Class 11th

Sum of elements $A\cap \left(B\cup C\right)=274*400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

$\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100*1093}{2}......\left(i\right)$          

Another possible number is 9k + 5 forms are {104, .995}

$\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}*1099.........\left(ii\right)$  

$\therefore Total=\frac{100}{2}*\left[1093+1099\right]=100*1096=274*4*100=274*400$           

$\therefore$  possible value of $\mathcal{l}$ = 5

T$1_{}=2\pi \sqrt[]{\frac{m}{k}}$

$k_{eq}=\frac{k*k}{k+k}=\frac{k}{2}$

$T_2=2\pi \sqrt[]{\frac{m}{\left(k/2\right)}}=2\pi \sqrt[]{\frac{2m}{k}}$

$\Rightarrow \frac{T_2}{T_1}=\sqrt[]{2}and\frac{k_2}{k_1}=\frac{1}{2}$

$I_{AB}=\frac{2}{5}m{a}^2*2+\left(\frac{2}{5}m{a}^2+m{b}^2\right)*2=\frac{8m{a}^2}{5}+2m{b}^2$

$\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$  

 Let sin x = t, t $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$  

 Minimum value of a for which solution exist = 9

  $z+\alpha \left|z-1\right|+2i=0$

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

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

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

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

From volume conservation

$n*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R^3=n{r}^3.........\left(1\right)$

Decrease in surface area = $n*4\pi r^2-4\pi R^2$

$\left(\Delta A\right)=4\pi \left[n{r}^2-R^2\right]=4\pi \left[\frac{n*r^3}{r}-R^2\right]=4\pi \left[\frac{R^3}{r}-R^2\right]=4\pi R^3\left[\frac{1}{r}-\frac{1}{R}\right]$

Energy released (W) = $T*\Delta A=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$

Heat produced (Q) = $\frac{W}{J}=\frac{4\pi T{R}^3}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$

$Now,Q=ms\Delta \theta$

$\Rightarrow \frac{4\pi R^{3}T}{J}\left[\frac{1}{r}-\frac{1}{R}\right]=\left(\frac{4}{3}\pi R^3\right)*\left|x\right|*\Delta \theta =\frac{3T}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$

F = m E cos θ

$$ $=m\left(\frac{GM}{R^3}*r\right)*\frac{x}{r}$ $ma=\frac{mg}{R}*\Rightarrow a=\frac{g}{R}x$

$\Rightarrow T=2\pi \sqrt[]{\frac{R}{g}}$

Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.

$F=\frac{k}{R^3}=\frac{m{v}^2}{R}\Rightarrow v=\sqrt[]{\frac{k}{m{R}^2}}=\sqrt[]{\frac{k}{m}}*\frac{1}{R}$

$T=\frac{2\pi R}{v}=\frac{2\pi R}{\sqrt[]{\frac{k}{m}*\left(\frac{1}{R}\right)}}\Rightarrow T\alpha R^2$