Physics

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

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

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 $\frac{dg}{dr}=0\Rightarrow 2-\frac{4r^3}{R^3}=0\Rightarrow r=\frac{R}{2^{1/3}}$ $\frac{}{}\frac{{}^{}}{{}^{}}\frac{}{{}^{}}$

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

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

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

Since plane is smooth hence motion is only translational

$a=gsin\theta =10*\frac{1}{2}=5m/s^2$              

$s=ut+\frac{1}{2}a{t}^2$            

$1.6=0+\frac{1}{2}*5*t^2$            

$t^2=\sqrt[]{\frac{3.2}{5}}=\sqrt[]{0.64}=0.8sec.$            

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

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

From Guass's Law

${\displaystyle ?\overrightarrow{E}.\overrightarrow{dA}=\frac{q_{in}}{{\in }_0}}$            

$E.4\pi r^2=\frac{{\displaystyle \underset{0}{\overset{r}{\int }}\left(4\pi r^2.dr\right)\left(\alpha r\right)}}{{\in }_0}$            

$E=\left(\frac{\alpha r^2}{4{\in }_0}\right)$