Class 12th

f : {1, 3, 5, 7, ….,99} {2, 4, 6, 8, …….100}

$f\left(3\right)\ge f\left(9\right)\ge f\left(15\right)......\ge f\left(99\right)$

$3+\left(n-1\right)6=99\Rightarrow n=17$

cases f (3) > f (9) > f (15) ……. > f (99)

$\therefore$ from the set {2, 4, 6, …., 100}

17 distinct numbers can be selected in 50C17 ways again remaining {1, 5, 7, 11, ….} can map in 33! ways

$\therefore$ total number of such required functions

${=}^{50}{C}_{17}*33!$

$=\frac{50!}{33!17!}*33!{=}^{50}{P}_{33}$

Liquation refining process is applicable for the metal having low m.p, but containing impurities have higher m.p

Higher adsorbtion of gas is corresponds to higher liquefaction and higher liquefaction is directly proportional to the higher critical temperature.

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill -3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill {\lambda }^2-\left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}9\\ -18\\ 16\end{array}\right)$

for no solution

$\frac{9}{10}\left({\lambda }^2-\left|\lambda \right|+3\right)-7=0$

$\Rightarrow 9{\lambda }^2-9\left|\lambda \right|-43=0$

$\Rightarrow \left|\lambda \right|=\frac{9+\sqrt[]{81+36*43}}{18}>0$

$\therefore$ only two possible value of $\lambda$ are there.

$\left|z-3\sqrt[]{2}\right|+\left|z-p\sqrt[]{2}i\right|$ is minimum for z,  $3\sqrt[]{2}\&p\sqrt[]{2}i$ are collinear.

$\iff {\left(3\sqrt[]{2}\right)}^2+{\left(p\sqrt[]{2}\right)}^2={\left(5\sqrt[]{2}\right)}^2$

$\Rightarrow 18+2p^2=50$

$2p^2=32$

$p=\pm 4$

According to definition of wave number, we can write

$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\Rightarrow 1-\frac{1}{n^2}=\frac{1}{\lambda R}$$$

$\Rightarrow \frac{1}{n^2}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}\Rightarrow n=\sqrt[]{\frac{\lambda R}{\lambda R-1}}$

$H=l^{2}Rt$

$\mathrm{\%}errorinH=\frac{\Delta H}{H}*100=2\left(\frac{\Delta l}{l}\right)*100+\left(\frac{\Delta R}{R}\right)*100+\left(\frac{\Delta t}{t}\right)*100=2*2+1+3=8\mathrm{\%}$

$R_{AB}=5+\left(10//5\right)+10=15+\frac{10*5}{10+5}=15+\frac{10}{3}=\frac{55}{3}k\Omega$

$V_{AB}=R_{AB}l=\left(\frac{55}{3}*10^3\right)*\left(15*10^{-3}\right)=275V$

f = $\frac{qB}{2\pi m}=\frac{1.6*10^{-19}*1.0*10^{-4}}{2*3.14*9*10^{-31}}=0.028*10^{8}Hz=2.8*10^{6}Hz$

This phenomenon works on resonance in AC circuit.