NCERT Solutions for Class 11 Physics Chapter 7 System Of Particles And Rotational Motion

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NCERT Solutions Class 11 Physics Chapter 7 –Download Free PDF

NCERT Solutions for Class 11 Physics Chapter 7 System Of Particles And Rotational Motion are considered to be both highly important and challenging within the Class 11 NCERT syllabus. On this particular page of Careers360, you will find comprehensive class 11 physics chapter 7 exercise solutions, meticulously crafted by subject experts. These NCERT Solutions intricately elucidate each step in a simplified manner, making the understanding of this complex topic more accessible. The physics class 11 chapter 7 NCERT solutions encompass a total of thirty-three questions, spanning from 7.1 to 7.21 in the exercise questions section, and from 7.22 to 7.33 in the additional exercise section.

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NCERT Solutions Class 11 Physics Chapter 7 –Download Free PDF
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion
Access Class 11 Physics Chapter 7 Exercise solutions
System of Particles Class 11 NCERT Solutions - Additional Exercis e
NCERT Solutions for Class 11 Physics Chapter Wise
Class 11 Physics Chapter 7 NCERT solutions - Download for Free
Highlights of system of particles and rotational motion class 11
Significance of NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion:

ch 7 class 11 physics System of particles and rotational motion is part of Ncert solution for class 11 physics . The chapter starts with the concepts of the rigid body. In the solutions of NCERT Cass 11 physics chapter 7 System of Particles and Rotational Motion, a detailed explanation of answers is given by considering the bodies to be rigid. Ideally speaking a rigid body is a body with a perfectly definite and unaltering shape. The distance between all pairs of particles of such a body does not change. According to this definition, the bodies that we see in our real life are not rigid because they will deform when a force is applied. But when this deformation is negligible we consider the bodies as rigid.

The CBSE NCERT solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion are important as far as board and competitive exams are considered. The two main topics that we come across in NCERT are the moment of inertia and the centre of mass. There are questions and answers based on these topics in the NCERT solutions for Class 11 Physics Chapter 7 system of particles and rotational motion. NCERT solutions for class 11 physics chapter 7 pdf help students in self-study and score good marks in exams.

NCERT Solutions for Class 11 Physics Chapter 7 pdf download for free.

**According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 7, " System of particles and rotational motion," has been renumbered as Chapter 6.

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

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Q7.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Answer:

For uniform mass density, the location of the centre of mass is the same as that of the geometric centre.

No, it is not necessary that the centre of mass lies inside the body. For example in the case of a ring the centre of mass outside the body.

Q7.2 In the HCl molecule, the separation between the nuclei of the two atoms is about $1.27\AA (1\AA =10^{-10}m)$ . Find the approximate location of the CM of the molecule, given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer:

Let us assume that the centre of mass of the molecule is x cm away from the chlorine atom.

So, we can write

$\frac{m\left ( 1.27\ -\ x \right )\ +\ 35.5mx}{m\ +\ 35.5 }\ =\ 0$

$x\ =\ \frac{-1.27}{\left ( 35.5-1 \right )}\ =\ -\ 0.037\AA$

Here negative sign indicates that the centre of mass lies leftward of the chlorine atom.

Q7.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Answer:

Since child and trolley are considered in a system thus speed of the centre of mass change only when an external force will act on the system. When the child starts to move on the trolley the force produced is the internal force of the system so this will produce no change in the velocity of the CM.

Q7.4 Show that the area of the triangle contained between the vectors $a\: and\: b$ is one half of the magnitude of $a\times b$ .

Answer:

Let a and b be two vectors having $\Theta$ angle between them.

Consider $\Delta$ MON,

$\sin \Theta \ =\ \frac{MN}{MO}$

or $\sin \Theta \ =\ \frac{MN}{\underset{b}{\rightarrow}}$

or $MN\ =\ b\sin \Theta$

= 2 (Area of $\Delta$ MOK)

Therefore the area of $\Delta$ MOK = One half of $\left | a\times b \right |$ .

Q7.5 Show that $a.(b\times c)$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors , $a$ , $b$ and $c$ .

Answer:

A parallelepiped is shown in the figure given below:-

1645181345147

Volume is given by : = abc

We can write :

$|b\times c|\ =\ |b||c|\sin \Theta\ \widehat{n}$ (The direction of $\widehat{n}$ is in the direction of vector a.)

$=\ |b||c|\sin 90^{\circ}\ \widehat{n}$

$=\ |b||c|\ \widehat{n}$

Now,

$a.(b\times c)\ =\ a.(bc)\ \widehat{n}$

$=\ abc \cos \Theta$

$=\ abc \cos 0^{\circ}$

$=\ abc$

This is equal to volume of parallelepiped.

Q7.6 Find the components along the $x,y,z$ axes of the angular momentum $1$ of a particle, whose position vector is $r$ with components $x,y,z$ and momentum is $p$ with components $p_{x}$ , $p_{y}$ and $p_{z}$ . Show that if the particle moves only in the $x-y$ plane the angular momentum has only a z-component.

