Physics

Sol. $g=\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}}$

$\Rightarrow \mathrm{}{\int }_v^0?dV=-{\int }_x^{\mathrm{\infty }}?gdx$

$\Rightarrow \mathrm{}O-V=-\left[\int \frac{Ax}{{\left(a^2+x^2\right)}^{3/2}}\right]$

Let, $a^2+x^2=t^2$

$\Rightarrow \mathrm{}2xdx=2tdt$

$\Rightarrow \mathrm{}\mathrm{x}\mathrm{d}x=\mathrm{t}\mathrm{d}\mathrm{t}$

$\Rightarrow \mathrm{}V=\int \frac{Atdt}{t^3}\Rightarrow -\frac{A}{t}\Rightarrow -{\left.\frac{A}{\sqrt[]{a^2+x^2}}\right|}_x^{\mathrm{\infty }}$

$\Rightarrow \mathrm{}V=\frac{A}{\sqrt[]{a^2+x^2}}$

Dimension

RC → [T]

$\frac{L}{R}\to \left[T\right]$

$\sqrt[]{Lc}\to \left[T\right]$

$\frac{L}{C}\to$ dimensionless

${\overrightarrow{B}}_{Net}=\frac{{\mu }_0}{4\pi }*2\left[\frac{i_1}{d_1}-\frac{i_2}{d_2}\right]\left(-\widehat{k}\right)$

$=10^{-7}*2\left[\frac{4}{4}*100-\frac{2}{6}*100\right]\left(-\widehat{k}\right)$

$=2*10^{-5}\left[1-\frac{1}{3}\right]-\widehat{k}$

${\overrightarrow{B}}_{Net}=\frac{4}{3}*10^{-5}\left(-\widehat{k}\right)$                        

q = 3πc

$\overrightarrow{v}=2\widehat{i}+3\widehat{j}$

$\overrightarrow{F}=q\left(\overrightarrow{v}*\overrightarrow{B}\right)$               

= $3\pi \left[\left(2\widehat{i}*3\widehat{j}\right)*\frac{4}{3}*10^{-5}\left(-\widehat{k}\right)\right]$

$\overrightarrow{F}=4\pi *10^{-5}\left(-3\widehat{i}+2\widehat{j}\right)N$  

x = 3

$F_{23}=\frac{k{q}_2{q}_3}{d^2}=\frac{k*2*2}{12}=4k$  

$F_{21}=4k$                                        

$F_{13}=4k$             

Net force on q₂ = F_R = $\sqrt[]{F_{21}^2+F_{13}^2+2F_{21}{F}_{23}cos\theta }$  

= $\sqrt[]{{\left(4k\right)}^2+{\left(4k\right)}^2+2*4kcos60°}$  

=  $k\sqrt[]{48}-\left(1\right)$

Force b/w (1) & (2) F₂₁ = 4k – (2)p

$\frac{F_R}{F_{21}}=\frac{k\sqrt[]{48}}{4k}=\frac{4\sqrt[]{3}}{4}=\sqrt[]{3}:1$

          

${?}_{flot}=\frac{q}{2{\in }_0}\left(1-cos\theta \right)$

When, 'q' is at the centre of the flat surface then, $\theta =50.$

${?}_{flat}=\frac{q}{2{\in }_0}\left(1-cos90°\right)$

$=\frac{q}{2{\in }_0}$

$x=sin\pi \left(t+\frac{1}{3}\right)$

$v=\frac{dx}{dt}=cos\left[\pi \left(t+\frac{1}{3}\right)\right]*\pi$

$=cos\left(\pi +\frac{\pi }{3}\right)*\pi$

$=-cos\frac{\pi }{3}*\left(\pi \right)$

$=-\frac{\pi }{2}=-1.57m/sec$              

Speed = 157 cm/sec

$\overline{x}$ (mean free path) = $\frac{1}{\sqrt[]{2}\pi d^2{n}_v}$  

$n_v=\frac{n{N}_A}{v},$ n → No. of moles in volume v N_A ® Avogadro's Number

$\frac{n}{v}=\frac{P}{R_T}$

$\overline{x}\propto v\propto \frac{1}{\rho (density}\&\overline{x}\alpha TatconstantP)$

^{$\rho \uparrow \iff \overline{x}\downarrow |\overline{x}\uparrow \iff T\uparrow$}                           

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is  $=\frac{1}{2}{k}_{B}T$ (for Mono atomic gases)

At constant Temp

$P\alpha \frac{1}{v}$

v₂ > v₁

⇒ P₁ > P₂

Conservation of Energy b/w (1) & (2)

$\frac{1}{2}m{v}^2+mg\mathcal{l}=\frac{1}{2}m{u}^2$

$v=\sqrt[]{u^2-2g\mathcal{l}}$

$\overrightarrow{\Delta v}=v_{\widehat{j}}-u_{\widehat{i}}$

$\left|\Delta \overrightarrow{v}\right|=\sqrt[]{{\left(u\right)}^2+{\left(\sqrt[]{u^2-2g\mathcal{l}}\right)}^2}$

$=\sqrt[]{u^2+u^2-2g\mathcal{l}}$

$=\sqrt[]{2}\left(u^2-g\mathcal{l}\right)$

x = 2