Class 12th

$E_0=2.25\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$ Speed of declromagnetic signal, $v=\frac{E_0}{B_0}=\frac{2.25v/m}{1.5*10^{-8}T}$  

    $B_0=1.5*10^{-8}T\Rightarrow v=\frac{225}{150}*10^8=1.5*10^{8}m/s$  

Time to reach each to the same

radar,   $t=\frac{2*3*10^{3}m}{1.5*10^{8}m/s}=4*10^{-5}S$  

Using Ampere's law for long hollow cylinder carrying current on its surface

B_in = 0

$B_{out}=\frac{{\mu }_{0}i}{2\pi r}$  

$E=\frac{kq}{2a^2}$

$E\text{'}=\frac{kq}{{\left(\sqrt[]{2}a\right)}^2}=\frac{kq}{2a^2}$
$E_{Net}=2Ecos45°+E\text{'}$

$=\frac{kq}{a^2}\left(\frac{1}{\sqrt[]{2}}+\frac{1}{2}\right)$

Kindly consider the following Image

A zwitterion is an ion that contains both negative and positive charges. This is a dipolar ion and is considered neutral.

It is a covalent chemical bond that connects amino acids to form polypeptide chains.

Different functions of biomolecules include storing genetic information (by nucleic acids), providing energy (by carbohydrates), and building and repairing tissue (by proteins).

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f =  $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5*10^{-3}\right)*\left(200*10^{-6}\right)}}$  

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5*10^{2}Hz$         $\left[Taking\pi =\sqrt[]{10}\right]$  

According to Young's double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1*10^{-3}*3*10^{-5}}{5*10^{-2}}=60*10^{-8}m=600nm$

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$  

Time required = 4 *  $t_{\frac{1}{2}}=20yrs$