Answer:

Linear momentum of particle is given by :

$\overrightarrow{p}\ =\ p_x \widehat{i}\ +\ p_y \widehat{j}\ +\ p_z \widehat{k}$

And the angular momentum is :

$\overrightarrow{l}\ = \overrightarrow{r}\times \overrightarrow{p}$

$= \left ( x \widehat{i}\ +\ y \widehat{j}\ +\ z \widehat{k} \right ) \times \left ( p_x \widehat{i}\ +\ p_y \widehat{j}\ +\ p_z \widehat{k} \right )$

$=\begin{vmatrix} i &j &k \\ x &y &z \\ P_x &P_y &P_z \end{vmatrix}$

$=\ \widehat{i}\left ( yp_z - zp_y \right )\ -\ \widehat{j}\left ( xp_z - zp_x \right )\ +\ \widehat{k}\left ( xp_y - yp_x \right )$

When particle is confined to x-y plane then z = 0 and p _z = 0.

When we put the value of z and p _z in the equation of linear momentum then we observe that only the z component is non-zero.

Q7.7 Two particles, each of mass $m$ and speed $v$ , travel in opposite directions along parallel lines separated by a distance $d$ . Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

Answer:

Assume two points (say A, B) separated by distance d.

So the angular momentum of a point about point A is given by :

$=\ mv \times d\ =\ mvd$

And about point B : $=\ mv \times d\ =\ mvd$

Now assume a point between A and B as C which is at y distance from point B.

Now the angular momentum becomes : $=\ mv \times \left ( d-y \right )\ +\ mv \times y$

$=\ mv d$

Thus it can be seen that angular momentum is independent of the point about which it is measured.

Q7.8 A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are $36.9^{0}$ and $53.1^{0}$ respectively. The bar is $2m$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

1645181374167

Answer:

The FBD of the given bar is shown below :

1645181390126

Since the bar is in equilibrium, we can write :

$T_1 \sin 36.9 ^{\circ}\ =\ T_2 \sin 53.1 ^{\circ}$

or $\frac{T_1}{T_2}\ =\ \frac{0.800}{0.600}\ =\ \frac{4}{3}$

or $T_1\ =\ \frac{4}{3}T_2$ .....................................................(i)

For the rotational equilibrium :

$T_1 \cos 36.9 ^{\circ}\times d\ =\ T_2 \cos 53.1 ^{\circ}\times \left ( 2-d \right )$ (Use equation (i) to solve this equation)

or $d\ =\ \frac{1.2}{1.67}\ =\ 0.72\ m$

Thus the center of gravity is at 0.72 m from the left.

Q7.9 A car weighs $1800 kg$ . The distance between its front and back axles is $1.8 m$ . Its centre of gravity is $1.05 m$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Answer:

The FBD of the car is shown below :

1645181411271

We will use conditions of equilibrium here :

$R_f\ +\ R_b\ =\ mg$

$=\ 1800\times 9.8\ =\ 17640\ N$ ....................................(i)

For rotational equilibrium :

$R_f\left ( 1.05 \right )\ =\ R_b(1.8-1.05)$

or $1.05R_f\ =\ 0.75R_b$

$R_b\ =\ 1.4R_f$ ..............................(ii)

From (i) and (ii) we get :

$R_f\ =\ \frac{17640}{2.4}\ =\ 7350\ N$

and $R_b\ =\ 17640-7350\ =\ 10290\ N$

Thus force exerted by the front wheel is = 3675 N

and force exerted by back wheel = 5145 N.

Q7.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be $2MR^{2}/5$ , where $M$ is the mass of the sphere and $R$ is the radius of the sphere.

Answer:

We know that moment of inertia of a sphere about diameter is :

$=\ \frac{2}{5}MR^2$

1645181433111 Using parallel axes theorem we can find MI about the tangent.

Moment of inertia of a sphere about tangent :

$=\ \frac{2}{5}MR^2\ +\ MR^2\ =\ \frac{7}{5}MR^2$

Q 7.10(b) Given the moment of inertia of a disc of mass $M$ and radius $R$ about any of its diameters to be $MR^{2}/4$ , find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer:

We know that moment of inertia of a disc about its diameter is :

$=\ \frac{1}{4}MR^2$

Using perpendicular axes theorem we can write :

Moment of inertia of disc about its centre:-

$=\ \frac{1}{4}MR^2\ +\ \frac{1}{4}MR^2\ =\ \frac{1}{2}MR^2$

Using parallel axes theorem we can find the required MI :

Moment of inertia about an axis normal to the disc and passing through a point on its edge is given by :

$=\ \frac{1}{2}MR^2\ +\ MR^2\ =\ \frac{3}{2}MR^2$

Q7.11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Answer:

The moment of inertia of hollow cylinder is given by $=\ mr^2$

And the moment of inertia of solid cylinder is given by :

$=\ \frac{2}{5}mr^2$

We know that : $\tau \ =\ I\alpha$

Let the torque for hollow cylinder be $\tau_1$ and for solid cylinder let it be $\tau_2$ .

According to the question : $\tau_1\ =\ \tau_2$

So we can write the ratio of the angular acceleration of both the objects.

$\frac{\alpha _2}{\alpha _1}\ =\ \frac{I_1}{I_2}\ =\ \frac{mr^2}{\frac{2}{5}mr^2}\ =\ \frac{2}{5}$

Now for angular velocity,

$\omega \ =\ \omega _o\ +\ \alpha t$

Clearly, the angular velocity of the solid sphere is more than the angular velocity of the hollow sphere. (As the angular acceleration of solid sphere is greater).

