Class 12th

$forA,t_{\frac{1}{2}}=4sec$  

$=m{A}_0{e}^{-}\frac{0.693}{4}*16$

$m_A=m_{A0}{e}^{-4*0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

$m_B=m_{B_0}{e}^{-2*0.693}$               -(2)

$\frac{m_A}{m_B}=\frac{m_{AO}}{m_{BO}}\frac{e^{-4}*0.693}{e^{-2}*0.693}$

$\frac{m_A}{m_B}=\frac{25}{100}=\frac{x}{100}x=25$

For maxima = y = (2n + 1) $\frac{\lambda }{2}\frac{D}{a}$

For 1^st maxima for l₁ wavelength (n = 1)

$y_1=\frac{3{\lambda }_1}{2}\frac{D}{a}$          - (1)

First maxima for l₂ wavelength

$y_2=\frac{3}{2}{\lambda }_2\frac{D}{a}$          - (2)

$\Rightarrow y_2-y_1=\frac{3}{2}\frac{D}{a}\left[{\lambda }_2-{\lambda }_1\right]$

$=\frac{3}{2}*\frac{2*5*10^{-9}}{0.5*10^{-3}}$

$=\frac{3}{10^{-3}}*10^{-8}$

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{rms}=\frac{l_0}{\sqrt[]{2}}$  

$V_{rms}=l_{rms}z$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(X_L\right)$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(2\pi *50*200*10^{-3}\right)$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(20\pi \right)$

$l_0=\frac{220\sqrt[]{2}}{20\pi }$

$l_0=\frac{11\sqrt[]{2}}{\pi }=\frac{\sqrt[]{a}}{\pi }$

a = 121 * 2

a = 242

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_B=\frac{V_B}{I_B}=\frac{10*10^{-3}}{10*10^{-6}}=10^3\Omega$  

$A_v=\left(\frac{\Delta l_C}{\Delta l_B}\right)*\left(\frac{R_C}{R_B}\right)$

=  $\frac{1.5*10^{-3}}{10*10^{-6}}*\frac{5*10^3}{10^3}$

A_v = 750

f₁ = 15 cm

$P_1=\frac{1}{15}*100=\frac{100}{15}D$

$P_3=\frac{100}{15}D$

$\frac{1}{f^2}=\left({\mu }_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$=\left(1.25-1\right)*\frac{-2}{R}$

$\frac{1}{f_2}=\frac{0.25*-2}{15}$

$\frac{1}{f_2}=\frac{-1}{30}c{m}^{-1}$

$\begin{array}{l}{p}_2=\frac{1}{f_2}=-\frac{100}{30}\\ P_R=P_1+P_2+P_3\\ P_R=\frac{100}{15}-\frac{100}{30}+\frac{100}{15}\\ P_R=10D\end{array}$

$P_R=\frac{1}{f_R}$

$f_R=\frac{1}{10}*100cm$

$f_R=10cm$

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)

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

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

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

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

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