Class 11th

$D=0\Rightarrow b^2=15a,\alpha =\frac{2b\pm 0}{2a}=\frac{b}{a}$  

  $\frac{b}{b^{\frac{2}{15}}}=\frac{15}{b}$

$\alpha +\beta =2b{\alpha }^2+{\beta }^2=4b^2-42$

$\alpha \beta =21$

$a{\left(x-\alpha \right)}^2=a\left(x^2-2\alpha x+{\alpha }^2\right)=a{x}^2-2bx+15$

a = 3                    a = -3

b = 7                    a = -7

a² + b² = 9 + 49 = 58

Kindly go through the solution

 $\begin{array}{l}-2<x+1<2\\ -3<x<1\end{array}|\begin{array}{cc}\hfill x-1\le -2orx-1\ge 2\hfill & \hfill x\le -1orx\ge 3\hfill \end{array}$

Total concentration of H+ after the mixing of HCl and H2SO4;

$\left[H^{+}\right]=\frac{\left(200*0.01\right)+\left(400*2*0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

Mass of pure carbon in coal = 0.6 * 1000gm * $\frac{60}{100}=360gm$

Mass of carbon converted into CO = 6 $\frac{60}{100}*360=216gm\Rightarrow Mol=\frac{216}{12}=18$ mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Mole = $\frac{144}{12}=12$ mole of carbon.

C (s) + O₂ (g) CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400 * 12 = 4800KJ

$C_{\left(s\right)}+\frac{1}{2}{O}_2\left(g\right)\underset{}{\overset{}{\to }}C{O}_{\left(g\right)}+100KJ$

Mole = 18 for CO production, Hence energy produced = 100 * 18 = 1800KJ

Total heat = 4800 + 1800 = 6600KJ

Sequence of sub-energy level decided by the rule of $\left(n+\mathcal{l}\right).$

$4HNO{I}_{3\left(\mathcal{l}\right)}+3KC{l}_{\left(s\right)}\underset{}{\overset{}{\to }}C{l}_{2\left(g\right)}+NOC{l}_{\left(g\right)}+2H_{2}O\left(g\right)+3KN{O}_3\left(g\right)$

$\downarrow$

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$\therefore$ 1 mole of KNO₃ produced by $\frac{4}{3}$ moles of HNO₃

and $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4*110}{3*101}$ moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole * mol.wt = 145 * 63 = 91.48 $\approx$ 91.5gm

For closed vessel

P a T     [v = constant]

->P = kT

$\frac{0.4}{100}=\frac{1}{1}$

T = 250 K

M₁ ->Mass of satellite (1)

M₂ -> Mass of satellite (2)

M_P -> Mass of planet    

Now NLM (2) on S₁

_{$\frac{G{M}_1{M}_P}{R_1^2}=\frac{m_1{v}_1^2}{R_1}$}

Similarly v₂ =   $\sqrt[]{\frac{G{M}_P}{R_2}}$

$\frac{v_1}{v_2}=\sqrt[]{\frac{R_2}{R_1}}=\sqrt[]{\frac{800}{3200}}=\frac{1}{2}=\frac{1}{x}$                

l₁ = M. l of solid sphere about its diameter

$=\frac{2}{5}MR{\text{'}}^2=\frac{2}{5}M{\left(2R\right)}^2=\frac{8}{5}M{R}^2$               

l₂ = M. I of solid cylinder about its axis

$=\frac{M{R}^{12}}{2}=\frac{M{\left(2R\right)}^2}{2}=2M{R}^2$               

I₃ = M. I of solid circular disc about its diameter

$=\frac{M{R}^{12}}{4}=\frac{M{\left(2R\right)}^2}{2}=M{R}^2$               

I₄ = M. I of this circular ring about its diameter

  $=\frac{M{R}^{12}}{2}=\frac{M{\left(2R\right)}^2}{2}=2M{R}^2$              

$\Rightarrow 6M{R}^2+2M{R}^2=\frac{x8M{R}^2}{5}$               

x = 5