Class 12th

$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22*2*4*10^{-7}}{9*108}$

B₀ = 42.9 * 10^-9

42.9 Ans

Plane containing both line $x=1$

Image of point $(\mathrm{2,1},3)$  in the plane $\frac{\alpha -2}{1}=\frac{\beta -1}{0}=\frac{\gamma -3}{0}=-2\left(1\right)$

$\alpha =0,\beta =1,\gamma =3;\alpha +\beta +\gamma =4$

$\lambda =\frac{h}{p}or\lambda =\frac{h}{mv}$

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ or 4 : 1

  $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

Equation of any plane passing through the line of intersection of planes (1) and (2)

$x-cy-bz+\lambda (cx-y+az)=0$

$(1+\lambda c)x-(c+\lambda )y+(-b+a\lambda )z=0$

If (3) &  (4) is same plane.

$\frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$

(i)

(ii) (iii)

By (i)  & (ii) $\lambda =-\frac{(a+bc)}{ac+b}$

By (ii) & (iii)

$\lambda =\frac{-(ab+c)}{1-a^2};a^2+b^2+c^2+2abc=1$

Amplitude is proportional to the slit – width, thus,

^{$\frac{A_1}{A_2}=3$}

^{$\frac{l_{max}}{l_{min}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}=\frac{{\left(3+1\right)}^2}{{\left(3-1\right)}^2}=4.$}

$h{f}_1=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R*\frac{8}{9}$

$h{f}_2=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R*\frac{3}{4}$

$\frac{f_1}{f_2}=\frac{8/9}{3/4}=\frac{32}{27}$

f2 = f1 * $\left(\frac{27}{32}\right)$

$=\left(2.92*10^{15}\right)*\left(\frac{27}{32}\right)$

$=2.46*10^{15}Hz$

Kindly consider the following figure

For mirror

Mean $=np$ , variance $=npq$

$np-npq=1;p+q=1$

$n^2{p}^2-n^2{p}^2{q}^2=11$

We get   $q=\frac{5}{6},p=\frac{1}{6},n=36\mathrm{}$  Probability of ' 3 ' success $={}^{36}{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{33}$

$\bar{a}+\bar{b}+\bar{c}+\bar{d}=0$

Magnitude of all vectors will be same as well as angle between these vectors will also be same. $\bar{a}+\bar{b}+\bar{c}=-\bar{d}$

$\begin{array}{c}|\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{b}+2\bar{b}\cdot \bar{c}+2\bar{c}\cdot \bar{a}=|d{|}^2\\ \mathrm{}1+1+1+2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =1\\ 6\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-2\\ \mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-\frac{1}{3}\\ \alpha ={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\left(\frac{-1}{3}\right)=\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\frac{1}{3}\end{array}$