Class 11th

$\frac{\Delta z}{z}=\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}$

  $V_{rms}=\sqrt[]{\frac{3RT}{M}}As.v_1=v_2$

$V_{rms}\alpha \sqrt[]{T}$ ${\eta }_2=4n_1$

$T_1=T_2\Rightarrow V_{rms-1}=V_{rms-2}$

$P_1=\frac{n_{1}RT}{v}{P}_z=\frac{n_{2}RT}{v}$

$\frac{P_1}{P_2}=\frac{n_1}{n_2}=\frac{1}{4}$            

$\eta =1-\frac{T_2}{T_1}=1-\frac{100}{400}=\frac{3}{4}=75\mathrm{\%}$

  $\frac{K_2}{k_1}=9\frac{A_1}{A_2}=2,\frac{L_1}{L_2}=2$

  $\frac{450-T}{\frac{L_1}{A_1{k}_1}}=\frac{T}{\frac{L_2}{k_2{A}_2}}\Rightarrow \left(450-T\right)\frac{L_2}{K_2{A}_2}=\left(T\right)\frac{L_1}{A_1{k}_1}$             

$\left(450-T\right)=T\left(\frac{L_1}{L_2}\frac{k_2{A}_2}{k_1{A}_1}\right)$                          

$\Rightarrow 450-T=T\left(2*9*\frac{1}{2}\right)=9T$               

10 T = 450

T = 45° C

$\frac{E_A}{E_B}=\frac{-\frac{GM{m}_A}{2*3r}}{\frac{-GM{m}_B}{2*4r}}=\frac{m_A}{m_B}*\frac{4}{3}=\frac{4}{3}*\frac{4}{3}=\frac{16}{9}$

  $\rho =10^{3}kg/m^3$

g = 10 m/s²

Final height in both vessels

$=\frac{100+150}{2}=125cm$

So, less in U =    ${\displaystyle \int A*0.25*g*0.25}$

= $10^3*625*10^{-4}*10*16*10^{-4}$  

= 625 * 16 * 10^-4

= 1J.

  $a=\mu g=0.4*10=4m/s^2$

Slipping stops when V_bag = V_conveyar

-> 0 + at = 2

-> t =    $\frac{2}{a}=\frac{2}{4}=.5sec$

So, x =    $\frac{1}{2}a{t}^2=\frac{1}{2}*4*\frac{1}{4}=0.5m$

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$

Let, Rain drop moving with terminal velocity vt in the air Fb (Bnoyancy force) = $\frac{4}{3}\pi r^3{\rho }_{air}g$ $\frac{}{}{}^{}{}_{}$

$mg=\frac{4}{3}\pi r^3{\rho }_{air}g$

$F_v=\left(visconsforce\right)=6\pi \eta r{v}_T$

$F_b+F_v=mg$

$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{air}g+6\pi \eta r{v}_T=\frac{4}{3}\pi r^3\rho g$

$v_T=\frac{2}{9}\frac{r^2}{\eta }\left(\rho -{\rho }_{air}\right)$

$v_T\propto r^2$