Physics

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

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

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

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

$y\left(x,t_0\right)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x}{b}\right)$

$y(x,0)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x+v{t}_0}{b}\right)$

$y(x,t)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x+v{t}_0-vt}{b}\right)$

$\therefore y(x,t)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left. [\frac{x-v\left(t-t_0\right)}{b}\right]$

$\mathrm{}{v}_{\mathrm{m}\mathrm{a}\mathrm{x}}=A\omega =A\sqrt[]{\frac{k}{m}}$

Momentum conservation   $m{v}_{\text{max}}+0=m{v}_A+2m{v}_B$

$v_A+2v_B=v_{\mathrm{m}\mathrm{a}\mathrm{x}}$

$v_2-v_1=u_1-u_2\Rightarrow v_B-v_A=v_{\mathrm{m}\mathrm{a}\mathrm{x}}$

By (1) & (2) $v_A=-\frac{A\omega }{3}$

$\left|v_A\right|=A_{\text{new}}\omega$

$\frac{A\omega }{3}=A_{\text{new}}\omega \mathrm{}\therefore \mathrm{}{A}_{\text{new}}=\frac{A}{3}\mathrm{};\mathrm{}f=\frac{1}{3}$

${\stackrel{?}{B}}_1=\frac{E_1}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t){\stackrel{?}{B}}_1$    where   ${\stackrel{?}{B}}_1={\hat{c}}_1*{\stackrel{?}{E}}_1=\hat{k}*\hat{i}={\stackrel{?}{B}}_1=\hat{j}$

${\stackrel{?}{B}}_2=\frac{E_2}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t){\stackrel{?}{B}}_2$ where ${\stackrel{?}{B}}_2={\hat{c}}_2*{\stackrel{?}{E}}_2=-\hat{k}*\hat{j}={\stackrel{?}{B}}_2=\hat{i}$

$\therefore \mathrm{}\stackrel{?}{B}={\stackrel{?}{B}}_1+{\stackrel{?}{B}}_2$

$\stackrel{?}{B}=\frac{E_1}{c}\hat{j}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t)+\frac{E_2}{c}\hat{i}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t)$