Class 12th

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

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

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

$\frac{E}{H}=\frac{Volt/metre}{Ampere/metre}=\frac{Volt}{Ampere}=ohm$

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

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

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

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

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

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{t{|}_{t_1|2}}$

$\Rightarrow N=10^{10}*{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{10^{10}}{\sqrt[]{2}}?7*10^9$

$V_{AB}=\frac{5}{5}*4=4V$

$\Rightarrow Q=CeqV=2x=8\mu c$

Emf = $\frac{-d?}{dt}$ $\frac{}{}$

When magnet passes through the centre rogion of loop, no emf is induced and the polarity of emf will be negative while entering the loop.

dA = 2r dr

$dq=\sigma *2\pi rdr$

$\Rightarrow E={\displaystyle \underset{0}{\overset{R}{\int }}\frac{kdqz}{{\left(z^2+r^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt[]{R^2+z^2}}\right]$

$\frac{dy}{dx}+\frac{e^y-2x}{2x^2}=0$  

$e^{-y}\frac{dy}{dx}-\frac{e^{-y}}{x}=-\frac{-1}{2x^2}$              

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

 I. F. $=e^{{\displaystyle \int \frac{dx}{x}}}=e^{lnx}=x$  

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$              

Biuret test is given by proteins and dipeptide & compounds having following groups

Hence positive test given by tripeptide & biuret.

Monomer of novolac is

Molecular mass of monomer unit

              = 12 * 7 + 1 * 8 + 16 * 2

              = 84 + 8 + 32

              = 124 g

Molecular mass of polymer = 963 g

$\therefore$ n * mass of monomer = mass of polymer + (n – 1) * 18

(n – 1) unit of H₂O is removed during formation

$\therefore 124*n=963+18\left(n-1\right)$

106n = 945

 $\therefore n=\frac{945}{106}=8.91\approx 9$