Class 12th

In Hall Heroult process electrolysis is carried out using carbon anode. The liberated

O2 at anode reacts with anode.

Cathode : $A{l}^{3+}\left(melt\right)+3ek\to A\mathcal{l}\left(\mathcal{l}\right)$

Anode : $\begin{array}{l}C\left(s\right)+O^{2-}\left(Melt\right)\to CO\left(g\right)+2e\\ C\left(s\right)+2O^{2-}\left(Melt\right)\to C{O}_2\left(g\right)+4e\end{array}$

Net Reaction 2Al2O3 + 3C $4A\mathcal{l}+3C{O}_2$

Penicillin is not a broad spectrum artibiotic and is used to treat only certain infections caused by streptococci and staphylococci such as pneumonia.

$Cellulose\underset{}{\overset{H^{+}/H_{2}O}{\to }}\beta -D-Glucose\underset{Water}{\overset{B{r}_2}{\to }}Gluconicacid$

$\frac{dy}{dx}+2ytanx=2sinx$  

  $I.F.=e^{2{\displaystyle \int tanxdx}}=se{c}^{2}x$             

$\therefore$  Solution y sec²x =   $2{\displaystyle \int sinxse{c}^{2}xdx=2{\displaystyle \int secxtanxdx}}$

y sec² x = 2sec x + c

y = 2 cos x + c cos²x passes $\left(\frac{\pi }{4},0\right)=B$     C =   $-2\sqrt[]{2}$

y = 2 cos x  $2\sqrt[]{2}co{s}^{2}x$ $\therefore {\displaystyle \underset{0}{\overset{\pi /2}{\int }}ydx}=2-\frac{\pi }{\sqrt[]{2}}$                              

Kindly consider the following figure

$T_{90\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{10}$

$T_{50\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{50}$

$\frac{T_{90\mathrm{\%}}}{T_{50\mathrm{\%}}}=\frac{log10}{log2}=\frac{1}{0.301}=3.32$

Kindly consider the following figure

Kindly consider the following figure

Based on theory 

 $\frac{l_1}{I_2}=\frac{9}{4}$ $\frac{l_{max}}{l_{max}}=\frac{l_1+l_2+2\sqrt[]{l_1{l}_2}}{l_1+l_2-2\sqrt[]{l_1{l}_2}}$

$=\frac{\frac{9}{4}+1+2\sqrt[]{\frac{9}{4}}}{\frac{9}{4}+1-2\sqrt[]{\frac{9}{4}}}$

$=\frac{\frac{13}{4}+\frac{2*3}{2}}{\frac{13}{4}-2*\frac{3}{2}}=\frac{25}{1}$