Physics

${\lambda }_{Red}>{\lambda }_{violet}$

$\beta =\frac{\lambda D}{d}\Rightarrow$ Fringe width

If the source of light used in a Young's double slit experiment is changed from red to violet : consecutive fringe lines will come closer.

(A) Source of microwave frequency   => Magnetron

(B) Source of infrared frequency                =>Vibration of atoms and molecules

(C) Source of Gamma rays                           => Radioactive decay of nucleus

(D) Source of x-rays                                      => inner shell electrons.

n = 1 mole

$W_{AB}=nRTln\frac{v_2}{v_1}=1*RTln\frac{2v_1}{v_1}=RTln2$

$W_{BC}=0$

$W_{CA}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{\frac{P_1}{4}*2V_1-P_1*V_1}{\gamma }$

$\Rightarrow W_{CA}=\frac{-RT}{2\left(\gamma -1\right)}$

$\therefore W_{total}=RT\left[\mathcal{l}n2-\frac{1}{2\left(\gamma -1\right)}\right]$

Let x = A sin $\omega t$ $$ & v = $A\omega cos\omega t$ $$

$\therefore v=\omega \sqrt[]{A^2-x^2}$

$v^2={\omega }^2{x}^2-{\omega }^2{A}^2$

$\frac{v^2}{{\omega }^2{A}^2}+\frac{x^2}{A^2}=1$

V = 1.24 * 10⁶ volt

${\lambda }_{min}=?$

${\lambda }_{min}=\frac{hc}{eV}=\frac{1242nm}{e\left(1.24*10^6\right)}=\frac{1000*10^{-9}}{10^6}\Rightarrow {\lambda }_{min}=10^{-3}nm$

Here, $K_{eq}=2k$ ${}_{}$

$T=2\pi \sqrt[]{\frac{m}{2k}}$

$\therefore f=\frac{1}{2\pi }\sqrt[]{\frac{2k}{m}}$

$\Delta E=hv\to \left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

(1) n₁ = 3, n₂ = 2

    $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{9-4}{36}=\frac{5}{36}$         

(2) n₁ = 4, n₂ = 3

$\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{16-9}{144}=\frac{7}{144}$        

(3) n₁ = 2, n₂ = 1

$\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{4-1}{4}=\frac{3}{4}$  

(4) n₁ = 5, n₂ = 4

$\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{25-16}{400}=\frac{9}{400}$

$F_{net}=\frac{2kee}{\left(d^2+x^2\right)}sin\theta =\left(\frac{2k{e}^2}{\left(d^2{+}^2\right)}\right)\theta$

[for small displacement i.e, x is small so|, sin $\approx$ $\approx$ $$ $$ tan ]

$\frac{{}^{}}{}$ $=\frac{2k{e}^2}{\left(d^2+x^2\right)}.\frac{x}{d}=2.\frac{1}{4\pi {\epsilon }_0}{q}^2\frac{x}{d^3}$   [q = e]

$=\frac{q^2}{2\pi {\epsilon }_0{d}^3}x=m{\omega }^{2}x$

$\therefore \omega =\sqrt[]{\frac{q^2}{2\pi {\epsilon }_0{d}^{3}m}}$

$F=-\alpha x^2\Rightarrow \frac{mvdv}{dx}=-\alpha x^2$

$\Rightarrow {\displaystyle \underset{v_0}{\overset{0}{\int }}vdv=-\frac{\alpha }{m}{\displaystyle \underset{0}{\overset{x}{\int }}{x}^{2}dx\Rightarrow \frac{0^2}{2}-\frac{v_0^2}{2}=\frac{-\alpha }{3m}{x}^3}}$

$\Rightarrow \frac{-v_0^2}{2}=-\frac{\alpha }{3m}{x}^3\Rightarrow x={\left(\frac{3v_0^{2}m}{2\alpha }\right)}^{1/3}$

A tristor is formed by doping a semi conductor water from one side by N-type dopant (high concentration while comparatively lower concentration on other side thus forming NPN tristor & vice-versa.

So, statement (I) is false & statement (II) is true.