Class 11th

Angular momentum also depends on how the mass is spread over the body and not just mass only. If you closely look at the formula of angular momentum:

L=r*p

You will observe that it is a cross product of two vector quantities. Hence, the formula won't be practical without direction being involved. 

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

Pollution due to oxides of sulphur gets enhanced due to hydrocarbons and oxidizing agents like O? & H? O?

105.8 * (2/4) = 52.9 pm ⇒ r_Li+ + r_X- = 52.9 * (3²/3) = 158.7 pm

M = (a³ * d * N_A) / Z = (3.608 * 10? )³ * 8.92 * 6.022 * 10²³) / 4
M = (46.96 * 10? ²? * 8.92 * 6.022 * 10²³) / 4 = 63 g/mole
the closest answer is choice (1)

P = (nRT) / V = (2 * 0.0831 * 300) / 10 = 4.986 bar

Formaldehyde 2-methyl propanal

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

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

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

3O? 2O?                                                                                                                                                  

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

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

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