Physics

(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.

When a soft ferromagnetic substance is placed in external magnetic field, the size of domain lying in the opposite direction of external magnetic field increases while size of domain lying in the opposite direction of field decreases if field is weak. However, if field is strong then the domain rotate in the direction of external magnetic field due to strong torque.