Class 12th

 $\beta =12mm=12*10^{-3}m$

$\begin{array}{l}\beta \text{'}=\frac{\beta }{\mu }=\frac{12*10^{-3}}{4/3}=9*10^{-3}m\\ \beta \text{'}=9mm\end{array}$

$\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist &   $a\in I.$

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$

exist

RHL =   $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$

Following method are useful:

1.Check re-evaluation or supplementary exam option 

2. Contact Shiksha peeth college and ask for refund if there is refund policy.

Or explain your situation and ask them to hold your seat until you pass supplementary exam 

There are botany offical cutoff released by college but in most colleges it is 50% to 60% is the eligibility criteria.

Hope for best

$E=IA$

$=100*1*10^{-4}$

$=10^{-2}W$

$E=nh\frac{c}{\lambda }$

$10^{-2}=\stackrel{?}{n}*\frac{6.64*10^{-34}*3*10^8}{900*10^{-9}}$

$\stackrel{?}{n}=\frac{10^{-2}*9*10^{-7}}{6.64*10^{-34}*3*10^8}$

$\stackrel{?}{n}=4.5*10^{16}$

$A=Ao{e}^{-\lambda t}$

$\Rightarrow 2250=4250e^{-\lambda t}$

$\Rightarrow \lambda *10*0.434=0.274$

$\Rightarrow \lambda =\frac{0.274}{10*0.434}=0.63mi{n}^{-1}$

${\lambda }_1=6MHz=6*10^{6}Hz$

${\lambda }_1=\frac{C}{{\upsilon }_1}=\frac{3*10^8}{6*10^6}=50m$

${\lambda }_2=\frac{C}{{\upsilon }_2}=10*10^6=\frac{3*10^8}{10*10^6}=30m$

Wavelength bandwidth

$=\left|{\lambda }_1-{\lambda }_2\right|=20m$

x + 2y + z = 2

$\alpha x+3y-z=\alpha$

$-\alpha x+y+2z=-\alpha$

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=1\left(6+1\right)-2\left(2\alpha -\alpha \right)+1\left(\alpha +3\alpha \right)=7+2a$

$\alpha =-\frac{7}{2}$

$Mn{O}_4^{-2}\to A+B$  

Oxidation state of Mn in B < A 

$Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

B is MnO₂

$\therefore$  Oxidation state of Mn = +4

${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

  $\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$  

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

$=3.87\approx 4$