Q7.12 A solid cylinder of mass $20 kg$ rotates about its axis with angular speed $100 \: rad\: s^{-1}$ . The radius of the cylinder is $0.25m$ . What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Answer:

Firstly we will calculate moment of inertia of the solid cyliner :

$I_c\ =\ \frac{1}{2}mr^2$

$=\ \frac{1}{2}(20)(0.25)^2\ =\ 0.625\ Kg\ m^2$

So the kinetic energy is given by :

$E_k\ =\ \frac{1}{2}I\omega ^2$

$=\ \frac{1}{2}\times (6.25)\times (100) ^2$

$=\ 3125\ J$

And the angular momentum is given by : $=\ I\omega$

$=\ 0.625\times 100$

$=\ 62.5\ Js$

Q7.13 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40\: rev/min$ . How much is the angular speed of the child if he folds his hands back and thereby reduces his
moment of inertia to $2/5 times$ the initial value? Assume that the turntable rotates without friction.

Answer:

We are given with the initial angular speed and the relation between the moment of inertia of both the cases.

Here we can use conservation of angular momentum as no external force is acting the system.

So we can write :

$I_1w_1\ =\ I_2w_2$

$w_2\ =\ \frac{I_1w_1}{I_2}$

$=\ \frac{I(40)}{\frac{2}{5}I}$

$=\ 100\ rev/min$

Q7.13(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Answer:

The final and initial velocities are given below :

$E_f\ =\ \frac{1}{2}I_2w_2^2$ and $E_i\ =\ \frac{1}{2}I_1w_1^2$

Taking the ratio of both we get,

$\frac{E_f}{E_i}\ =\ \frac{\frac{1}{2}I_2w_2^2}{\frac{1}{2}I_1w_1^2}$

or $=\ \frac{2}{5}\times \frac{100\times 100}{40\times40}$

or $=\ 2.5$

Thus the final energy is 2.5 times the initial energy.

The increase in energy is due to the internal energy of the boy.

Q7.14 A rope of negligible mass is wound round a hollow cylinder of mass $3kg$ and radius $40cm$ . What is the angular acceleration of the cylinder if the rope is pulled with a force of $30\: N$ ? What is the linear acceleration of the rope ? Assume that there is no slipping.

Answer:

The moment of inertia is given by :

$I\ =\ mr^2$

or $=\ 3\times (0.4)^2\ =\ 0.48\ Kg\ m^2$

And the torque is given by :

$\tau \ =\ r\times F$

$=\ 0.4\times 30\ =\ 12\ Nm$

Also, $\tau\ =\ I\alpha$

So $\alpha \ =\ \frac{\tau }{I}\ =\ \frac{12}{0.48}\ =\ 25\ rad/s^{-2}$

And the linear acceleration is $a\ =\ \alpha r\ =\ 0.4\times 25\ =\ 10\ m/s^{-2}$

Q7.15 To maintain a rotor at a uniform angular speed of $200\: rad\: s^{-1}$ , an engine needs to transmit a torque of $180\: N\: m$ . What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is $100^{0}/_{0}$ efficient.

Answer:

The relation between power and torque is given by :

$P\ =\ \tau \omega$

or $=\ 180\times 200\ =\ 36000\ W$

Hence the required power is 36 KW.

Q7.16 From a uniform disk of radius $R$ , a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Answer:

Let the mass per unit area of the disc be $\sigma$ .

So total mass is $=\ \Pi r^2\sigma\ =\ m$

1645181462431

Mass of the smaller disc is given by :

$=\ \Pi \left ( \frac{r}{2} \right )^2\sigma\ =\ \frac{m}{4}$

Now since the disc is removed, we can assume that part to have negative mass with respect to the initial condition.

So, the centre of mass of the disc is given by the formula :

$x\ =\ \frac{m_1r_1\ +\ m_2r_2}{m_1\ +\ m_2}$

or $=\ \frac{m\times 0\ -\ \frac{m}{4}\times \frac{r}{2}}{m\ -\ \frac{m}{4}}$

$=\ \frac{-r}{6}$

Hence the centre of mass is shifted $\frac{r}{6}$ leftward from point O.

Q7.17 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5\: g$ are put one on top of the other at the $12.0 cm$ mark, the stick is found to be balanced at $45.0 cm$ . What is the mass of the metre stick?

Answer:

The centre of mass of meter stick is at 50 cm.

Let the mass of meter stick be m.

Now according to the situation given in the question, we will use the rotational equilibrium condition at the centre point of the meter stick.

$10g(45 - 12)\ -\ mg(50 - 45)\ =\ 0$

or $m\ =\ \frac{10\times 33}{5}\ =\ 66\ g$

Thus the mass of the meter stick is 66 g.

Q7.18 (a) A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

(a) Will it reach the bottom with the same speed in each case?

Answer:

Let the height of the plane is h and mass of the sphere is m.

Let the velocity at the bottom point of incline be v.

So the total energy is given by :

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2$

Using the law of conservation of energy :

$\frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2\ =\ mgh$

For solid sphere moment of inertia is :

$I\ =\ \frac{2}{5}mr^2$

So,

$\frac{1}{2}mv^2\ +\ \frac{1}{2}\left ( \frac{2}{5}mr^2 \right )w^2\ =\ mgh$

Put v = wr and solve the above equation.

