Class 12th

k/6 = ∫? ^ (π/6) (x²)/ (1-x²)³/² dx x = sinθ dx = cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (1-sin²θ)³/² * cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (cos³θ) * cosθdθ
=> k/6 = ∫? ^ (π/6) tan²θdθ = ∫? ^ (π/6) (sec²θ-1)dθ
=> k/6 = [tanθ - θ]? ^ (π/6) = (1/√3 - π/6) = (2√3-π)/6
=> k = 2√3 - π

$\mathrm{}a\mathrm{c}\mathrm{o}\mathrm{s}?\theta =b\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)=c\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)=k$

$a=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta },b=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)},c=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}$

$ab+bc+ca=k^2\frac{\left[\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\right]}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)\mathrm{c}\mathrm{o}\mathrm{s}?\theta \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)}$

$={\mathrm{k}}^2\left[\frac{\mathrm{c}\mathrm{o}\mathrm{s}?\theta +2\mathrm{c}\mathrm{o}\mathrm{s}?(\theta +\pi )\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\frac{\pi }{3}\right)}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}\right]$

$={\mathrm{k}}^2\left[\frac{\mathrm{c}\mathrm{o}\mathrm{s}?\theta -2\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \frac{1}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}\right]=0$

$\mathrm{c}\mathrm{o}\mathrm{s}??=\frac{(a\hat{i}+b\hat{j}+c\hat{k})\cdot (b\hat{i}+c\hat{j}+a\hat{k})}{\sqrt[]{a^2+b^2+c^2}\cdot \sqrt[]{b^2+c^2+a^2}}=ab+bc+ca=0$

$?=\frac{\pi }{2}$

According to Faraday law

  $\mathrm{W}=\mathrm{Z}\mathrm{I}\mathrm{t}*\eta \mathrm{}$ Where $\eta =$ efficiency

or $W=\frac{E}{96500}*1*t*\eta$

Putting values

$.104=\frac{(52/3)*2*8*60*\eta }{96500}\Rightarrow \eta =0.6$

So percentage efficiency $=60\mathrm{\%}$

For isotonic solution

${\mathrm{i}}_1{\mathrm{C}}_1={\mathrm{i}}_2{\mathrm{C}}_2\mathrm{}$ ffor protein $\left.\mathrm{i}=1\right\}$

${\mathrm{C}}_1={\mathrm{C}}_2$

$\frac{0.73*1000}{M_A*250}=\frac{1.65}{M_B*1}$

$\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}=\frac{0.73*4}{1.65}=1.77=177*{10}^{-2}$

Phosphinic acid is hypo phosphorous acid $\left({\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2\right)$ .

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+{\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2?{\mathrm{N}\mathrm{a}\mathrm{H}}_2{\mathrm{P}\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

For neutrization

${\left({\mathrm{N}}_1{\mathrm{V}}_1\right)}_{\text{acid}}={\left({\mathrm{N}}_2{\mathrm{V}}_2\right)}_{\text{base}}$

$0.1*10=0.1*{\left({\mathrm{V}}_{\mathrm{m}\mathrm{L}}\right)}_{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}$

For a given reaction, rate $=-\frac{1}{2}\frac{d{n}_A}{dt}=-\frac{1}{3}\frac{d{n}_B}{dt}=-\frac{2}{3}\frac{d{n}_C}{dt}$

$\text{rate}=\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{A}}}{\mathrm{d}\mathrm{t}}=\frac{2}{3}\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}=\frac{4}{3}\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{C}}}{\mathrm{d}\mathrm{t}}$

Due to inter molecular $\mathrm{H}$ -Bonding in $\mathrm{B}$ , than $\mathrm{A},\mathrm{B}$  is more soluble and having more B.P point than A.

$P_{\text{gas}}=\frac{{\mathbf{n}}_{\text{gas}}RT}{V}$

as $n,T$ & $V$  constant So

${\mathrm{P}}_{{\mathrm{H}}_2}={\mathrm{P}}_{{\mathrm{O}}_2}={\mathrm{P}}_{{\mathrm{H}}_{\theta }}=2\mathrm{a}\mathrm{t}\mathrm{m}$

So, ${\mathrm{P}}_{\text{Total}}={\mathrm{P}}_{{\mathrm{H}}_2}+{\mathrm{P}}_{{\mathrm{O}}_2}+{\mathrm{P}}_{\mathrm{H}\mathrm{e}}=6\mathrm{a}\mathrm{t}\mathrm{m}$   

Conc. ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$  acts as dehydrating agent.

Molar mass of given complex $=266.5\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .

On treating with conc. ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$  the mass

lost by the complex $==\frac{13.5}{100}\left(266.5\right)\approx 36\mathrm{g}=2$ moles of ${\mathrm{H}}_2\mathrm{O}$

Formula of the complex $=\left. [\mathrm{C}\mathrm{r}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{C}\mathrm{l}}_2\right]\mathrm{C}\mathrm{l}.2{\mathrm{H}}_2\mathrm{O}$