Class 12th

In the integral J, substitute x + 1 = t

$\Rightarrow dx=dtand{x}^2+2x=\left(t^2-1\right)$            

Now   $J={\displaystyle \underset{1}{\overset{e}{\int }}\frac{e^{\frac{t^2-2}{2}}}{t}}dtandK={\displaystyle \underset{1}{\overset{e}{\int }}t}lnt{e}^{\frac{t^2-2}{2}}dt$

Hence   $\left(J+K\right)={\displaystyle \underset{1}{\overset{e}{\int }}{e}^{\frac{t^2-2}{2}}}\left(\frac{1}{t}+tlnt\right)dt$

$={\left(e^{\frac{t^2-2}{2}}lnt\right)}_{t=1}^{t=e}=e^{\frac{e^2-2}{2}}={\left(\sqrt[]{e}\right)}^{e^2-2}$

TSA  of cube $=6a^2$

$\frac{d}{dt}\left(6a^2\right)=4.8;12a\frac{da}{dt}=4.8;a\frac{da}{dt}=0.4;\frac{dv}{dt}=\frac{d}{dt}\left(a^3\right)\Rightarrow 3a\left(a\frac{da}{dt}\right)$

$3*15*0.4=3*15*\frac{04}{10}\Rightarrow 18$

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)