Physics

Viscosity = Pascal . Second

$P^X{A}^Y{T}^Z=\left[M^1{L}^{-1}{T}^{-1}\right]$        

=  ${\left[M^1{L}^{+1}{T}^{-1}\right]}^x\left[L^2\right]y{\left[T^1\right]}^z=M^{-1}{L}^{-1}{T}^{-1}$

= x = 1,     x + 2y = -1,      -x + z = 1

y = -1,              Z = 0

Viscosity = [P¹A^-1T⁰]

Final range = 3 Initial Range

$\Rightarrow d_f=3d_i$ $\begin{array}{l}\Rightarrow \sqrt[]{2h\text{'}R}=3\sqrt[]{2hR}\\ \Rightarrow h\text{'}=9h=900m\end{array}$

$A=\frac{A_0}{16}=\frac{A_0}{2^{30/t_{1/2}}}\Rightarrow 2^4=2^{30/t_{1/2}}$

$\Rightarrow \frac{A_0}{16}=\frac{A_0}{2^{30/t_{1/2}}}\Rightarrow 2^4=2^{30/t_{1/2}}$

$\Rightarrow t_{1/2}=\frac{30}{4}=7.5$ years

Given that   $\overrightarrow{v}=v_{0}iand\overrightarrow{E}=-E_0\widehat{i}$

  $\overrightarrow{v}\left(t\right)=v_0\widehat{i}+\frac{eEt}{m}\widehat{i}$              

So, de Broglie wave length,   $\lambda =\frac{h}{mv}$

$\Rightarrow \lambda =\frac{h}{m{v}_0\left(1+\frac{q{E}_0}{m{v}_0}t\right)}=\frac{{\lambda }_0}{\left(1+\frac{q{E}_0}{m{v}_0}t\right)}$

Resolving power

$E_y=900sin\omega \left(t-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)$

$\frac{F_E}{F_B}=\frac{qE}{\left|q\overline{V}*\overline{B}\right|}=\frac{C}{V}=\frac{3*10^8}{3*10^7}=10$

$\Rightarrow \frac{F_E}{F_B}=\frac{10}{1}$

In DC :-                In AC:-

              i₁ = 4A                 i₂ = 4 sin   $\omega t$

              R₁ = 3   $\Omega$             R₂ =   $2\Omega$

              In same time internal of 't' - $\frac{H_1}{H_2}=\frac{i_1^2{R}_{1}t}{{\left({\mathcal{l}}_{2^{-rms}}\right)}^2{R}_{2}t}=\frac{16*3t}{{\left(\frac{4}{\sqrt[]{2}}\right)}^2*2*t}=\frac{48}{\left(\frac{32}{2}\right)}=3:1$

Let angle of dip =

Then B cos 45° = B' cos 60°- (i)

B' sin 60 = B sin - (ii)

$\left(i\right)÷\left(ii\right)\Rightarrow cost60=cot\alpha cos45°$

$cot\alpha =\frac{cot60°}{cos45°}=\frac{1\sqrt[]{2}}{\sqrt[]{3}x}=\sqrt[]{\frac{2}{3}}$

$tan\alpha =\sqrt[]{\frac{3}{2}}$

$\alpha =ta{n}^{-1}\sqrt[]{\frac{3}{2}}$

T₁ = 3 sec T₂ = 4sec

$T=2\pi \sqrt[]{\frac{I}{\mu B}}$

$\frac{I_1}{I_2}=\frac{3}{2}\Rightarrow \frac{T_1}{T_2}=\sqrt[]{\frac{l_1{\mu }_2}{{\mu }_1{l}_2}}$

$\Rightarrow {\left(\frac{3}{4}\right)}^2=\frac{I_1}{I_2}\left(\frac{{\mu }_2}{{\mu }_1}\right)\Rightarrow \frac{{\mu }_2}{{\mu }_1}=\frac{l_2}{l_1}*\frac{9}{16}=\frac{2}{3}*\frac{9}{16}=\frac{3}{8}$ $\Rightarrow \frac{{\mu }_1}{{\mu }_2}=\frac{8}{3}$