We obtain : $\frac{1}{2}v^2\ +\ \frac{1}{5}v^2\ =\ gh$

or $v\ =\ \sqrt{\frac{10}{7}gh}$

Thus the sphere will reach the bottom at the same speed since it doesn't depend upon the angle of inclination.

Q7.18 (b) A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

(b) Will it take longer to roll down one plane than the other?

Answer:

Let us assume the angle of inclination to be $\Theta_1$ and $\Theta_2$ . $\left ( \Theta_2\ > \Theta_1 \right )$

And the acceleration of the sphere will be $g \sin \Theta_1$ and $g \sin \Theta_2$ respectively.

Thus $a_2 > a_1$ (since $\sin \Theta _2 > \sin \Theta _1$ )

Since their acceleration are different so the time taken to roll down the inclined plane will be different in both the cases.

Q7.18 (c) A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

Answer:

The equation of motion gives :

$v\ =\ u\ +\ at$

So $t \propto \frac{1}{a}$

Since we know that $a_2 > a_1$

So $t_2 < t_1$

Hence the inclined plane having a smaller angle of inclination will take more time.

Q7.19 A hoop of radius $2m$ weighs $100\: kg$ . It rolls along a horizontal floor so that its centre of mass has a speed of $20cm/s$ . How much work has to be done to stop it?

Answer:

Total energy of hoop is given by :

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2$

Also the moment inertia of hoop is given by : $I\ =\ mr^2$

We get,

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}(mr^2)w^2$

And v = rw

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}mv^2\ =\ mv^2$

So the work required is :

Q7.20 The oxygen molecule has a mass of $5.30\times 10^{-26}kg$ and a moment of inertia of $1.94\times 10^{-46}kg\: m^{2}$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500m/s$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Answer:

We are given the moment of inertia and the velocity of the molecule.

Let the mass of oxygen molecule be m.

So the mass of each oxygen atom is given by : $\frac{m}{2}$

Moment of inertia is :

$I\ =\ \frac{m}{2}r^2\ +\ \frac{m}{2}r^2\ =\ mr^2$

or $r\ =\ \sqrt{\frac{I}{m}}$

or $r\ =\ \sqrt{\frac{1.94\times 10^{-46}}{5.36\times 10^{-26}}}\ =\ 0.60\times 10^{-10}\ m$

We are given that :

$E_{rot}\ =\ \frac{2}{3}E_{tra}$

or $\frac{1}{2}I\omega ^2\ =\ \frac{2}{3}\times \frac{1}{2}mv^2$

or $\omega \ =\ \sqrt{\frac{2}{3}}\times \frac{v}{r}$

or $=\ 6.80\times 10^{12}\ rad/s$

Q7.21 (a) A solid cylinder rolls up an inclined plane of angle of inclination $30^{0}$ . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of $5m/s$ .

(a) How far will the cylinder go up the plane?

Answer:

The rotational energy is converted into the translational energy.(Law of conservation of energy)

$\frac{1}{2}I \omega ^2\ =\ \frac{1}{2}mv^2\ =\ mgh$

Since the moment of inertia for the cylinder is :

$I\ =\ \frac{1}{2}mr^2$

Putting the value of MI and v = wr in the above equation, we get :

$\frac{1}{4}v^2\ +\ \frac{1}{2}v^2\ =\ gh$

or $h\ =\ \frac{3}{4}\times \frac{v^2}{g}$

or $=\ \frac{3}{4}\times \frac{5\times 5}{g}\ =\ 1.91\ m$

Now using the geometry of the cylinder we can write :

$\sin \Theta \ =\ \frac{h}{d}$

or $\sin 30^{\circ}\ =\ \frac{h}{d}$

or $d\ =\ \frac{1.91}{0.5}$

$=\ 3.82\ m$

Thus cylinder will travel up to 3.82 m up the incline.

Q7.21 (b) A solid cylinder rolls up an inclined plane of angle of inclination $30^{0}$ . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of $5m/s$ .

b) How long will it take to return to the bottom?

Answer:

The velocity of cylinder is given by :

$v\ =\ \left ( \frac{2gh}{1\ +\ \frac{k^2}{r^2}} \right )^ \frac{1}{2}$

or $v\ =\ \left ( \frac{2gd\sin \Theta }{1\ +\ \frac{k^2}{r^2}} \right )^ \frac{1}{2}$

We know that for cylinder :

$K^2\ =\ \frac{R^2}{2}$

Thus $v\ =\ \left ( \frac{4}{3} gd \sin \Theta \right )^\frac{1}{2}$

Required time is :

$t\ =\ \frac{d}{v}$

or $=\ \left ( \frac{11.46}{19.6} \right )^\frac{1}{2}\ =\ 0.764\ s$

Hence required time is 0.764(2) = 1.53 s.

System of Particles Class 11 NCERT Solutions - Additional Exercis e

Q 7.22 As shown in Fig.7.40, the two sides of a step ladder $BA$ and $CA$ are $1.6m$ long and hinged at A. A rope $DE$ , $0.5m$ is tied half way up. A weight $40 kg$ is suspended from a point $F$ , $1.2m$ from $B$ along the ladder $BA$ . Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. $(Take g=9.8m/s^{2})$ .
(Hint: Consider the equilibrium of each side of the ladder separately.)

1645181494043

Answer:

The FBD of the figure is shown below :

1645181509526

Consider triangle ADH,

$AH\ =\ \sqrt{\left ( AD^2\ -\ DH^2 \right )}$ (AD and DH can be found using geometrical analysis.)

