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Let $\phi_{0}$ be the work function of the surface of the material. Then,

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

$$

K_{\max _{1}}=\frac{h c}{\lambda}-\phi_{0}

$$

and that in the second case is

$$

\begin{array}{c}

K_{\max _{2}}=\frac{h c}{\lambda \frac{\lambda}{2}}-\phi_{0}=\frac{2 h c}{\lambda}-\phi_{0} \\

\text { But } K_{\max _{2}}=3 K_{\max _{1}} \text { (given) } \\

\therefore \quad \frac{2 h c}{\lambda}-\phi_{0}=3\left(\frac{7_{1 c}}{\lambda}-0_{0}\right) \\

\frac{2 h c}{\lambda}-\phi_{0}=\frac{3 h c}{\lambda}-30_{0} \\

3 \phi_{0}-\phi_{0}=\frac{3 h c}{\lambda}-\frac{2 h c}{\lambda} \\

2 \phi_{0}=\frac{h c}{\lambda} \text { or } \phi_{0}=\frac{h_{c}}{2 \lambda}

\end{array}

$$

2. (b): Here,

Input signal, $V_{i}=2 \cos \left(15 t+\frac{\pi}{3}\right)$
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2 Answers

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(a): Electric field at distance $x$, $E=-\frac{\lambda}{2 \pi \varepsilon_{0} x}+\frac{Q}{4 \pi \varepsilon_{0}} \frac{x}{\left(R^{2}+x^{2}\right)^{3 / 2}}$

(considering right direction as positive) $=-\frac{\lambda}{2 \pi \varepsilon_{0}} \frac{1}{x}+\frac{x 4 \sqrt{2 \lambda} R}{4 \pi \varepsilon_{0}\left(R^{2}+x^{2}\right)^{3 / 2}}$

$=\frac{\lambda}{2 \pi \varepsilon_{0}}\left[-\frac{1}{x}+\frac{2 \sqrt{2} x R}{\left(R^{2}+x^{2}\right)^{3 / 2}}\right]$

Initially $x=\sqrt{3} R$

$\therefore \quad E=\frac{\lambda}{2 \pi \varepsilon_{0} R}\left[-\frac{1}{\sqrt{3}}+\frac{2 \sqrt{2} \sqrt{3}}{8}\right]$

$=\frac{\lambda}{2 \pi \varepsilon_{0} R}\left[\frac{-2 \sqrt{2}+3}{\sqrt{3}(2 \sqrt{2})}\right]=\frac{\lambda}{2 \pi \varepsilon_{0} R}\left[\frac{3-2 \sqrt{2}}{2 \sqrt{6}}\right]$

$\therefore$ Acceleration, $a=\frac{(-e)(E)}{m}=-\frac{e \lambda}{\pi \varepsilon_{0} m R}\left(\frac{3-2 \sqrt{2}}{4 \sqrt{6}}\right)$
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$\begin{array}{l} \text { Given, } Q=a t-b t^{2} \\ \therefore \quad I=\frac{d Q}{d t}=a-2 b t \\ \text { At } t=0, Q=0 \Rightarrow I=0\end{array}$

Also, $I=0$ at $t=a / 2 b$

$\therefore \quad$ Total heat produced in resistance $R$, $\begin{aligned} H &=\int_{0}^{a / 2 b} I^{2} R d t=R \int_{0}^{a / 2 b}(a-2 b t)^{2} d t \\ &=R \int_{0}^{a / 2 b}\left(a^{2}+4 b^{2} t^{2}-4 a b t\right) d t \\ &=R\left[a^{2} t+4 b^{2} \frac{t^{3}}{3}-4 a b \frac{t^{2}}{2}\right]_{0}^{a / 2 b} \\ &=R\left[a^{2} \times \frac{a}{2 b}+\frac{4 b^{2}}{3} \times \frac{a^{3}}{8 b^{3}}-\frac{4 a b}{2} \times \frac{a^{2}}{4 b^{2}}\right] \\ &=\frac{a^{3} R}{b}\left[\frac{1}{2}+\frac{1}{6}-\frac{1}{2}\right]=\frac{a^{3} R}{6 b} \end{aligned}$
by (7.7k points)

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