A closed vessel contains 0.1 mole of a monatomic ideal gas at $200\mathrm{K}$ . If 0.05 mole of the same gas at $400\mathrm{K}$ is added to it, the final equilibrium temperature (in $\mathrm{K}$ ) of the gas in the vessel will be close to

Process-AB $\Rightarrow$ Isobaric,                                                           

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

from newtan’s law of cooling :

$\frac{dT}{dt}=-k\left(T-T_0\right)$

Using average value :

$\frac{75-65}{5}=-k\left[\frac{\left(75+65\right)2}{2}-25\right]$

$\frac{\left(65-T\right)}{5}=-k\left[\frac{\left(65+T\right)}{2}-25\right]$

$\Rightarrow \frac{10}{65-T}=\frac{45}{\frac{65+T}{2}-25}$

$\Rightarrow T=57°C$

Efficiency $\eta =50\mathrm{\%}$ $$

$\Rightarrow \eta =\left(1-\frac{T_2}{T_1}\right)\times 100=50$

$\frac{{}_{}}{}$ $\Rightarrow 1-\frac{T_2}{T_1}=\frac{50}{100}\Rightarrow 1-\frac{T_2}{T_1}=\frac{1}{2}$  -(i)

After the change :- $1-\frac{T_2^1}{T_1}=\frac{50\times \left(1.3\right)}{100}=\frac{65}{100}=.65$ $\frac{{}_{}^{}}{{}_{}}\frac{}{}\frac{}{}$ -(ii)

$\Rightarrow 1-\left(\frac{T_2-40}{T_1}\right)=.65$ --(ii)

$\Rightarrow \frac{40}{T_1}=.65-.50=.15\Rightarrow T_1=\frac{4000}{15}=266.7k$

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$=\left(l_L+l_C\right)$

Therefore current through R circuit at resonance will be zero

 $r=\frac{mv}{qB}$

$r_{\alpha }=\frac{m_{\alpha }v}{q_{\alpha }B}$

$r_P=\frac{m_{P}v}{q_{P}B}$

$\frac{r_{\alpha }}{r_P}=\frac{m_{\alpha }{q}_P}{m_P{q}_{\alpha }}=\frac{m_{\alpha }}{m_P}\left(\frac{q_P}{q_{\alpha }}\right)$

$=\frac{4m}{m}\left(\frac{q}{2q}\right)=2:1$