Class 11th

Ka for C₃H₇COOH = 2 * 10^-5

$pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$                

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

    $pH=\frac{pKa-\mathcal{l}ogC}{2}$            

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27*10^{-1}$          

$\therefore$ Ans 27

$B_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1={\pi }_{2py}^1\to$ Paramagnetic

$\begin{array}{l}L{i}_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2\to Diamagnetic\\ C_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2\to Diamagnetic\\ C_2^{-}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^1\to Paramagnetic\end{array}$

$O_2^{-2}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*2}\equiv {\pi }_{2py}^{*2}$ Diamagnetic

$O_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^0\to$  Paramagnetic

$H{e}_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*1}\to$ Paramagnetic

$\therefore$ Paramagnetic molecules are ${}_{}{}_{}^{}{}_{}^{}{}_{}^{}$ $=B_2,C_2^{-},O_2^{+},H{e}_2^{+}$

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$\therefore n_x=1$          

  $n_y=1$              

$n_z=\frac{0.05}{3}=0.0167$ here z is limiting reagent.

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$\therefore$ mass of xyz₃ = n * molecular mass

=  $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$

= 0.5 * 4 = 2g

$I_{OO\text{'}}=\left[\frac{1}{4}M{\left(\frac{a}{2}\right)}^2\right]*2+\left[\frac{5}{4}m{\left(\frac{a}{2}\right)}^2\right]*2$

$=\frac{6M{a}^2}{8}=\frac{3}{4}M{a}^2$

x=3

loss in k = Gain in U spring

$\Rightarrow \frac{3E}{4}=\frac{1}{2}k{\left(\frac{25}{100}\right)}^2$

$\Rightarrow \frac{3E}{4}=\frac{k}{32}\Rightarrow k=\frac{3F}{4}*32=24E$                

d₁ = 2mm = 2r

$\rho =1750kg/m^3$ coeff of viscosity H

  $\frac{d}{dt}\left(2r\right)=0.35cm/s$              

here,   $\left(\frac{4}{3}\pi r^3\right)\rho g=6\pi nrv$

$\Rightarrow \eta =\frac{4r^2\rho g}{3*6v}=\frac{2r^2\rho g}{9v}$

$\frac{350}{9*35}=\frac{10}{9}?1$  

An excellent example of pure rotational motion is the motion of the blades of a ceiling fan when it is switched on. In this type of motion, the object rotates about a fixed axis, and all particles of the object move in circles centred on that axis. Windmills and turbines also follow the same rotational mechanics. 

${\upsilon }_0=320Hz.v_{train}=36km/h=\frac{36*1000}{60*60}=10m/s$

${\upsilon }_{\left(reflectedSound\right)}={\upsilon }_0\left(\frac{v}{v-v_{train}}\right)$

$=320\left(\frac{330}{330-10}\right)=330Hz$

$\Rightarrow {\upsilon }_{echo}={\upsilon }_{reflected}\left(\frac{v+v_{train}}{V}\right)$

$=330\left(\frac{330+10}{330}\right)=340Hz.$