Class 12th

$\underset{h\to 0}{Lt}\frac{2*2\left[\frac{\sqrt[]{3}}{2}sin\left(\frac{\pi }{6}+h\right)-\frac{1}{2}cos\left(\frac{\pi }{6}+h\right)\right]}{2\sqrt[]{3}h\left[\frac{\sqrt[]{3}}{2}cosh-\frac{1}{2}sinh\right]}$

$=\underset{h\to 0}{Lt}4\frac{sin\left(\frac{\pi }{6}+h-\frac{\pi }{6}\right)}{2\sqrt[]{3}hsin\left(\frac{\pi }{3}-h\right)}=\frac{2}{\sqrt[]{3}}*1*\frac{1}{\frac{\sqrt[]{3}}{2}}=\frac{4}{3}$

 ${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx}}}}$

${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x-\left[x\right]}dx+{\displaystyle \underset{1}{\overset{2}{\int }}{e}^{x-\left[x\right]}dx}}+{\displaystyle \underset{2}{\overset{3}{\int }}{e}^{x-\left[x\right]}dx+.....+{\displaystyle \underset{99}{\overset{100}{\int }}{e}^{x-\left[x\right]}dx}}$

$=e-1+\frac{1}{e}\left(e^2-e\right)+\frac{1}{e^2}\left(e^3-e^2\right)+......+\frac{1}{e^{99}}\left(e^{100}-e^{99}\right)$

$=e-1+\left(e-1\right)+\left(e-1\right)+......+\left(e-1\right),100times$

$=100\left(e-1\right)$

2nd method

${\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx={\displaystyle \sum _{n=1}^{100}{\displaystyle \underset{n-1}{\overset{n}{\int }}{e}^{\left\{x\right\}}dx=}}{\displaystyle \sum _{n=1}^{100}\left({\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}\right)}}}={\displaystyle \sum _{n=1}^{100}\left(e-1\right)}=100\left(e-1\right)$

Blue cupric metaborate is reduced to cuprous metaborate in a luminous flame.

  $2Cu{\left(B{O}_2\right)}_2+2NaB{O}_2+C\underset{}{\overset{Luminous}{\to }}2CuB{O}_2+N{a}_2{B}_4{O}_7+CO$            

Cupric metaborate is obtained by heating boric anhydride & copper sulphate in a non-luminous flame as

$CuS{O}_4+B_2{O}_3\underset{}{\overset{Non-LuminiousFlame}{\to }}\underset{Blue}{Cu{\left(B{O}_2\right)}_2}+S{O}_3.$            

The points on the curve are (0, 0), (2, 2) and $\left(3,\frac{21}{2}\right)$

$\therefore \frac{dy}{dx}=2x^3-15x^2+36x-19$

$at\left(0,0\right),\frac{dy}{dx}=-19,at\left(2,2\right),\frac{dy}{dx}=9at\left(3,\frac{21}{2}\right),\frac{dy}{dx}=8$

Hence maximum slope at (2, 2) is 9.

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

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