Class 12th

Kindly go through the solution

Answer (12.00)

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having manufacturing of N-based fertizers)

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

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

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

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

$\therefore \frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt[]{I_{S_4}}+\sqrt[]{I_{S_3}}\right)}^2}{{\left(\sqrt[]{I_{S_4}}-\sqrt[]{I_{S_3}}\right)}^2}={\left(\frac{2+1}{2-1}\right)}^2=\frac{9}{1}$

${\lambda }_1=\frac{h}{\sqrt[]{2m{E}_1}}={\lambda }_2=\frac{h}{\sqrt[]{2m{E}_2}}$

$=E_2=\frac{4}{9}{E}_1=4eV$

$E_2=E_1-e{V}_0=V_0=5V$

Energy of 1 photon = $\frac{hc}{\lambda }$  

Energy of 1 mole of photon =  $N_A*\frac{hc}{\lambda }$  

$=\frac{6.022*10^{23}*6.63*10^{-34}*3*10^8}{300*10^{-9}}$  

=0.399 * 10⁶ J = 399KJ

$i=\frac{v_0}{R}\left(1-e^{\frac{tR}{L}}\right)$

$U=\frac{1}{2}L{i}^2$

$\frac{dU}{dt}=Li\frac{di}{dt}$

$2L*\frac{R}{L}*\frac{v_0^2}{R^2}\left(1-e^{\frac{tR}{L}}\right)e^{\frac{tR}{L}}=\frac{v_0^2}{R}\left(1-e^{\frac{tR}{L}}\right)$

$t=\frac{L}{R}In2$

$f\left(A,B\right)=\frac{1-sinB+sinA-sinAsinB+cosAcosB}{cosA\left(1-sinB\right)}$

$\begin{array}{l}\Rightarrow f\left(A,B\right)\left(cosA-cosAsinB-cosB+sinAcosB\right)\\ =2sinA-2sinB\end{array}$

$\Rightarrow f\left(A,B\right)=\frac{2\left(sinA-sinB\right)}{cosA-cosB+sin\left(A-B\right)}=2g\left(A,B\right)$

$x^2\left(y+z\right)y^2\left(z+x\right)z^2\left(x+y\right)=a^3{b}^3{c}^3=x^3{y}^3{z}^3$

$\Rightarrow \left(x+y\right)\left(y+z\right)\left(z+x\right)=xyz$

$\begin{array}{l}\Rightarrow x^2\left(y+z\right)+y^2\left(z+y\right)+z^2\left(x+y\right)+xyz=0\\ \Rightarrow a^3+b^3+c^3+abc=0\end{array}$

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

$?f\text{'}(3)=0$

So f' (x) = 0 has minimum 7 roots in $x?\left[0,6\right]?f\text{'}\text{'}\left(x\right)$  has min 6 roots in   $x?\left[0,6\right]$

h (x) = f' (x) . f' (x)

h' (x) = (f' (x)² + f' (x) f' (x)

h (x) = 0 has 13 roots in x $?$   [0, 6]

h' (x) = 0 has 12 roots in x $?$ [0, 6]