Class 11th

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

For equilibrium along the incline plane is given by,

F cos 60° = mg sin 60°

  $F=\frac{0.2*10*\sqrt[]{3}}{2*1}*2$              

  $=2\sqrt[]{3}N=\sqrt[]{12}N$              

  $\underset{\left(acid\right)}{H_{2}S}+\underset{Base}{H_{2}O}?H_3{O}^{+}+H{S}^{-}$

$\underset{\left(acid\right)}{H_{2}O}+\underset{\left(Base\right)}{N{H}_3}?N{H}_{4}OH$

With H₂S water acts like base and with NH₃ it acts like acid.

$f=\frac{m{v}^2}{R}$            

$\Rightarrow \frac{m{v}^2}{R}=\mu N$

 

$\frac{m{v}^2}{R}=\mu mg$

$\Rightarrow 30=\sqrt[]{\mu *75*10}$                

  $\mu =\frac{30*30}{750}=1.2$              

              Now, R = 48 m (new Radius of curvature)

    $\mu =1.2$

$V_{max}=\sqrt[]{\mu Rg}$

$=\sqrt[]{1.2*48*10}$            

$=24m/sec$               

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

$\frac{\Delta z}{z}*100=3\frac{\Delta x}{x}*100+\frac{1}{2}\frac{\Delta y}{y}*100$

$=3*4+\frac{1}{2}*12$

$=\left(12+6\right)\mathrm{\%}$

$=18\mathrm{\%}$

m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{rms}=\sqrt[]{\frac{3RT}{M}}=\sqrt[]{\frac{3NkT}{Nm}}=\sqrt[]{\frac{3kT}{m}}=\sqrt[]{\frac{3*1.38*10^{-23}*300}{5*10^{-17}}}$

$=\sqrt[]{18*13.8*10^{-6}}$

$=15.7*10^{-3}m/s$

$=15mm/s$

W_total = W_D→E + W_E→F = Area of triangle DEF

= $\frac{1}{2}*3*300=450J.$