Class 12th

Intersity a number of photons kinetic Energy a f

P(W) = $\frac{1}{3};$  P(B) =   $\frac{2}{3}$

$\Rightarrow p=\frac{1}{3};q=\frac{2}{3}$           and r = 4 or 5 and n = 5

Use   $P\left(r\right){=}^n{C}_r{p}^r{q}^{n-r}$

 P(4) + P(5)

${=}^5{C}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^1{+}^5{C}_5{\left(\frac{1}{3}\right)}^5$            

$=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}=\frac{11}{243}$  

$R=\frac{\sqrt[]{2Em}}{B.q}$

$\Rightarrow R*\frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_1}{R_2}=\frac{\sqrt[]{4}}{2}*\frac{3}{\sqrt[]{16}}=\frac{3}{4}$

$\Rightarrow R_2=\frac{4R_1}{3}$

$sin\theta =\frac{d}{R}\Rightarrow \theta \alpha \frac{1}{2}\Rightarrow {\theta }_2<{\theta }_1$

$\lambda >h\Rightarrow \lambda >400m$

$Q_1=C_{1}V=2V\mu C$

$Q_2=Q_3=\frac{C_2{C}_3}{C_2+C_3}V=\frac{6*12}{6+12}V=4V\mu C$

$Q_1:Q_2:Q_3=1:2:2$

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$dN=\left(\frac{N}{b-a}\right)dr$

$\therefore {\left(dB\right)}_{atcentre}=\frac{{\mu }_{0}ldN}{2r}$

$B=\frac{{\mu }_{0}lN}{2\left(b-a\right)}\mathcal{l}n\frac{b}{a}$

$\boxed{\begin{array}{lll}\text{12.(D)}& {\int }_0^{\frac{\pi }{6}}??\frac{4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \frac{2}{5}*2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta }{\left(4-4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \right)}=\frac{16}{5*8}\int \frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \mathrm{c}\mathrm{o}\mathrm{s}?d\theta }{{\mathrm{c}\mathrm{o}\mathrm{s}}^3?\theta }& x^{5/2}=2\mathrm{s}\mathrm{i}\mathrm{n}?\theta \\ & & \frac{5}{2}{x}^{3/2}dx=2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta \\ & & =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??{\mathrm{t}\mathrm{a}\mathrm{n}}^2?\theta d\theta =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??\left({\mathrm{s}\mathrm{e}\mathrm{c}}^2?\theta -1\right)d\theta =\frac{2}{5}[\mathrm{t}\mathrm{a}\mathrm{n}?\theta -\theta {]}_0^{\pi /6}=\frac{2}{5}\left(\frac{1}{\sqrt[]{3}}-\frac{\pi }{6}\right)\end{array}}$

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

$\Delta V=\left(10sin\omega t-3\right)volt$  

But in reverse biased condition of diode,

$\Delta V=0$ (across diode)

$\left(a-1\right)x+0y+z=\alpha$        ……(1)

$x+\left(b-1\right)y+0z=\beta$                       ……(2)

$0x+y+\left(c-1\right)z=\gamma$                       ……(3)

$\left|\begin{array}{ccc}\hfill \left(a-1\right)\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \left(b-1\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \left(c-1\right)\hfill \end{array}\right|=0$           For no unique solution D = 0

$\left(a-1\right)\left(b-1\right)\left(c-1\right)+1=0$            

$\therefore a=2;b=2;c=0$            

Hence, |a + b + c| = 4

V = 10 $\left(1?e^{?t/RC}\right)$ ${}^{}$

2 = 20 $\left(1?e^{?t/RC}\right)$ ${}^{}$

$\frac{1}{10}=1?e^{?t/RC}$

$e^{t/RC}=\frac{10}{9}$

$\frac{t}{RC}=ln\left(\frac{10}{9}\right)=0.105$

$C=\frac{t}{R*0.105}=\frac{10^{?6}}{10*0.105}=0.95?F$