Class 11th

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

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

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

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

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

Transverse waves can travel through solids or any medium that can withstand shearing strain. Fluids, including liquids and gases, cannot sustain shear force, so such types of mechanical waves cannot travel through them. 

Longitudinal waves can travel through all elastic media that can sustain compressive strain. That means, these mechanical waves can travel through solids, liquids, and gases.