Physics

$U=a{P}^4$

$\frac{PV}{\gamma -1}=a{P}^4$

$P^{-3}V=a\left(\gamma -1\right)$

$P{V}^{-1/3}=constant$

$ComparingwithP{V}^x=constant$

$x=-\frac{1}{3}$

$C=\frac{R}{\gamma -1}+\frac{R}{1-x}=\frac{R}{\frac{5}{3}-1}+\frac{R}{1+\frac{1}{3}}=\frac{3R}{2}+\frac{3R}{4}$

$C=\frac{9R}{4}$

$b=a\sqrt[]{1-e^2}$

$\frac{dA}{dt}=\frac{L}{2m}$

$\frac{A}{T}=\frac{L}{2m}$

$T=\frac{2mA}{L}=\frac{2m\pi ab}{L}=\frac{2m\pi a^2\sqrt[]{1-e^2}}{L}$

Angular retardation,

Let S be the distance between Two ends a be the constant acceleration

As we know $v^2-u^2=2aS$

or, $aS=\frac{v^2-u^2}{2}$

Let $v_c$  be velocity at mid point.

Therefore, $v_c^2-u^2=2a\frac{S}{2}$

$v_c^2=u^2+aS$

$v_c^2=u^2+\frac{v^2-u^2}{2}$

$v_c=\sqrt[]{\frac{u^2+v^2}{2}}$

P? V indicator diagram for isobaric

P? V indicator diagram for isochoric process

P? V indicator diagram for isothermal process

When the ring rotates about its axis with a uniform frequency fHz, the current flowing in the ring is

I=q/T=qf

Magnetic field at the centre of the ring is

When Two rods are connected in series

$Q=\frac{A\left(T_1-T_2\right)t}{\frac{d_1}{K_1}+\frac{d_2}{K_1}}=\frac{A\left(T_1-T_2\right)t}{\left(d_1+d_2\right)/K}$

$\frac{d_1+d_2}{K}=\frac{d_1}{K_1}+\frac{d_2}{K_2}$

$K=\frac{\left(d_1+d_2\right)}{\frac{d_1}{K_1}+\frac{d_2}{K_1}}$

$T\downarrow \left(300Kto70K\right)$

$\begin{array}{l}T\downarrow R_{meta1}\downarrow R\uparrow \\ \left(Al\right)\left(Si\right)\end{array}$ semi? conductor

$V_{in}=\frac{-GM}{2R}\left[3-{\left(\frac{r}{R}\right)}^2\right],$

$V_{surface}=\frac{-GM}{R},V_{out}=\frac{-GM}{r}$

Case I:

$T-N=40a$

and $20g-T=20a$  

Also $N=20a$  

After simplifying, we get

$a=\frac{g}{4}$

Acceleration of block $B,=\sqrt[]{2}a=\frac{g}{2\sqrt[]{2}}.$

Case II:

$T=40a$

and $20g-T=20a$

After simplifying above equation, we get

$a=g/3$

Ratio $=\frac{g/2\sqrt[]{2}}{g/3}=\frac{3}{2\sqrt[]{2}}.$