Class 11th

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)*8.31*\frac{284}{3*10^{-3}}*10^{-5}bar=0.84287bar$  

$P=84.28*10^{-2}bar$

$\approx 84*10^{-2}bar$  

$\kappa =\frac{1}{R}*\frac{\mathcal{l}}{A}=\left[\left(\frac{1}{1500}\right)*1.14\right]Sc{m}^{-1}=\frac{1.14}{1500}Sc{m}^{-1}$

${\lambda }_m=\frac{\kappa }{M}*1000Sc{m}^{2}mo{l}^{-1}$

${\lambda }_m=1000*\frac{\left(\frac{1.14}{1500}\right)}{0.001}Sc{m}^{2}mo{l}^{-1}=760Sc{m}^{2}mo{l}^{-1}$

Minimum temperature at which reaction becomes spontaneous is,

  $T_{min}=\frac{\Delta H^0}{\Delta S^0}$             

$\Delta {H^o}_{min}=\left[\Delta H_f^o\left(Fe\right)+\Delta H_f^o\left(CO\right)\right]-\left[\Delta H_f^o\left(FeO\right)+\Delta H_f^o\left(C\right)\right]$

$=\left[0-110.5\right]-\left[-266.3+0\right]$

$=155.8KJ/mol$

$\Delta S°=\left[\Delta S°\left(Fe\right)+\Delta S^o\left(CO\right)\right]-\left[\Delta S°\left(FeO\right)+\Delta S°\left(C\right)\right]$

$=\left(27.28+197.6\right)-\left(57.49+5.74\right)J/molK$

$=161.65J/molK$

$T_{min}=\frac{155.8*10^3}{161.65}K=963.8K\approx 964K$                        

Moles of NH₄HS initially taken = $\frac{5.1}{51}=0.1mol$  

$N{H}_{4}HS\left(S\right)+N{H}_3\left(g\right)+H_{2}S\left(g\right)$                             

t =  $\infty$ 0       0.1 mol               0            0

t =   $\infty$  0.1 (1 - 0.2)   0.1 * 0.2   0.1 * 0.2

$P_{N{H}_3}=\frac{nRT}{V}=\frac{0.1*0.2*0.082*300}{2}=0.246atm=P_{H_{2}S}$

$K_P=P_{N{H}_3}*P_{H_{2}S}={\left(0.246\right)}^2=0.060516=6.05*10^{-2}$

x = 6 (Nearest integer).

at point P,

$=K{E}_P+P{E}_P$

$=\frac{1}{2}m{v}_P^2+\left\{\frac{-G{M}_{1}m}{\left(r/2\right)}-\frac{G{M}_{2}m}{r/2}\right\}$

at infinity (ie for escaping from both masses)

$\Rightarrow v_P=\sqrt[]{\frac{4G\left(M_1+M_2\right)}{r}}$

$h=\frac{1}{2}g{t}^2$

or $x=v\sqrt[]{\frac{2h}{g}}$

So, x = vt

or, $x=v\text{exception 25:}$

So, dist of man from helicopter is

$\sqrt[]{h^2+x^2}=\sqrt[]{h^2+v^2\frac{2h}{g}}=\sqrt[]{\frac{2v^{2}h}{g}+h^2}$

The speed of a transverse wave depends on the intrinsic properties of the medium it propagates through. These could be elastic and inertial properties. For a transverse wave on a string, we need to know factors such as tension (a common force) in the string, and its linear mass density. 

W = 10 kg m/s²

A = 100 cm²

^{$\mathcal{l}=20cm$}

Y = 2 * 10¹¹ N/m²

^{$Y=\frac{F\mathcal{l}}{A\Delta \mathcal{l}}\Rightarrow \Delta \mathcal{l}=\frac{F\mathcal{l}}{AY}$}

For elemental mass of length, dx

Change in its length

$\Delta \mathcal{l}=5*10^{-10}m$

$\to \left[W\right]=\left[\tau \theta \right]=\left[M{L}^2{T}^{-2}\right]\to correct$

$\to \left[h\right]=\left[L\right]=\left[\frac{2scos\theta }{\rho rg}\right]\to correct$

$\left[v\right]=\left[L^3\right]\left[\frac{\pi p{a}^4}{8\eta L}\right]=\left[L^3{T}^{-1}\right]\to$ Incorrect

$\to J=\epsilon \frac{\partial E}{\partial t}\to$ correct

$\left[A\stackrel{\downarrow }{L}{}^{-2}\right]\left[A\stackrel{\downarrow }{L}{}^{-2}\right]$