Class 12th

$\frac{si{n}^{-1}x}{a}=\frac{co{s}^{-1}x}{b}=\frac{ta{n}^{-1}y}{c}=\lambda$

$\therefore \frac{si{n}^{-1}x}{a}=\lambda$

$\therefore si{n}^{-1}x=a\lambda$

$x=sina\lambda .........\left(i\right)$

$x=cosb\lambda ..........\left(ii\right)$

$y=tanc\lambda ..........\left(iii\right)$

From (i) & (ii), we get, $sina\lambda =cosb\lambda =sin\left(\frac{\pi }{2}-b\lambda \right)$

$\therefore a\lambda =\frac{\pi }{2}-b\lambda \Rightarrow \left(a+b\right)\lambda =\frac{\pi }{2}\Rightarrow \lambda =\frac{\pi }{2\left(a+b\right)}...........\left(iv\right)$

From (iii) y = tan c $\lambda$

$\Rightarrow cos\left(\frac{2c\pi }{2\left(a+b\right)}\right)=\frac{1-y^2}{1+y^2}\Rightarrow cos\left(\frac{\pi c}{a+b}\right)=\frac{1-y^2}{1+y^2}$

$y=\overline{\overline{A}+B}=A.\overline{B}$

All the charge given to a conducting sphere resides on outer surface.

  $e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$            

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$        

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$       

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2*\frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}*\frac{\sqrt[]{3}}{2}}=\frac{1}{2}$

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6*\frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6*\frac{24}{25}eV..........\left(1\right)$

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6*\frac{24}{25}*1.6*10^{-19}}{3*10^8*1.67*10^{-27}}=4.17m/s$

$R_i=\frac{\rho \mathcal{l}}{A}$

$R_f=\frac{\rho \left(1.25\mathcal{l}\right)}{\left(A/1.25\right)}={\left(1.25\right)}^2*\frac{\rho \mathcal{l}}{A}$

$\Rightarrow R_f=1.5625*R_i$

$\Rightarrow \frac{R_f-R_l}{R_l}*100=56.25\mathrm{\%}$

$l=\frac{1}{2}c{\epsilon }_0{E}_0^2=\eta *\left(\frac{P}{4\pi r^2}\right)$ Here $\eta$ is the efficiency

$E_0=\sqrt[]{\frac{\eta P}{2\pi r^{2}c{\epsilon }_0}}=\sqrt[]{\frac{1.25}{100}*\frac{1000}{2*3.14*4*3*10^8*8.85*10^{-12}}}$

$\Rightarrow E_0=\sqrt[]{\frac{12.5}{8*3.14*3*8.85*10^{-4}}}=13.69Vm=136.9*10^{-1}V/m$

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$           

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$        

$\therefore f\left(x\right)$ is continuous $\forall x\in R$  

Bv = B sin 60°

$B_v=2.5*10^{-4}*\frac{\sqrt[]{3}}{2}$

$Emf=B_v*v*\mathcal{l}=2.5*10^{-4}*\frac{\sqrt[]{3}}{2}*180*\frac{5}{18}*1=108.25*10^{-3}volts$