqrt[]{2K_{\alpha }{m}_{\alpha }}$

k_P = eV P_P = $\sqrt[]{2K_P{m}_P}$

$\therefore \frac{P_{\alpha }}{P_P}=\sqrt[]{\frac{K_{\alpha }{m}_{\alpha }}{K_P{m}_P}}=\sqrt[]{\frac{2*\left(4m_P\right)}{m_P}}\begin{array}{cc}\hfill \sqrt[]{8}:1\hfill & \hfill 2\sqrt[]{2}:1\hfill \end{array}$

$I=I_{0}co{s}^{2}30°$

$=I_0{\left(\frac{\sqrt[]{3}}{2}\right)}^2=\frac{3}{4}{I}_0$ 

Find total (5 digit) number of divisors of 7. Find total (5 digit) number of divisors of 35

v = 100 sin $\omega t$

$i_0=\frac{100}{R_x}=5\Rightarrow R_x=20\Omega$

i = $5sin\left(\omega t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

i = 100 5sin $\omega t$

$5=\frac{100}{R_y}\Rightarrow R_y=\frac{100}{5}=20\Omega$

When x and y both are connected in series :-

v = 100sin $\omega t$

tan = 1 = 45°

R₀ = $\sqrt[]{R_x^2+R_y^2}=20\sqrt[]{2}\Omega$

$\therefore \stackrel{o}{{\mathcal{l}}_0}=\frac{v_0}{R_0}=\frac{100}{20\sqrt[]{2}}=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\sqrt[]{2}$}\right.A.$

$\therefore {\mathcal{l}}_{ms}=\frac{{\mathcal{l}}_0}{\sqrt[]{2}}$

$=\frac{5}{\sqrt[]{2}\cdot \sqrt[]{2}}$

$=\frac{5}{2}A=\frac{5}{2}A$

$\left(\overrightarrow{a}*\overrightarrow{b}\right)*\widehat{i}=\left(\overrightarrow{a}.\widehat{i}\right)\overrightarrow{b}-\left(\overrightarrow{b}.\widehat{i}\right)\overrightarrow{a}$

$\left(\left(\overrightarrow{a}*\overrightarrow{b}\right)*\widehat{i}\right).\widehat{k}=\left(\overrightarrow{a}.\widehat{i}\right)\left(\overrightarrow{b}.\widehat{k}\right)-\left(\overrightarrow{b}.\widehat{i}\right)\left(\overrightarrow{a}.\widehat{k}\right)$

F_x-y = $\frac{{\mu }_0{i}_1{i}_2}{2\pi r}*\left(.5\right)$

$=\frac{2*10^{-7}*6*0.5}{0.05}$

$=\frac{6}{5}*10^{-5}N=1.2*10^{-5}$

Force on Y = 1.2 * 105N towards 'x'

  $\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill \beta \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 4\hfill \end{array}\right|=\left(4+5\beta \right)\widehat{i}+\left(3\beta -4\alpha \right)\widehat{j}+\left(-5\alpha -3\right)\widehat{k}$                

Compare with  $\overrightarrow{a}*\overrightarrow{b}=-\widehat{i}+9\widehat{j}+12\widehat{k}$  

b = -1, a = -3

Projection of  $\overrightarrow{b}-2\overrightarrow{a}on\overrightarrow{b}+\overrightarrow{a}$  

$=\frac{\left(\overrightarrow{b}-2\overrightarrow{a}\right).\left(\overrightarrow{b}+\overrightarrow{a}\right)}{\left|\overrightarrow{b}+\overrightarrow{a}\right|}$   

$\frac{dy}{dx}-y=x,$  linear differential equation.

IF = e^-x

$y.e^{-x}={\displaystyle \int x{e}^{-x}dx=-e^{-x}\left(x+1\right)+C}$                

$y=-\left(x+1\right)+C.e^x$                

$y_2\left(x\right)=-\left(x+1\right)+2e^x$                

y₂ > y₁, no solution.

${\left(x+2\sqrt[]{3}\right)}^2+y^2=4^2$                

$y^2=8\left(x+\frac{1}{2}\right)$

Point of intersection are (0, 2) and (0, -2)

  $2{\displaystyle \underset{0}{\overset{2}{\int }}\left[\left(\sqrt[]{16-y^2}-2\sqrt[]{3}\right)-\frac{\left(y^2-4\right)}{8}\right]dy}$              

  $=\frac{1}{3}\left(8\pi +4-12\sqrt[]{3}\right)$

Let f(x) = 8 sin x - sin 2x

$f\text{'}\left(x\right)=8cosx-2cos2x$              

$f\text{'}\text{'}\left(x\right)=-8sinx+4sin2x$               

= -8 (sin x) (1 – cos x)

f''(x) < 0, f'(x) is a decreasing function

  ^{$f\text{'}\left(\frac{\pi }{3}\right)<f\text{'}\left(x\right)<f\text{'}\left(\frac{\pi }{a}\right)$}               

  $5<f\text{'}\left(x\right)<\frac{8}{\sqrt[]{2}}$

${\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}dx<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{f\left(x\right)}{x}<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{8}{\sqrt[]{2}}}}}$

$5\pi 12<l<\frac{8}{\sqrt[]{2}}.\frac{\pi }{\sqrt[]{2}}$