Class 12th

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}*2$

$\Rightarrow 300*10^{-6}=\frac{4\pi *10^{-7}}{2\pi *4*10^{-2}}*2$                                                                           

$\Rightarrow$ I = 30 A

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Let a and b are the roots of $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$  

  $\therefore \alpha +\beta >0and\alpha \beta >0$ Also, it has a common root with x² + 2x – 8 = 0

$\therefore$  The common root between above two equations is 4.

  $\Rightarrow 16\left(p^2+q^2\right)-8q\left(p+r\right)+q^2+r^2=0\Rightarrow \left(16p^2-8pq+q^2\right)+\left(16q^2-8qr+r^2\right)=0$  

      $\Rightarrow {\left(4p-q\right)}^2+{\left(4q-r\right)}^2=0\Rightarrow q=4pandr=16p\therefore \frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272$  

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$

$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$

The light wave contains two lights of different frequencies, so

$E_1=h{\upsilon }_1=\frac{4.14*10^{-15}*6*10^{15}}{2\pi }=3.96eV,and$  

$E_2=h{\upsilon }_2=\frac{4.14*10^{-15}*9*10^{15}}{2\pi }=5.92eV$  

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

$tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}*\frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

$S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$  

Þ $si{n}^2\theta =co{s}^2\theta$               $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$  

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5*10^{10}}{2.0*10^{10}}=\frac{3}{4}$                    

According to question, we can write

$\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

$0\le \frac{2+3p}{6}\le 1\Rightarrow p\in \left[-\frac{2}{3},\frac{4}{3}\right],0\le \frac{2-p}{8}\le 1\Rightarrow p\in \left[-6,2\right]and0\le \frac{1-p}{2}\le 1\Rightarrow p\in \left[-1,1\right]$  

  $0<P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)\le 10\le \frac{13}{12}-\frac{p}{8}\le 1\Rightarrow p\in \left[\frac{2}{3},\frac{26}{3}\right]$ Taking intersection to all $p\in \left[\frac{2}{3},1\right]$  

$\therefore p_1+p_2=\frac{5}{3}$