Class 12th

Cu is the only element of 3d – series whose M²⁺ / M value is positive because of fact that low hydration enthalpy and high sublimation & ionization enthalpies.

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 = $\frac{q^2}{8\pi {\epsilon }_{0}r}$ $\frac{{}^{}}{{}_{}}$

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

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

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

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

Band Width = 2 * n * the highest modulation frequency

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

Maximum Distance = 3 (x + x) = 6x = 3 * 2x = 3 * 50 = 150 cm

 $l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{co{s}^{2}x}{1+3^x}dx..........\left(i\right)}$

Using properties, ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{co{s}^{2}x}{1+3^{-x}}dx}={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{3^{x}co{s}^{2}x}{1+3^x}dx.......\left(ii\right)}$

Adding (i) and (ii), we get

$2l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{\left(1+3^x\right)co{s}^{2}x}{1+3^x}}dx={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}co{s}^{2}xdx=2}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}co{s}^{2}xdx}$

$\therefore l=\frac{\pi }{4}$

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

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

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

$\Rightarrow v_1:v_2=1:3$

x = 1

 $\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a+2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a+3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a+4\hfill & \hfill 1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|,$ [using properties]

$=0+\left(a+1\right)\left(a+2\right)\left(-1\right)+2\left(a+3\right)\left(a+4\right)0\left(a+2\right)\left(a+3\right)+4\left(\left(a+2\right)\left(a+3\right)-3\left(a+3\right)\left(a+4\right)\right)$

$=2\left(a+2\right)-2\left(a+3\right)=2\left(a+2-a-3\right)=-2$