Physics

Well, the centre of mass is a single point. It does not rotate by itself. But an object or a system of particles can rotate about its centre of mass. The individual particles of a system can move around the centre of mass, and these particles can make the centre of mass move in a straight line at a constant velocity unless there is a net external force. This is pretty much Newton's First Law, because we are talking about motion in general.  

Centre of mass becomes easier to calculate when you are confident with concepts such as weighted average and mass distribution. But this concept can be challenging when applying to complex systems and in combination with rotational motion. Either way, you will need to build a solid foundation with kinematics, Newton's Laws, and work-energy principles before approaching the centre of mass concept in the Systems of Particles and Rotational Motion chapter in Physics Class 11.   

Yes, it's an important concept to tackle JEE Main questions on rotational mechanics, rigid bodies, and collisions. Learning about the centre of mass simplifies complex motions of objects by treating the entire mass into a single point.  In exams such as JEE, questions on centre of mass are also interrelated with other advanced concepts in physics. So, a thorough conceptual understanding of it is essential.   

3C (Graphite) + 4H₂ (g) -> C₃H₈ (g)

$\Delta H_f\left(C_3{H}_8\right)=\left[3*\Delta H_{comb}\left(C\right)\right]+\left[4*\Delta H_{comb}\left(H_2\right)\right]-\left[\Delta H_{comb}\left(C_3{H}_8\right)\right]$        

= -10.3.7 kJ/mole

$d=\sqrt[]{2Rh}$          

$d=\sqrt[]{2*6400*h*10^{-3}}\left(hinm\right)$

Area =  $\pi d^2$

$=\left(\pi *2*6400*h*10^{-3}\right)$ km²

h = 150m

$E_9=\frac{hc}{\lambda }=\frac{1242}{\lambda \left(nm\right)}=\frac{1242}{400}=3.105$

Each wile has resistance = $\frac{\rho 4\mathcal{l}}{\pi d^2}=r$ $\frac{}{{}^{}}$

Eight wire in parallel, then equivalent resistance is

$\frac{r}{8}=\frac{\rho \mathcal{l}}{2\pi d^2}$

Single copper wire of length $2\mathcal{l}$ $$ has resistance

$R=\rho \frac{2\mathcal{l}*4}{\pi d_1^2}=\frac{\rho \mathcal{l}}{2\pi d^2}$

d1 = 4d

No. of electric field line per unit area = electric field

$E=\frac{\rho r}{3{\in }_0}forr=R$

$E=45*10^{10}N/C$

$v_S=0,v_{O_b}=5m/s$

$f_{direct}=\left(\frac{320-5}{320}\right)640=630Hz$

$f_{beat}=\left(650-630\right)=20Hz$