System of Particles and Rotational Motion – NCERT Solutions for Class 11 Physics (Ch 6) PDF

Ans.7.1 All the structures specified are symmetric bodies with uniform mass density. For all these bodies, their centre of mass will lie in their geometric centres.

Not necessarily, the centre of gravity of a circular ring is at the imaginary centre of the ring.

Ans.7.2

Let us assume that H atom and Cl atom are at a distance of x and y respectively from the CM (Centre of Mass).

If mass of the H atom = m, mass of the Cl atom = 35.5m

Given x + y = 1,27 À

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore,

We can have,: (my+35.5mx)/(m+35.5m) = 0

 mx + 35.5my = 0

x = 35.5 (1.27 – x)

x = 1.24 À

So the centre of mass lies 1.24 À from H atom

Ans.7.3 The child is sitting on the trolley and there is no external force, hence it is a single system. The velocity of the centre of mass will not change, irrespective of any internal motion.

Ans.7.4

Let AB is equal to the vector a and AC be equal to the vector b.

Consider two vectors  $\sin \theta$ =  $\frac{CN}{AC}$ =  $\frac{CN}{\left|\vec{B}\right|}$

$\overrightarrow{AC}$ =  $\vec{b}$ inclined at an angle  $\theta$

MN =  $\vec{b}\sin \theta$

|  $\vec{a}$  $\times \vec{b}$ | = |  $\vec{a}$ | |  $\vec{b}|\sin \theta =AB\times AC$

The area of ΔABC, we can write the relation

Area of Δ ABC =  $\frac{1}{2}$ AB  $\times AC$ =  $\frac{1}{2}\left[\vec{a}\times \vec{b}\right]$