Class 12th

$\text{27.(32)}\begin{array}{rr}& I=2{\int }_0^{\pi /2}??\left({\mathrm{c}\mathrm{o}\mathrm{s}}^3?x+{\mathrm{c}\mathrm{o}\mathrm{s}}^4?x+{\mathrm{s}\mathrm{i}\mathrm{n}}^4?x\right)dx\\ & I=2\left[{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^3?xdx+{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^4?xdx+{\int }_0^{\pi /2}??{\mathrm{s}\mathrm{i}\mathrm{n}}^4?xdx\right]\\ & I=2\left[\frac{2}{3}*1+\frac{3*1}{4*2}*\frac{\pi }{2}+\frac{3*1}{4*2}*\frac{\pi }{2}\right]\\ & I=2\left[\frac{2}{3}+\frac{3\pi }{8}\right]\Rightarrow I=\frac{4}{3}+\frac{6\pi }{8}\Rightarrow 24I=32+18\pi \\ & 24I-18\pi =32\end{array}$

Kindly consider the following figure

Kindly consider the following figure

Clearly all equations can be $\lambda x-2y-6z=0$  form where $\lambda =1,\frac{4}{3},\frac{6}{5}$  

So for $x=0y=-3z$

So $x^2+y^2+z^2=10z^2$  (as  ) $x=0\mathrm{}y=-3z$

ATQ $10\le 10z^2\le 200$

$1\le z^2\le 20\Rightarrow z=\pm 1,\pm 2,\pm 3,\pm 4$

we know, = $\frac{\lambda }{d}$ $\frac{}{}$ $\mu =\frac{C}{V}=\frac{C}{\upsilon \lambda }$ $\frac{}{}\frac{}{}$

${}_{}\frac{}{{}_{}{}_{}}$  ${\lambda }_1=\frac{C}{{\mu }_1{\upsilon }_1}$ [with change in medium frequency remain constant]

${\lambda }_2=\frac{C}{{\mu }_2{\upsilon }_2}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mu }_1}{{\mu }_2}=\frac{5}{7}$ ${\lambda }_2=\frac{5}{7}{\lambda }_1$

${\theta }_1=\frac{{\lambda }_1}{d_1}$

2= $\frac{{\lambda }_2}{d_2}$

$\frac{{\theta }_2}{{\theta }_1}=\frac{{\lambda }_2}{{\lambda }_1}[?d_1=d_2]$

2 $=\frac{5}{7}*{\theta }_1$

${\theta }_2=\frac{5}{7}*0.35$

Equation of tangent of at $y=\mathrm{l}\mathrm{o}\mathrm{g}?x$  at $\left(e^{\lambda },\lambda \right)$  is $y-\lambda =\frac{1}{e^{\lambda }}\left(x-e^{\lambda }\right)$

It cuts y   axis at $(0,\lambda ,-1)$

Also normal to $y^2=8x$    at $\left(\mathrm{7,4}\right)$    is $y-4=-(x-2)$  

  $y=-x+6\mathrm{}$  Cuts y axis at $\left(\mathrm{0,6}\right)\mathrm{}$  SO ATQ $6=-1+\lambda \Rightarrow \lambda =7$

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

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

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

B₀ = 42.9 * 10^-9

42.9 Ans

Plane containing both line $x=1$

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

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