Class 12th

A = 30p cm² = 30p * 10^-4 m²

d = 1mm            = 10^-3 m

E = (dielectric strength for breakdown)

= 3.6 * 10⁷ v/m

Q = 7 * 10^-6 C

$\frac{\sigma }{{\in }_0}=E$

        $\Rightarrow \frac{Q}{kA{\in }_0}=E$

$k=\frac{Q}{A{\in }_{0}E}$        

$=\frac{7*10^{-6}*4\pi *9*10^9}{30\pi *10^{-4}*1*3.6*10^7}$

$=\frac{7*4\pi *9}{30\pi *3.6}$

$=\frac{7*4*9*10}{30*36}$

$=\frac{70}{30}$

$k=\frac{7}{3}=2.33$

CN is a strong field ligand, so pairing occurs.

1. ${\left[Fe{\left(CN\right)}_6\right]}^{4-}$

So; it is diamagnetic

2. ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$

So; it is paramagnetic.

3. ${\left[Ti{\left(CN\right)}_6\right]}^{3-}$

Ti³⁺ = 4s⁰3d¹

 It is paramagnetic

4. ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

It is diamagnetic

5. ${\left[Co{\left(Cn\right)}_6\right]}^{3-}$

Hence,  ${\left[Fe{\left(CN\right)}_6\right]}^{3-}and{\left[Ti{\left(CN\right)}_6\right]}^{3-}$ ${{}_{}}^{}{{}_{}}^{}$ are paramagnetic.

Baseband Signal frequency 3.5 MHz

& Carrier signal frequency = 3.5 GHz

${\upsilon }_C=3.5GHz=3.5*10^9$

$\lambda =\frac{C}{{\upsilon }_c}=\frac{3*10^8}{3.5*10^9}=\frac{3}{3.5}*10$

$=\frac{30}{35}*10$

= $\frac{60}{7}$ $\frac{}{}$

Size of antenna = $\frac{\lambda }{4}=\frac{60}{7*4}$ $\frac{}{}\frac{}{}$

= 21.4 mm

$t_{1/2}=200days=\frac{0.693}{k}$

$t=\frac{2.303}{k}lo{g}_{10}\frac{N_0}{N}$

83 =  $\frac{2.303}{0.693}*200log\frac{N_0}{N}$

$83=\frac{200}{0.3010}log\frac{N_0}{N}$

0.125 = $log\frac{N_0}{N}$

$\frac{N_0}{N}=antilog\left(0.125\right)$

= 1.333

Activating remaining = $\frac{N}{N_0}*100$

$=\frac{\frac{N_0}{1.333}}{N_0}*100$

= 75%

$A{\to }^{105}B{+}^{115}C$

$\Rightarrow Q=\left[105*6.4+115*6.4\right]-\left[220*5.6\right]Mev$

Q = 176 MeV

According to Rutherford, e- revolves around in nucleus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e- should loose energy and finally should collapse in the nucleus.

Velocity of wave in a medium = $\frac{1}{\sqrt[]{{\mu }_m{\in }_m}}$ $\frac{}{\text{exception 25:}}$

$\begin{array}{l}v=\frac{1}{\sqrt[]{{\mu }_0{\mu }_r{\in }_r{\in }_0}}\\ v=\frac{3*10^8}{\sqrt[]{1.61*6.44}}=0.931*10^{8}m/sec\end{array}$

$\begin{array}{l}B={\mu }_{m}H\\ ={\mu }_r{\mu }_{0}H\\ =1.61*4\pi *10^{-7}*4.5*10^{-2}\\ =1.61*4\pi *4.5*10^{-9}\end{array}$

$\frac{E}{B}=v$

E = vB   = 0.931 * 10⁸ * 1.61 * 4p * 4.5 * 10^-9

= [0.931 * 1.61 * 4p * 4.5] * 10^-1

= 8.476

» 8.476 v m^-1

$E_{S{n}^{2+}/Sn}^0=-0.140V=E_2^0$

$E_{S{n}^{4+}/Sn}^0=+0.010V=E_3^0$

Here, $\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$

$-4F{E}_3^0=-2F{E}_1^0-2F{E}_2^0$

$E_3^0=\frac{E_1^0+E_2^0}{2}$

$+0.010=\frac{E_1^0+\left(-0.140\right)}{2}$

$E_1^0=E_{S{n}^{4+}/S{n}^{2+}}^0=+0.16$

= 16 * 10^-2 V

Let l = l₀ cos wt

l = l₀, at t₁ = 0

&  $l=\frac{l_0}{\sqrt[]{2}},at\omega t=\frac{\pi }{4}\Rightarrow t_2=\frac{\pi }{4\omega }$

$t_2=\frac{\pi }{4\omega }=\frac{\pi }{4}*\frac{T}{2\pi }=\frac{T}{8}$           

$t_2=\frac{1}{8}*\frac{1}{\upsilon }\left(T=\frac{1}{\upsilon }\right)$           

$t_2=\frac{1}{8*50}$           

^{$t_2=\frac{1}{4*10^{+2}}$}             

= 0.25 * 10^-2

t₂ = 2.5 ms

qE = mg

$\Rightarrow q=\frac{mg}{E}=\frac{0.1}{1000}*\frac{9.8}{4.9*10^5}$         

q = 2 * 10^-9 C