Class 11th

Bond strength $\propto$  Bond order

$N_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2$  

$O_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*1}$

$C_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2$  

$B_2\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1\equiv {\pi }_{2py}^1$  

NO ® Number of electron = 7 + 8 = 15

B.O. Similar to  $N_2^{-}$  

$N_2^{-}\to {\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^{*}$  

B.O. of N₂ = 3     B.O of C₂ = $\frac{8-4}{2}=2$  

Removal of e^- form antibonding molecular orbital increases bond order.

In NO & O₂ has valance e^- in p orbital.

Using conversation of momentum in direction perpendicular to the original direction of motion,

mv₁ sin 30° = mv₂ sin 30°

$\Rightarrow \frac{v_1}{v_2}=1$

f = W = 0.5 * 10 = 5 N

N = F

For block not to slide,

  $f\le \mu N$            

$\begin{array}{l}\Rightarrow 5\le 0.2F\\ \Rightarrow F\ge 25N\end{array}$

Ratio of masses on two pistons of the hydraulic lift equals to that of their cross- section area.

$\frac{A_1}{A_2}=\frac{100}{m}$            

Now,   $\frac{4^2{A}_1}{A_2/4^2}=\frac{M}{m}\Rightarrow M=256\frac{A_1}{A_2}.m=25600kg.$  

If   $\mu$ is Poisson's ratio,

Y = 3K (1 - 2  $\mu$ )      ……… (1)

and Y = 2  $n\left(1+\mu \right)$ ……… (2)

With the help of equations (1) and (2), we can write

$\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$            

$\overrightarrow{F}=10\widehat{i}+5\widehat{j}$

$\overrightarrow{a}=\frac{10}{m}\widehat{i}+\frac{5}{m}\widehat{j}$

$=\frac{10}{0.1kg}\widehat{i}+\frac{5}{0.1}\widehat{j}$

$\overrightarrow{a}=100\widehat{i}+50\widehat{j}$

$a_x=100,a_y=50$

$S_x=ut+\frac{1}{2}a{t}^2$

$=0*2+\frac{1}{2}*100*2*2$

Sx = 200 m

$\frac{a}{b}=\frac{200}{100}=2$

There are three main processes Isothermal, adiabatic and cyclic process. In isothermal, the system is thermally conductive and the temperature remains constant. In adiabatic process, the system is thermally isolated and there is no change in heat temperature. The system returns to its initial stage in the cyclic process.

$W_{AB}=nRTln\frac{2V_1}{V_1}=nRTln2.$

$W_{BC}=P_2\left(V_1-2V_1\right)=-P_2{V}_1=-\frac{1}{2}nRT$                         

[At B, 2P₂ V₁ = nRT]

$W_{CA}=0$   [CA is isochoric process].

$W_{ABCA}=W_{AB}+W_{BC}+W_{CA}=nRT\left(\mathcal{l}n\left(2\right)-\frac{1}{2}\right)$            

2²? ¹ - 1 - 1 = 126
2²? ¹ = 2?
2n + 1 = 7
n = 3