Class 11th

The law of inertia, also known as Newton's first law of motion, states that an object at rest will stay at rest. Likewise, an object in motion will remain in motion with the same speed and in the same direction unless there is a net external force. 

Now in both ideal states, the net external force on an object is zero.

From the perspective of classical mechanics, there isn't a significant distinction between rest and uniform motion. They can be seen as the same state of motion viewed from different reference frames. An object is considered to be in a state of equilibrium whether it is at rest or in uniform motion.

All given oxide have nitrogen – nitrogen bond except N₂O₅ as ;

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

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