Class 12th

${\left(n_i\right)}^2=n_e*n_h$ where $n_e=\frac{5*{10}^{28}}{{10}^6}$

$\begin{array}{rr}\therefore \mathrm{}{n}_h& =\frac{{\left(n_i\right)}^2}{n_e}=\frac{\left(1.5*{10}^{16}\right)\left(1.5*{10}^{16}\right)*{10}^6}{5*{10}^{28}}\\ n_h& =4.5*{10}^9/{\mathrm{m}}^3\end{array}$

${}_7^{14}\mathrm{N}$ contains 7 protons and 7 neutrons.

Mass defect,  $\left({}_{}{}_{}\right)\left(\begin{array}{c}\\ \end{array}\right)$ $\mathrm{\Delta }m=\left(7m_H+7m_n\right)-m\left(\begin{array}{c}14\\ 7\end{array}\right)=(7*1.00783u)+(7*1.00867u)-14.00307u$

$=0.11243u$

Binding energy $=0.11243*931.5\mathrm{M}\mathrm{e}\mathrm{V}=104.7\mathrm{M}\mathrm{e}\mathrm{V}$

$i_i=\frac{84}{140}=0.6A$

$i_2=\frac{84}{60}=1.4A$

p.d across capacitor = 20 v

Q = CV = 5 * 20 = 100    $?C$

After S is open

Q = Q . e^-t/RC

=100 e^-4

If n₂ -> n₁ in H (Z = 1) gives $\lambda$ then $z{n}_2\to z{n}_1$  give same $\lambda$  H-like ion for He⁺ ion (z = 2)

  $\frac{P}{Q}=\frac{R}{X}{X}_{max}=\frac{R_{max}.Q_{max}}{P_{min}}$

$=9000*\frac{1000}{10}=9*10^{5}ohm$

$x_{min}=\frac{R_{min}.Q_{min}}{P_{max}}=100*\frac{10}{1000}=1ohm$  

$R=8\mathrm{c}\mathrm{m},A=2{\mathrm{m}\mathrm{m}}^2,B_0=0$  at   $t=0\mathrm{s}\mathrm{e}\mathrm{c}$ $$

$$ $B=2$  Tat $$ $t=0.2\mathrm{s}\mathrm{e}\mathrm{c}$

$$ $\left|\epsilon \right|=\frac{2-0}{0.2}\pi *64*{10}^{-4}$  volts

$\frac{}{}$ $i=\frac{\left|\epsilon \right|}{r}=\frac{64\pi *{10}^{-3}}{5*{10}^{-3}}$  ampere

$$ $i=\frac{64\pi }{5}$  ampere

$$ $dF=idlB=$  magnetic force

$dm{\omega }^{2}R=$ centrifugal force

$\begin{array}{ll}\therefore & idlB=dm{\omega }^{2}R\\ \Rightarrow & idl\cdot B=\frac{m}{2\pi R}\cdot dl{\omega }^{2}R\\ \Rightarrow & \omega =\sqrt[]{\frac{2\pi iB}{m}}=\sqrt[]{\frac{2\pi *64\pi *1}{5*9*{10}^{-3}}}\\ & \omega =\frac{16\pi }{3}*10\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}=170\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}\end{array}$

c = -5, a = 2, b = 1

1 + x? - x? = a? (1+x)? + a? (1+x) + a? (1+x)² . + a? (1+x)?
Differentiate
4x³ - 5x? = a? + 2a? (1+x) + 3a? (1+x)².
12x² - 20x³ = 2a? + 6a? (1+x).
Put x = -1
12 + 20 = 2a? ⇒ a? = 16

A² - 4A + 4I = 0
(A - 2I)² = 0 ⇒ A-2I ≠ 0
B = 297A - 594I
= 297 (A - 2I)

? C? +? C? =? C?
Σ (from r=0 to 3)? C? =? C? +? C? +? C? +? C?
=? C? +? C? +? C? +? C?
= ¹²C? = (12*11*10)/6 = 220