$=\ \sqrt{\left ( 0.8^2\ -\ 0.25^2 \right )}\ =\ 0.76\ m$

Now we will use the equilibrium conditions :

(i) For translational equilibrium :

$N_c\ +\ N_b\ =\ mg\ =\ 40\times 9.8\ =\ 392\ N$ ......................................(i)

(ii) For rotational equilibrium :

$-N_b\times BI\ +\ mg\times FG\ +\ N_c \times CI \ +\ T\times AG\ -\ T\times AG\ =\ 0$

or $(N_c\ -\ N_b)\times 0.5\ =\ 49$

or $N_c\ -\ N_b\ =\ 98$ ............................................................(ii)

Using (i) and (ii) we get :

$N_c\ =\ 245\ N$ and $N_b\ =\ 147\ N$

Now calculate moment about point A :

$-N_b \times BI \ +\ mg\times FG\ +\ T\times AG\ =\ 0$

Solve the equation : $T\ =\ 96.7\ N$

Q 7.23 A man stands on a rotating platform, with his arms stretched horizontally holding a $5kg$ weight in each hand. The angular speed of the platform is $30$ revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from $90 cm$ to $20 cm$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $7.6 kg m^{2}$ .

(a) What is his new angular speed? (Neglect friction.)

Answer:

Moment of inertia when hands are stretched :

$=\ 2\times mr^2\ =\ 2\times (0.5)9^2$

$=\ 8.1\ Kg\ m^2$

So the moment of inertia of system (initial ) = 7.6 + 8.1 = 15.7 Kg m ² .

Now, the moment of inertia when hands are folded :

$=\ 2\times mr^2\ =\ 2\times 5(0.2)^2\ =\ 0.4\ Kg\ m^2$

Thus net final moment of inertia is : = 7.6 + 0.4 = 8 Kg m ² .

Using conversation of angular momentum we can write :

$I_1\omega_1\ =\ I_2 \omega_2$

or $\omega_2 \ =\ \frac{15.7\times 30}{8}\ =\ 58.88\ rev/min$

Q 7.23 A man stands on a rotating platform, with his arms stretched horizontally holding a $5kg$ weight in each hand. The angular speed of the platform is $30$ revolutions per minute. The man then brings his arms close to his body with the distance of each
weight from the axis changing from $90 cm$ to $20 cm$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $7.6 kg m^{2}$ .

(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Answer:

No, the kinetic is not constant. The kinetic energy increases with decrease in moment of inertia. The work done by man in folding and stretching hands is responsible for this result.

Q7.24 A bullet of mass $10g$ and speed $500m/s$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0m$ wide and weighs $12kg$ . It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint: The moment of inertia of the door about the vertical axis at one end is $ML^{2}/3$ .)

Answer:

The imparted angular momentum is given by :

$\alpha \ =\ mvr$

Putting all the given values in the above equation we get :

$=\ (10\times 10^{-3})\times500\times \frac{1}{2}$

$=\ 2.5\ Kg\ m^2\ s^{-1}$

Now, the moment of inertia of door is :

$I\ =\ \frac{1}{3}ML^2$

or $=\ \frac{1}{3}(12)1^2\ =\ 4\ Kg\ m^2$

Also, $\alpha\ =\ I \omega$

or $\omega\ =\ \frac{\alpha}{I}\ =\ \frac{2.5}{4}\ =\ 0.625\ rad\ s^{-1}$

Q7.25 (a) Two discs of moments of inertia $I_{1}$ and $I_{2}$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds $\omega _{1}$ and $\omega _{2}$ are brought into contact face to face with their axes of rotation coincident.

(a) What is the angular speed of the two-disc system?

Answer:

Let the moment of inertia of disc I and disc II be I ₁ and I ₂ respectively.

Similarly, the angular speed of disc I and disc II be w ₁ and w ₂ respectively.

So the angular momentum can be written as :

$L_1\ =\ I_1 \omega_1$ and $L_2\ =\ I_2 \omega_2$

Thus the total initial angular momentum is : $=\ I_1 \omega_1\ +\ I_2 \omega_2$

Now when the two discs are combined the angular momentum is :

$L_f\ =\ (I_1\ +\ I_2)w$

Using conservation of angular momentum :

$I_1\omega_1\ +\ I_2\omega_2 =\ (I_1\ +\ I_2)w$

Thus angular velocity is :

$\omega \ =\ \frac{I_1\omega_1\ +\ I_2\omega_2}{I_1\ +\ I_2}$

Q7.25 (b) Two discs of moments of inertia $I_{1}$ and $I_{2}$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds $\omega _{1}$ and $\omega _{2}$ are brought into contact face to face with their axes of rotation coincident.

(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take $\omega _{1}\neq \omega _{2}$ .

Answer:

The initial kinetic energy is written as :

$K_i\ =\ \frac{1}{2}I_1\omega_1^2\ +\ \frac{1}{2}I_2\omega_2^2$

or $K_i\ =\ \frac{1}{2}\left ( I_1\omega_1^2\ +\ I_2\omega_2^2 \right )$

Now the final kinetic energy is :

$K_f\ =\ \frac{1}{2}\left ( I_1\ +\ I_2 \right )\omega^2$

Put the value of final angular velocity from part (a).

