Physics

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

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

  $\frac{1}{2}\mu v_r^2=\frac{1}{2}k{x}^2$  

  $x={V_{rer}}^2\sqrt[]{\frac{\mu }{k}}$          

$=2*\sqrt[]{\frac{8/6}{10}}=\sqrt[]{\frac{8}{15}}$            

From conservation of momentum

2 * 4 + 4 * 2 = 2v₁ + 4v₂

8 = v₁ + 2v₂                           ….(1)

From conservation of energy

  $\frac{1}{2}*2*4^2+\frac{1}{2}*4*2^2=\frac{1}{2}*2*v_1^2+\frac{1}{2}*4*v_2^2$           ….(ii)

On solving   $v_1=\frac{4}{3}m/s,v_2=\frac{10}{3}m/s$

${}_7^{14}\mathrm{N}$ 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

From conservation of Angular momentum about hinged point

$mv\mathcal{l}=\left(\frac{m{\mathcal{l}}^2}{3}+2m.{\mathcal{l}}^2\right)\omega$            

$\omega =\left(\frac{3}{7}\frac{v}{\mathcal{l}}\right)$            

Now from conservation of Energy.

$\frac{1}{2}*\frac{7}{3}m{\mathcal{l}}^2*{\omega }^2=mgl+2mg.2\mathcal{l}$            

$v=\sqrt[]{\frac{70}{3}g\mathcal{l}}$            

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)

Let $2x$ $$ length of rod is immersed in water.

$$ ${\tau }_{\text{Hinge}}($  net $)=0$ $$

$\Rightarrow \mathrm{}\mathrm{m}\mathrm{g}?\frac{b}{2}\mathrm{s}\mathrm{i}\mathrm{n}?\theta -F_B(b-x)\mathrm{s}\mathrm{i}\mathrm{n}?\theta =0$

$\Rightarrow \mathrm{}mgb=2(b-x)F_B$

$$ $\therefore \mathrm{}{F}_B=\frac{mgb}{2(b-x)}=\frac{\left(Abd\right)b}{2(b-x)}g$  , where $d=$ $$ density of material of rod

${}_{}$  $F_B=$ Buoyant force $\mathrm{}\frac{}{}\frac{}{}$

Equate $F_B,\frac{A{b}^{2}d}{2(b-x)}=2Ax\rho$ ${}_{}\frac{{}^{}}{}$

$\Rightarrow \mathrm{}{b}^2=4x(b-x)\frac{9}{5}$

$\mathrm{}{}^{}\frac{{}^{}}{}\frac{}{}$  $\Rightarrow \mathrm{}{x}^2-bx+\frac{5b^2}{36}=0\Rightarrow x=\frac{b}{6}\therefore$ immersed $=\frac{b}{3}$

Let the acceleration of string = a

For 2 kg 20 – T₁ = 2a ………… (i)

For monkey T₂ – 80 – T₁ = 8 (2 – a)  …. (ii)

For 10 kg T₂ – 100 = 10a    ……. (iii)

$\therefore a=0.8m/s^2$

Acceleration of monkey w.r. to ground

= 2 – 0.8 = 1.2 m/s²

  $s=ut+\frac{1}{2}a{t}^2$        

$2.4=0+\frac{1}{2}*1.2*t^2$            

t = 2 sec

$\mathrm{}g=\frac{G{M}_{\text{in}}}{r^2}$  , where $M_{\text{in}}={\int }_0^r?\rho 4\pi x^2\cdot dx$

$\begin{array}{rr}& M_{\text{in}}=4\pi {\rho }_0{\int }_0^r??\left(1-\frac{x^3}{R^3}\right)x^{2}dx=4\pi {\rho }_0\left(\frac{r^3}{3}-\frac{r^6}{6R^3}\right)\\ & \therefore \mathrm{}g=\frac{4\pi G{\rho }_0}{6}\left(2r-\frac{r^4}{R^3}\right)\end{array}$ ${}_{}{}_{}^{}{}^{}$

For $\mathrm{g}$ $$ to be maximum or minimum or constant $\frac{dg}{dr}=0\Rightarrow 2-\frac{4r^3}{R^3}=0\Rightarrow r=\frac{R}{2^{1/3}}$ $\frac{}{}\frac{{}^{}}{{}^{}}\frac{}{{}^{}}$

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

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

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