Physics

For a conducting sphere, the electric field E = σ/ε? and potential V = σR/ε?
When two spheres are connected by a wire, their potentials become equal: V? = V?
σ? R? /ε? = σ? R? /ε?
σ? R? = σ? R? ⇒ σ? /σ? = R? /R?

Kindly go through the solution

The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 * 8.5 + 120 * 8.5) - (240 * 7.6) MeV
Q = (2 * 120 * 8.5) - (240 * 7.6) MeV = 2040 - 1824 = 216 MeV.

In n-type semiconductor majority charge carriers are e- and P type semiconductor majority charge carriers are holes.

$\mathrm{I}={\mathrm{n}\mathrm{e}\mathrm{A}\mathrm{V}}_{\mathrm{d}}=$ neA $\left(\mu \mathrm{E}\right)$

${\mu }_{\mathrm{e}}>{\mu }_{\mathrm{h}}\Rightarrow {\mathrm{I}}_{\mathrm{e}}>{\mathrm{I}}_{\mathrm{h}}$

${\mathrm{P}}_{\text{in}}=\frac{\mathrm{m}\mathrm{g}\mathrm{h}}{\mathrm{t}}=\frac{15*10*60}{1}=9000\mathrm{w}$

${\mathrm{P}}_{\text{out}}=90\mathrm{\%}$ of ${\mathrm{P}}_{\text{in}}$

$\Rightarrow 8.1\mathrm{k}\mathrm{w}$

Gravitational constant $=\left. [{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]$

Gravitational potential energy $=\left. [{\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$

Gravitational potential $=\left. [{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$

Gravitational intensity $=\left. [{\mathrm{L}\mathrm{T}}^{-2}\right]$

Root mean square speed of gas molecules

${\nu }_{\text{rms}}=\sqrt[]{\frac{3RT}{M}}$

Pressure exerted by ideal Gas

$\mathrm{P}=\frac{1}{3}\rho v_{\mathrm{r}\mathrm{m}\mathrm{s}}^2$

$\mathrm{P}=\frac{1}{3}\mathrm{m}\mathrm{n}{\nu }^2$

$\rho =\mathrm{m}\mathrm{n},{\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\nu }^{-2}$

Average kinetic energy of a molecular

$\mathrm{K}\mathrm{E}=\frac{3}{2}\mathrm{K}\mathrm{T}$

Total internal energy of 1 mole of a diatomic gas

$\mathrm{U}=\frac{\mathrm{f}}{2}\mu \mathrm{R}\mathrm{T}$

$\mathrm{U}=\frac{5}{2}\mathrm{R}\mathrm{T}$ (For 1 mole diatomic gas)

Mass $=\mathrm{M}$

Density of ball $=\mathrm{d}$

Density of glycerine $=\frac{\mathrm{d}}{2}$

${\mathrm{F}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{s}}{\rho }_{\mathcal{l}}\mathrm{g}=\mathrm{V}\frac{\mathrm{d}}{2}\mathrm{g}$

${\mathrm{F}}_{\mathrm{g}}=\mathrm{M}\mathrm{g}=\mathrm{v}\mathrm{d}\mathrm{g}$

For constant velocity, ${\mathrm{F}}_{\text{net}}=0$

$\therefore {\mathrm{F}}_{\mathrm{B}}+{\mathrm{F}}_{\mathrm{v}}=\mathrm{M}\mathrm{g}$

${\mathrm{F}}_{\mathrm{V}}=\mathrm{M}\mathrm{g}-{\mathrm{F}}_{\mathrm{B}}=\mathrm{V}\mathrm{d}\mathrm{g}-\frac{\mathrm{V}\mathrm{d}\mathrm{g}}{2}=\frac{\mathrm{V}\mathrm{d}\mathrm{g}}{2}=\frac{\mathrm{M}\mathrm{g}}{2}$

$\begin{array}{rr}& \mathrm{V}=\left(\text{no. of moles}\right)\left(22.4\text{litre}\right)\\ & =\frac{\text{mass}}{\text{molar mass}}\left(22.4*{10}^{-3}{\mathrm{m}}^3\right)\\ & =\frac{4.5*{10}^3}{18}*22.4*{10}^{-3}{\mathrm{m}}^3\\ & =5.6{\mathrm{m}}^3\end{array}$