We need to find :

$K_i\ -\ K_f\ =\ \frac{1}{2}\left ( I_1\omega_1^2\ +\ I_2\omega_2^2 \right )\ -\ \frac{1}{2}\left ( I_1\ +\ I_2 \right )\omega^2$

Solve the above equation, we get :

$=\ \frac{I_1 I_2}{2(I_1\ +\ I_2)}(\omega_1^2\ +\ \omega_2^2\ -\ 2\omega_1\omega_2)$

or $=\ \frac{I_1 I_2}{2(I_1\ +\ I_2)}(\omega_1\ -\ \omega_2)^2$

$>\ 0$ (Since none of the quantity can be negative.)

Thus initial energy is greater than the final energy. (due to frictional force).

Q7.26 (a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point $(x,y)$ in the $x-y$ plane from an axis through the origin and perpendicular to the plane is $x^{2}+y^{2}$ ).

Answer:

Consider the figure given below:

1645181540750

The moment of inertia about x axis is given by :

$I_x\ =\ mx^2$

And the moment of inertia about y-axis is :

$I_y\ =\ my^2$

Now about z-axis :

$I_z\ =\ m\left ( \sqrt{\left ( x^2\ +\ y^2 \right )} \right )^2$

or $I_z\ =\ I_x\ +\ I_y$

Q 7.26 (b) Prove the theorem of parallel axes.
(Hint : If the centre of mass of a system of n particles is chosen to be the origin $\sum m_{i}r_{i}=0$ ).

Answer:

Consider the figure given below :

1645181561197 The moment of inertia about RS axis :-

$I_{RS}\ =\ \sum m_ir_i^2$

Now the moment of inertia about QP axis :-

$I_{QP}\ =\ \sum m_i(a+r_i)^2$

or $=\ \sum m_i(a^2 +r_i^2 + 2ar_i)$

or $=\ I_{RS} + \sum m_i a^2 + 2 \sum m_iar_i^2$

Thus $I_{QP} =\ I_{RS} + M a^2$ $\left ( \because 2\sum m_iar_i^2 \right )\ =\ 0$

Hence proved.

Q7.27 Prove the result that the velocity $v$ of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height $h$ is given by

$v^{2}=\frac{2gh}{(1+k^{2}/R^{2)}}$
using dynamical consideration (i.e. by consideration of forces and torques). Note $k$ is the radius of gyration of the body about its symmetry axis, and $R$ is the radius of the body. The body starts from rest at the top of the plane.

Answer:

Consider the given situation :

1645181579336

The total energy when the object is at the top (potential energy) = mgh.

Energy when the object is at the bottom of the plane :

$E_b\ =\ \frac{1}{2}I\omega^2\ +\ \frac{1}{2}mv^2$

Put $I\ =\ mk^2$ and $v\ =\ \omega r$ , we get :

$E_b\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )$

By the law of conservation of energy we can write :

$mgh\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )$

$v\ =\ \frac{2gh}{\left ( 1\ +\ \frac{k^2}{r^2} \right )}$ .

Q7.28 A disc rotating about its axis with angular speed $\omega _{0}$ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is $R$ . What are the linear velocities of the points $A$ , $B$ and $C$ on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?

1645181599740 Answer:

Let the angular speed of the disc is $\omega$ .

So the linear velocity can be written as $v\ =\ \omega r$

(a) Point A:-

The magnitude of linear velocity is $\omega r$ and it is tangentially rightward.

(b) Point B:-

The magnitude linear velocity is $\omega r$ and its direction is tangentially leftward.

The magnitude linear velocity is $\frac{ \omega r}{2}$ and its direction is rightward.

The disc cannot roll as the table is frictionless.

Q7.29 (a) Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated

(a) Give the direction of frictional force at $B$ , and the sense of frictional torque, before perfect rolling begins.

Answer:

Since the velocity at point B is tangentially leftward so the frictional force will act in the rightward direction.

The sense of frictional torque is perpendicular (outward) to the plane of the disc.

Q7.29 (b) Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

(b) What is the force of friction after perfect rolling begins?

Answer:

Perfect rolling will occur when the velocity of the bottom point (B) will be zero. Thus the frictional force acting will be zero.

Q7.30 A solid disc and a ring, both of radius $10cm$ are placed on a horizontal table simultaneously, with initial angular speed equal to $10\Pi rad\: s^{-1}$ . Which of the two will start to roll earlier ? The co-efficient of kinetic friction is $\mu _{k}=0.2$ .

Answer:

Friction is the cause for motion here.

So using Newton's law of motion we can write :

$f\ =\ ma$

$\mu _k mg\ =\ ma$

or $a\ =\ \mu _k g$

Now by the equation of motion, we can write :

$v\ =\ u\ +\ at$

or $v\ =\ \mu _k gt$

The torque is given by :

$\tau\ =\ I \alpha$

or $r\times f \ =\ -I \alpha$

or $\mu _k mgr \ =\ -I \alpha$

or $\alpha\ =\ -\frac{\mu _k mgr}{I}$

Now using the equation of rotational motion we can write :

$\omega\ =\ \omega _o\ +\ \alpha t$

or $\omega\ =\ \omega _o \ +\ -\frac{\mu _k mgr}{I}t$

Condition for rolling is $v\ =\ \omega r$

So we can write :

$v =\ r\left ( \omega _o \ +\ -\frac{\mu _k mgr}{I}t \right )$

For ring the moment of inertia is : $mr^2$

So we have :

$t\ =\ \frac{r \omega _o}{2 \mu _k g}$

or $=\ \frac{0.1\times 10\times 3.14}{2\times 0.2 \times 10}\ =\ 0.80\ s$

Now in case of the disc, the moment of inertia is :

$I\ =\ \frac{1}{2}mr^2$

Thus $t\ =\ \frac{r \omega _o}{3 \mu _k g}$

or $=\ \frac{0.1\times 10\times 3.14}{3 \times 0.2 \times 9.8}\ =\ 0.53\ s$

Hence disc will start rolling first.

