Class 11th

${90}^{?}$

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.

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

$90.5\mathrm{J}$
$$

$\frac{-{\mathrm{G}\mathrm{M}}_{\mathrm{e}}\mathrm{M}}{\mathrm{R}}+\frac{1}{2}{\mathrm{M}\mathrm{u}}^2=\frac{-{\mathrm{G}\mathrm{M}}_{\mathrm{e}}\mathrm{M}}{2\mathrm{R}}+\frac{1}{2}{\mathrm{M}\mathrm{v}}^2$

$\text{exception 25:}$  $v=\sqrt[]{u^2-\frac{G{M}_e}{R}}$ Transverse velocity of rocket

${}_{}$ $V_T\to$  Radial velocity of rocket

$V_R\to$

$V-\sqrt[]{\frac{G{M}_e}{2R}}$

Kinetic energy $\frac{M}{10}{V}_T=\frac{9M}{10}\sqrt[]{\frac{G{M}_e}{2R}}\mathrm{};\frac{M}{10}{V}_r=M\sqrt[]{u^2-\frac{G{M}_e}{R}}$ $\frac{}{}{}_{}\frac{}{}\text{exception 25:}\mathrm{}\frac{}{}{}_{}\text{exception 25:}$

$=\frac{1}{2}\frac{M}{10}\left(V_T^2+V_r^2\right)=\frac{M}{20}\left(81\frac{G{M}_e}{2R}+100u^2-100\frac{G{M}_e}{R}\right)$

$=\frac{\mathrm{M}}{20}\left(100{\mathrm{u}}^2-\frac{119\mathrm{G}\mathrm{M}}{2\mathrm{R}}\right)$

  $\mathrm{\Delta }\mathcal{l}=0.03\mathrm{m}\mathrm{m}$ $$

${\mathrm{P}}_1=1\mathrm{a}\mathrm{t}\mathrm{m},{\mathrm{T}}_1=273\mathrm{K}$

Now work done $\begin{array}{rr}& {}_{}{}_{}^{}{}_{}{}_{}^{}\\ \end{array}$ $\begin{array}{rr}& {\mathrm{P}}_1{\mathrm{V}}_1^{\gamma }={\mathrm{P}}_2{\mathrm{V}}_2^{\gamma }\\ & {\mathrm{P}}_2={\mathrm{P}}_1{\left[\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right]}^{\gamma }\\ & =1\mathrm{a}\mathrm{t}\mathrm{m}{\left(\frac{1}{3}\right)}^{1.4}\end{array}$

Closes answer is $\frac{{}_{}{}_{}{}_{}{}_{}}{}$ $=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=88.7\mathrm{J}$ .

  $\mathrm{}v=\sqrt[]{\frac{T}{\mu }}$ $\mathrm{}\text{exception 25:}$

$\mathrm{T}=\mu {\mathrm{v}}^2\mathrm{}\frac{\mu {\mathrm{v}}^2}{\mathrm{A}}=\mathrm{Y}\frac{\mathrm{\Delta }\mathcal{l}}{\mathcal{l}}$

after substituting value of $\mathrm{\Delta }\mathcal{l}=\frac{\mu {\mathrm{v}}^2\mathcal{l}}{\mathrm{A}\mathrm{Y}}$ $\frac{{}^{}}{}$ and $\mu ,\mathrm{v},\mathcal{l},\mathrm{A}$ $$ we get

$\mathrm{Y}$

  $\mathrm{}{\gamma }_{\text{mixture}}=\frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1{C}_{V_1}+n_2{C}_{V_2}}=\frac{n_1\frac{{\gamma }_{1}R}{{\gamma }_1-1}+n_2\frac{{\gamma }_{2}R}{{\gamma }_2-1}}{\frac{n_{1}R}{{\gamma }_1-1}+\frac{n_{2}R}{{\gamma }_2-1}}$ $\mathrm{}{}_{}\frac{{}_{}{}_{{}_{}}{}_{}{}_{{}_{}}}{{}_{}{}_{{}_{}}{}_{}{}_{{}_{}}}\frac{{}_{}\frac{{}_{}}{{}_{}}{}_{}\frac{{}_{}}{{}_{}}}{\frac{{}_{}}{{}_{}}\frac{{}_{}}{{}_{}}}$

on rearranging we get

$\frac{{\mathrm{n}}_1+{\mathrm{n}}_2}{{\gamma }_{\text{mix}}-1}=\frac{{\mathrm{n}}_1}{{\gamma }_1-1}+\frac{{\mathrm{n}}_2}{{\gamma }_2-1};\frac{5}{{\gamma }_{\text{mix}}-1}=\frac{3}{1/3}+\frac{2}{2/3}$

$\frac{5}{{\gamma }_{\text{mix}}-1}=9+3=12\mathrm{}\Rightarrow {\gamma }_{\text{mixture}}=\frac{17}{12}+1+\frac{5}{12};{\gamma }_{\text{mix}}=1.42$

${\mathrm{L}}_{\mathrm{o}\mathrm{i}}={\mathrm{L}}_{\mathrm{o}\mathrm{f}}$

$\begin{array}{rr}& \frac{\mathrm{m}\mathrm{v}}{\sqrt[]{2}}*\frac{\mathcal{l}}{2}=\left. [\frac{4\mathrm{m}{\mathcal{l}}^2}{12}+\frac{\mathrm{m}{\mathcal{l}}^2}{4}\right]*\omega \\ & \omega =\frac{6\mathrm{v}}{7\sqrt[]{2}\mathcal{l}}=\frac{3\sqrt[]{2}\mathrm{v}}{7\mathcal{l}}\end{array}$