Class 11th

$N{a}_{2}O+H_{2}O\to 2NaOH$

w = 20 g

$V_{H_{2}O}=500mL$        

Mole of Na₂O=   $\frac{20}{62}$

$?$ 1 mole of Na₂O gives 2 mole of NaOH

$\therefore \left(\frac{20}{62}\right)$ mole of Na₂O gives $\left(2*\frac{20}{62}\right)$ moles of NaOH

$=\frac{20}{31}mole$    

Molarity of NaOH solution

$=\frac{\frac{20}{31}*100}{500}M=\frac{20}{31}*2M=\frac{40}{31}M$

= 1.29 M

=  $12.9*10^{-1}M\approx 13*10^{-1}M$

Ans. = 13

Zn (30) = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$  

$Z{n}^{+}=\left[Ar\right]4s^{1}3d^{10}$      

Outermost electron is 4s electron,

$n=4,\mathcal{l}=0,m=0,s=\pm 1/2$         

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H₂O

30 * 1 mmol      50 * 1 mmol     

0 mmol               20 mmol

${\left[H^{+}\right]}_{mix}=\frac{20}{50+30}M=\frac{20}{80}M=\frac{1}{4}M=0.25M$

$x*10^{-4}=6021*10^{-4}$    

$\therefore$ x = 6021

Ans. = 6021

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$=0-\frac{1}{2}m{V}^2$              

As 90% of K.E. is losed by friction so that

$-\frac{90}{100}\left(\frac{1}{2}m{V}^2\right)-\frac{1}{2}K{x}^2=-\frac{1}{2}m{V}^2$                

-K -> -16 * 10⁵

K = 16 * 10⁵

$O_2^{-2}=8*2+2=18e^{-}$

${\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2={\pi }_{2py}^2{\pi }_{2px}^{*2}={\pi }_{2py}^{*2}$      

Number of unpaired e^- = 0

Ans. = 0

$x?\left(x\right)={\displaystyle \underset{}{\overset{}{\int }}\left(3t^2-2?\left(t\right)\right)dt,x>-2,By}$  Leibniz theorem

$?\left(x\right)+x?\text{'}\left(x\right)=3x^2-2?\text{'}\left(x\right)\Rightarrow ?\text{'}\left(x\right)+\frac{?\left(x\right)}{x+2}=\frac{3x^2}{x+2}..............\left(i\right)$

$\therefore I.F.=e^{{\displaystyle \int \frac{dx}{x+2}}}=x+2$

Multiplying (i) by I.F. we get,

$\therefore ?\left(x\right)=\frac{x^3+8}{x+2}$

$\therefore ?\left(2\right)=4$

$l=\frac{m{\mathcal{l}}^2}{3}$  

Energy conservation Low

$mg\mathcal{l}=\frac{1}{2}\frac{m{\mathcal{l}}^2}{3}{\omega }^2.....\left(i\right)$                

And speed V =    $\omega r=\omega \mathcal{l}$

then    $V=\sqrt[]{6g\mathcal{l}}=\sqrt[]{6*10*.6}$

->6 m/s

for smooth surface

$a=gsin30°=\frac{g}{2}$

$S_1=ut+\frac{1}{2}a{t}^2$         

$S_1=\frac{1}{2}\frac{g}{2}{t}^2=\frac{g}{4}{t}^2.......\left(i\right)$               

for rough Surface

$S=\frac{1}{2}\frac{g}{2}\left(1-\mu \sqrt[]{3}\right){\alpha }^2{t}^2........\left(ii\right)$               

By (i) and (ii)

$\mu =\frac{1}{\sqrt[]{3}}\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)x=\sqrt[]{3}$            

$t_r=\left(3r+4\right)\left(2r+6\right)=6r^2+26r+24$

$\therefore {\displaystyle \sum _{r=1}^{10}{t}_r}=\frac{6*10*11*21}{6}+\frac{26*10*11}{2}+24*10=10*398$

$Mean=\frac{10*398}{10}=398$

For A satellite T₁ = 1hour

$So{\omega }_1=2\pi red/hour$                

for B satellite T₂ = 8 hour

given R₁ = 2 * 10³ Km

Relative  $\omega =\frac{V_1-V_2}{R_2-R_1}=\frac{2\pi *10^3}{6*10^3}$  

$\Rightarrow \frac{\pi }{3}$ rad/hour

 x = 3