Q7.31 (a) A cylinder of mass $10kg$ and radius $15cm$ is rolling perfectly on a plane of inclination $30^{0}$ . The coefficient of static friction $\mu _{s}=0.25$ .

(a) How much is the force of friction acting on the cylinder?

Answer:

Consider the following figure :

1645181627992

Moment of inertia of cylinder is :

$I\ =\ \frac{1}{2}mr^2$

Thus acceleration is given by :

$a\ =\ \frac{mg \sin \Theta}{m\ +\ \frac{1}{2}\times \frac{ mr^2}{r^2}}$

or $=\ \frac{2}{3}\ g \sin 30^{\circ}\ =\ 3.27\ m/s^2$

Now using Newton's law of motion :

$F\ =\ ma$

or $mg \sin 30^{\circ}\ -\ f\ =\ ma$

or $f\ =\ mg \sin 30^{\circ}\ -\ ma$

or $=\ 49\ -\ 32.7\ =\ 16.3\ N$

Hence frictional force is 16.3 N.

Q7.31 (b) A cylinder of mass $10kg$ and radius $15cm$ is rolling perfectly on a plane of inclination $30^{0}$ . The coefficient of static friction $\mu _{s}=0.25$ .

(b) What is the work done against friction during rolling ?

Answer:

We know that the bottommost point of body (which is in contact with surface) is at rest during rolling. Thus work done against the frictional force is zero.

Q7.31 (c) A cylinder of mass $10kg$ and radius $15cm$ is rolling perfectly on a plane of inclination $30^{0}$ . The coefficient of static friction $\mu _{s}=0.25$ .

(c) If the inclination $\Theta$ of the plane is increased, at what value of $\Theta$ does the cylinder begin to skid, and not roll perfectly?

Answer:

In case of rolling without any skidding is given by :

$\mu\ =\ \frac{1}{3}\tan \Theta$

Thus $\tan \Theta\ =\ 3 \mu$

or $=\ 3 \times (0.25)\ =\ 0.75$

or $\Theta \ =\ 37.87 ^{\circ}$

Q7.32 (a) Read each statement below carefully, and state, with reasons, if it is true or false;

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the $CM$ of the body.

Answer:

False . Friction also opposes the relative motion between the contacted surfaces. In the case of rolling, the cm is moving in backward direction thus the frictional force is directed in the forward direction.

Q7.32 (b) Read each statement below carefully, and state, with reasons, if it is true or false;

(b) The instantaneous speed of the point of contact during rolling is zero.

Answer:

True. This is because the translational speed is balanced by rotational speed.

Q7.32 (c) Read each statement below carefully, and state, with reasons, if it is true or false;

Answer:

False. The value of acceleration at contact has some value, it is not zero as the frictional force is zero but the force applied will give some acceleration.

Q7.32 (d) Read each statement below carefully, and state, with reasons, if it is true or false;

(d) For perfect rolling motion, work done against friction is zero.

Answer:

True. As the frictional force at the bottommost point is zero, so the work done against it is also zero.

Q7.32 (e) Read each statement below carefully, and state, with reasons, if it is true or false;

(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Answer:

True. Since it is a frictionless plane so frictional force is zero thus torque is not generated. This results in slipping not rolling.

Q7.33 (a) Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :

(a) Show $P=P_{i}^{^{'}}+m_{i}V$
where $P_{i}$ is the momentum of the ith particle (of mass $m_{i}$ )and $P_{t}^{'}=m_{t}V_{t}$ Note $V_{t}^{'}$ is the velocity
of the ith particle relative to the centre of mass.Also, prove using the definition of the centre of mass $\sum P_{t}'=0$

Answer:

The momentum of i _th particle is given by : $p_i\ =\ m_iv_i$

The velocity of the centre of mass is V.

Then the velocity of ith particle with respect to the center of mass will be : $v_i'\ =\ v_i\ -\ v$

Now multiply the mass of the particle to both the sides, we get :

$mv_i'\ =\ mv_i\ -\ mv$

or $p_i'\ =\ p_i\ -\ p$ (Here p' _i is the momentum of ith particle with respect to center of mass.)

or $p_i\ =\ p_i'\ +\ p$

Now consider $p_i'$ :

$\sum p_i'\ =\ \sum m_iv_i'\ =\ \sum m_i\frac{dr_i}{dt}$

But as per the definition of centre of mass, we know that :

$\sum m_ir_i'\ =\ 0$

Thus $\sum p_i\ =\ 0$

Q7.33 (b) Separation of Motion of a system of particles into the motion of the centre of mass and motion about the centre of mass :

(b) show $K=K^{'}+1/2MV^{2}$ where $K$ is the total kinetic energy of the system of particles, $K^{'}$ is the total kinetic energy of the
system when the particle velocities are taken with respect to the centre of mass and $MV^{2}/2$ is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.

