Class 11th

Kindly consider the following figure

$\mathrm{}{\mathrm{N}}_2+{\mathrm{O}}_2?2\mathrm{N}\mathrm{O}$

Oxidation state changes.

$=\frac{{10}^{14}}{{10}^3}={10}^{11}$

$\mathrm{}\underset{2\mathrm{x}-2=0}{{\mathrm{K}}_2\mathrm{O}?\mathrm{x}=+1}$

$\underset{2\mathrm{x}-2=0}{{\mathrm{K}}_2{\mathrm{O}}_2}?\mathrm{x}=+1$

Theory based.                                                                                                                                                                             

${90}^{?}$

Sol. $\mathrm{}|\stackrel{?}{\mathrm{P}}+\stackrel{?}{\mathrm{Q}}|=\left|\stackrel{?}{\mathrm{P}}\right|$

${\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{P}\mathrm{Q}\mathrm{c}\mathrm{o}\mathrm{s}?\theta ={\mathrm{P}}^2$

$\Rightarrow \mathrm{}Q+2P\mathrm{c}\mathrm{o}\mathrm{s}?\theta =0$

$\Rightarrow \mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}?\theta =-\frac{\mathrm{Q}}{2\mathrm{P}}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\alpha =\frac{2\mathrm{P}\mathrm{s}\mathrm{i}\mathrm{n}?\theta }{2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}}=\mathrm{\infty }$

$?\mathrm{}[2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}=0]$

$\alpha ={90}^{?}$

$\rho ={\rho }_0\left(1-\frac{r^2}{R^2}\right)\mathrm{}0<r\le R$

$\begin{array}{rr}& \mathrm{m}\mathrm{g}=\mathrm{B}\\ & \int \rho \left(4\pi r^{2}dr\right)={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3;\mathrm{}{\int }_0^{\mathrm{R}}??{\rho }_0\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)4\pi r^{2}dr={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3\\ & {\int }_0^{\mathrm{R}}??{\rho }_{0}4\pi \left({\mathrm{r}}^2-\frac{{\mathrm{r}}^4}{{\mathrm{R}}^2}\right)\mathrm{d}\mathrm{r}={\rho }_{0}4\pi {\left(\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{5{\mathrm{R}}^2}\right)}_0^{\mathrm{R}}={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3;\mathrm{}\frac{2}{5}{\rho }_0={\rho }_{\mathrm{L}}\end{array}$

$\stackrel{?}{E}=\left(9\mathrm{s}\mathrm{i}\mathrm{n}?\left(1.6*{10}^{3}x+48*{10}^{10}\mathrm{t}\right)\stackrel{\mathrm{ˆ}}{\mathrm{k}}?/\mathrm{m}\right)$

$4000*\mathrm{V}+\mathrm{m}\mathrm{g}*\mathrm{V}=\mathrm{P}$

$\begin{array}{rr}& \frac{60*746}{2000*4000}=V\\ & V=1.86\mathrm{m}/\mathrm{s}\approx 1.9\mathrm{m}/\mathrm{s}\end{array}$

$\mathrm{}{M}_{\text{ice}}{L}_f+m_{\text{ice}}(40-0)C_w=m_{\text{steam}}{L}_v+m_{\text{steam}}(100-40)C_w$

$\Rightarrow 200[80+40(1\left)\right]=\mathrm{M}[540+60(1\left)\right]$

$\Rightarrow 200\left(120\right)=\mathrm{M}\left(600\right)$

$\mathrm{M}=40\mathrm{g}\mathrm{m}$

$\mathrm{}50.00$

For no toppling

$F\left(\frac{a}{2}+b\right)\le mg\frac{a}{2}$

$\mu \frac{\mathrm{a}}{2}+\mu \mathrm{b}\le \frac{\mathrm{a}}{2}$

$0.2a+0.4b\le 0.5a$

$0.4b\le 0.3a$

$\mathrm{b}\le \frac{3\mathrm{a}}{4}$

$\mathrm{b}\le 0.75\mathrm{a}\mathrm{}$ (in limiting case)

But is not possible as maximum value of  can be equal to  only.

$\therefore {\left(100\frac{\mathrm{b}}{\mathrm{a}}\right)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=50.00$