Class 12th

Capacitor makes potential difference constant.

Using faraday's law magnetic field should be outward and decreasing with time

$LHS=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to \infty }{lim}\frac{-1}{h}ln\left(\frac{1-\frac{h}{a}}{1+\frac{h}{b}}\right)=\underset{h\to 0}{lim}\frac{ln\left(1-\frac{h}{a}\right)}{a\left(\frac{-h}{a}\right)}+\underset{h\to 0}{lim}\frac{ln\left(1+\frac{h}{b}\right)}{b\left(\frac{h}{b}\right)}=\frac{1}{a}+\frac{1}{b}.............\left(i\right)$

$RHS=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{cos2x-1}{x^2+1-1}\left(\sqrt[]{x^2+1}+1\right)=\underset{x\to 0}{lim}\frac{-2si{n}^{x}x}{x^2}*2=-4...............\left(ii\right)$

$and\underset{x\to 0}{lim}f\left(x\right)=k.............\left(iii\right)$

$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=-4=k$

From (i), (ii) and (iii) we get $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

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Energy of electron in first existed state will be -3.4 eV.

So total energy difference will be (2.6 + 3.4) eV.

Wavelength $\left(\lambda \right)=\frac{1242eV-nm}{6eV}=207nm$

$Frequency=\frac{c}{\lambda }=\frac{3*10^8}{207*10^{-9}}=1.45*10^{9}MHz$

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De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6*10^{-34}}{\sqrt[]{3*9*10^{-31}*1.38*10^{-23}*300}}$

$\begin{array}{l}\lambda =6.26*10^{-9}m\\ \lambda =6.26nm\end{array}$