Class 11th

Orbital angular momentum = $\frac{nh}{2\pi }$ $\frac{}{}$

= $\frac{3h}{2\pi }\left(?n=3\right)$

CaC₂ + 2H₂O Ca (OH)2 + C2H2

Number of molecules   $=\frac{w\left(g\right)}{M\left(g\right)}*N_A$

=  $\frac{56}{28}*N_A$

= 2 N_A

Water level in both section A and B will be goes up due being of centrifugal force water will goes away and it acquire the force from upside so the water will does up so water level in both the section A and B goes up.

By given $U=54joulem=3kg$

By conservation of mechanical energy law

$\left(K+U\right)$ Earth surface $=\left(K+U\right)\infty$

$\frac{1}{2}m{v}^2+\left(-54J\right)=0+0$

$\frac{1}{2}m{v}^2=54$

$3v^2=108$

$v^2=\frac{108}{3}=36$

v= 6m/s

From first law of thermodynamics

$?W=?Q-?U$

$?Q=n{C}_p?T$

$?U=n{C}_v?T$

$?W=n(C_p-C_v)?T$

$\frac{?Q}{?W}=\frac{\gamma }{\gamma -1}\left. [\frac{C_p}{C_p}=\gamma \right]$

From F.B. Ds of

$9kg\to 90-T_1=9a\dots .\left(1\right)$

$6kg\to T_1-T_2=6a\dots ..\left(2\right)$

$5kg\to T_2-50=5a\dots ..\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)$

$40=20a?a=2m/s^2$

$a_{cp}=\frac{v^2}{r}=\frac{{\left(\sqrt[]{3gl}\right)}^2}{l}=\frac{3gl}{l}=3g$