Class 11th

K.E. energy of electron = eV

Translational K.E. of N₂ =   $\frac{3}{2}KT$

So eV = $\frac{3}{2}KT$  

$1.6*10^{-19}*0.1=\frac{3}{2}*1.38*10^{-23}*T$                

T = 773 – 273 = 500°C

${\left(\frac{x}{4}-\frac{12}{x^2}\right)}^{12}$

According to question, 12 – 3r = 0 r = 4

$\Rightarrow \frac{3^6}{4^4}*55=\frac{3^6}{4^4}*k\Rightarrow k=55$

Since process is isochoric

So    $\Delta U=n{C}_v\Delta T$

$\Delta U=n\left(\frac{5}{2}R\right)\Delta T--(i)\left[C_V=\frac{5}{2}R\right]$      

And external work

$\Delta W=nR\Delta T--\left(ii\right)$    

$\frac{5}{2}=\frac{x}{10}\Rightarrow x=25.00$                

Force  = Aya   $\Delta T$

Force $=\left(10*10^{-4}\right)*\left(2*10^{11}\right)*10^{-5}*400$  

$F=8*10^{5}N$              

  $x*10^5=8*10^5$              

x = 8

VOWELS

vowels – 2, constants – 4

all the consonants never come together = 6! – 3! 4! =720 – 144 = 576

Since centres C₁ (1, 1) and C₂ (9, 1) lies opposite sides of the line3x + 4y = α

$((3+4-\alpha )((27+4-\alpha )<0\Rightarrow \alpha \in \left(7,31\right)..........\left(i\right)$

Also length of perpendicular from centre of the circle is greater than radius of the circle.

$\frac{\left|3+4-\alpha \right|}{5}\ge 1\Rightarrow \alpha \le 2or\alpha \ge 12............\left(ii\right)$

and $\frac{\left|27+4-\alpha \right|}{5}\ge 2\Rightarrow \alpha \le 21or\alpha \ge 41...........\left(iii\right)$

from (i), (ii) and (iii) we get $$ $\in$ [12, 21]

$\therefore$ sum of all integers = 165 

$l={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dx+}{\displaystyle \underset{-1/2}{\overset{1}{\int }}\left|2x\right|dx}$

let  $l_1={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dxput2x=t\Rightarrow dx=\frac{dt}{2}}$

$\therefore l=l_1+l_2=0+\frac{5}{8}=\frac{5}{8}\Rightarrow 8l=5$

$Y=\frac{Mg{L}^3}{4b{d}^3\delta }$

For significant error in Y

$=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta }{\delta }$              

$=\frac{1*10^{-3}}{2}+\frac{3*10^{-3}}{1}+\frac{10^{-2}}{4}+3*\frac{0.01*10^{-1}}{0.4}+\frac{10^{-2}}{5}$             

= 0.0155

$Y=\frac{Mg{L}^3}{4b{d}^3?}$              

For significant error in Y

$=\frac{?M}{M}+\frac{3?L}{L}+\frac{?b}{b}+3\frac{?d}{d}+\frac{??}{?}$           

$=\frac{1*10^{?3}}{2}+\frac{3*10^{?3}}{1}+\frac{10^{?2}}{4}+3*\frac{0.01*10^{?1}}{0.4}+\frac{10^{?2}}{5}$              

= 0.0155

$tan\theta =\frac{3x}{18}=\frac{x}{6}........\left(i\right)$ and

$tan2\theta =\frac{10x}{8}=\frac{5x}{9}..........\left(ii\right)$

Solving (i) and (ii)  $\frac{2tan\theta }{1-ta{n}^2\theta }=\frac{5x}{9}wegetx=\sqrt[]{\frac{72}{5}}$  

Height of pole = 10x =  $12\sqrt[]{10}$