Class 12th

With the help of conservation of volume, we can write

  $27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$                            

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U₁ =   $\frac{q^2}{8\pi {\epsilon }_{0}r}$

Potential energy of bigger drop =   $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27*27*q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$

$\Rightarrow \frac{U_2}{U_1}=243$           

$l=\frac{90?30}{4000}=15mA$

$I_1=\frac{30}{5000}=6mA\&l_2=9mA$

$A=\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill y\hfill \end{array}\right],\left|A\right|=3xyz-\left(x^3+y^3+z^3\right)=-\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right]$  

A² = l

A. A' = l    (as A = A')

$\therefore x^2+y^2+z^2=1andxy+yz+zx=0$

$x^3+y^3+z^3=3*2+1*\left(1-0\right)=7$

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$

$\overrightarrow{r}*\overrightarrow{a}=\overrightarrow{c}*\overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n=\frac{90k{H}_z}{2*5kHz}=9$

$L_1=\frac{x-\lambda }{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$

$\therefore SD=\frac{\left|-2\lambda +3\left(2\lambda +\frac{1}{2}\right)+\lambda \right|}{\sqrt[]{14}}=\frac{\left|5\lambda +\frac{3}{2}\right|}{\sqrt[]{14}}$

$5\lambda +\frac{3}{2}=-\frac{7}{2}\Rightarrow 5\lambda =-5\Rightarrow \lambda =-1$

$\therefore \left|\lambda \right|=1$

${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$

$PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$

Case I :- $2?-?=\frac{1}{2}m{v}_1^2........\left(i\right)$  

Case II :-     $10?-?=\frac{1}{2}m{v}_2^2........\left(ii\right)$

$\Rightarrow \frac{1}{9}=\frac{v_1^2}{v_2^2}$                                        

$\Rightarrow v_1:v_2=1:3$                            

->x = 1