Class 12th

Yes, students can get admission into University of Jammu BTech courses with 60% in Class 12. The minimum marks required are 50%. Anybody with 50% or above in Class 12 with PCM are eligible to apply.

  $z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

 for $\alpha \in Ry+2=0\Rightarrow y=-2$  

$x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$      = 0

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$

$af\left(x\right)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta }{x}............\left(i\right)$  

Replace x by   $\frac{1}{x}$

$af\left(\frac{1}{x}\right)+\alpha f\left(x\right)=\frac{b}{x}+\beta x..............\left(ii\right)$  

$a\left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)+\alpha \left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)=b\left(x+\frac{1}{x}\right)+\beta \left(x+\frac{1}{x}\right)$           

$\therefore \frac{f\left(x\right)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{b+\beta }{a+\alpha }=\frac{2}{1}=2$           

The best conductor of electricity is Silver (Ag) at room temperature. It is not used in normal wiring even though being the best conductor because of its high cost.

• Electrical Conductivity ($\sigma$): $6.3 \times 10^ {7} \ \text {S/m}$

• Electrical Resistivity ($\rho$): $1.59 \times 10^ {-8} \ \Omega \cdot \text {m}$ (the lowest among all metals)

α - sulphur & β - sulphur – Diamagnetic, S₂ – form is paramagnetic due to presence of unpaired electron in π* orbital like O₂.

  $\frac{2^{21}{\left(\frac{-1+i\sqrt[]{3}}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1-i}{\sqrt[]{2}}\right)}^{24}}+\frac{2^{21}{\left(\frac{1+\sqrt[]{3}i}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1+i}{\sqrt[]{2}}\right)}^{24}}=k$

=   ${\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+5-1\right)={\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+4\right)}}$

= ${\displaystyle \sum _{j=0}^5\left(j^2+9\widehat{j}+20\right)=\frac{5*6*11}{6}+\frac{9*5*6}{2}+20*6}$  

55 + 135 + 120 = 310

  $\Delta T_f=k_f.m$

$T_f^0-T_f=5.12*\frac{\left(\frac{10}{58}\right)}{\left(\frac{200}{1000}\right)}$

    5.5 – T_f = $\frac{5.12*5*10}{58}$  

  $T_f=1.086°C=\left(1.086\right)°C\cong 1°C$             

$3CH\equiv CH\left(g\right)\to C_6{H}_6\left(\mathcal{l}\right)$       

$\Delta G^0=-RT\mathcal{l}nK.........\left(i\right)$

$\Delta G^0=\sum \Delta {G^0}_P-\sum \Delta {G^0}_R..........\left(ii\right)$    

Equating (i) & (ii)

-2.303 RTlogk = 4.88 * 10⁵

$logK=\frac{-4.88*10^5}{2.303*R*T}=\frac{-488000}{5705.848}=-85.52=-855*10^{-1}$

So; magnitude of log K = 855 * 10^-1

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

$E_1=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}{\left[\frac{1}{H^{+}}\right]}^8$  

[H⁺] = 1M

$E_2=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}+\frac{0.059}{5}log10^{-32}$   

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 * 10^-4 V

$\frac{1.86}{93}=\frac{w}{135}$

W = 2.70 g

Mass produced actual = 2.70 $*\frac{90}{100}=2.43=243*10^{-2}g$