Answer:

From the first part we can write :

$\sum mv_i'\ =\ \sum mv_i\ -\ \sum mv$

or $\sum mv_i\ =\ \sum mv_i'\ +\ \sum mv$

Squaring both sides (in vector form taking dot products with itself), we get :

$\sum mv_i. \sum mv_i\ =\ \sum m(v_i'\ +\ v)\ .\ \sum m(v_i'\ +\ v)$

or $M^2 \sum v_i^2\ =\ M^2 \sum v_i'^2\ +\ M^2v^2$

or $\frac{1}{2}M \sum v_i^2\ =\ \frac{1}{2}M \sum v_i'^2\ +\ \frac{1}{2}Mv^2$

Hence $K\ =\ K'\ +\ \frac{1}{2}Mv^2$

Q7.33 (c) Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :

(c) Show $L=L^{'}+R\times MV$
where $L=\sum r_{i}^{'}\times P_{i}^{'}$
is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $r_{i}^{'}=r_{i}-R$ ; rest of the notation is the standard notation used in the chapter. Note $L^{'}$ and $MR\times V$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

Answer:

The position vector of the ith particle (with respect to the centre of mass) is given by :

$r_i'\ =\ r_i\ -\ R$

or $r_i\ =\ r_i'\ +\ R$

From the first case we can write :

$p_i\ =\ p_i'\ +\ p$

Taking cross product with position vector we get ;

$\sum r_i\times p_i\ =\ \sum r_i\times p_i'\ +\ \sum r_i\times p$

or $L\ =\ L'\ +\ \sum R\times p_i'\ +\ \sum r_i\times m_iv\ +\ \sum R\times m_iv$

or $L=L^{'}+R\times MV$

7.33 (d) Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :

(d) Show $\frac{dL^{'}}{dt}=\sum r_{i}^{'}\times \frac{dp^{'}}{dt}$
Further, show that

$\frac{dL^{'}}{dt}=\tau ^{'}_{ext}$

where $\tau ^{'}_{ext}$ is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and third law of motion. Assume the
internal forces between any two particles act along the line joining the particles.)

Answer:

Since we know that :

$L'\ =\ \sum r_i' \times p_i'$

Differentiating the equation with respect to time, we obtain :

$\frac{dL'}{dt}\ =\ d\frac {\left ( \sum r_i' \times p_i' \right )}{dt}$

or $=\ \frac{d\left ( \sum m_ir_i' \right )}{dt}\times v_i'\ +\ \sum r_i'\ \times \frac{d}{dt}p_i'$

or $\frac {dL'}{dt} =\ \sum r_i'\ \times m_i \frac{d}{dt}v_i'$ $\left ( \because \sum m_ir_i\ =\ 0 \right )$

Now using Newton's law of motion we can write :

$\sum r_i'\ \times m_i \frac{d}{dt}v_i'\ =\ \tau _{ext}'$

Thus $\frac{dL^{'}}{dt}=\tau ^{'}_{ext}$

Chapter 7 of NCERT Physics Solutions consists of a total of thirty-three questions. Questions 22-33 belong to the additional exercise section. The solutions provided in the NCERT Solutions are explained step by step and are easy to understand. Students can also download the solutions in PDF format and use them according to their convenience. It is always advisable to attempt the problems on their own first and then refer to the solutions provided to check their understanding and identify areas of improvement.

NCERT Solutions for Class 11 Physics Chapter Wise

Chapter 1	Physical world
Chapter 2	Units and Measurement
Chapter 3	Motion in a straight line
Chapter 4	Motion in a Plane
Chapter 5	Laws of Motion
Chapter 6	Work, Energy and Power
Chapter 7	System of Particles and Rotational motion
Chapter 8	Gravitation
Chapter 9	Mechanical Properties of Solids
Chapter 10	Mechanical Properties of Fluids
Chapter 11	Thermal Properties of Matter
Chapter 12	Thermodynamics
Chapter 13	Kinetic Theory
Chapter 14	Oscillations
Chapter 15	Waves

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Get a grip on Physics with our helpful class 11 physics chapter 7 exercise solutions. They come with important formulas (provided below), easy-to-understand diagrams, and a useful eBook link. Boost your understanding of the System of Particles and Rotational Motion and do great in chapter 7 physics class 11 numericals.

Center of mass

Motion of COM

1693275990889

$\vec{v}_{\mathrm{cm}}=\frac{\sum m_i \vec{v}_i}{M}, \quad \vec{p}_{\mathrm{cm}}=M \vec{v}_{\mathrm{cm}}, \quad \vec{a}_{\mathrm{cm}}=\frac{\vec{F}_{\mathrm{ext}}}{M}$

Angular Velocity & Linear Velocity

Average angular velocity =∆θ/∆t

Instantaneous angular velocity, ω=dθ/dt

Moment of Inertia of few useful configurations

Class 11 Physics Chapter 7 NCERT solutions - Download for Free

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Physics class 11 chapter 7 NCERT solutions system of particles and rotational motion are designed for comprehensive learning.
ch 7 class 11 physics physics solutions are concise and well-explained, ensuring that students can understand them easily.
All the essential aspects of the physics class 11 chapter 7 are covered in these solutions.
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Overall, the system of particles class 11 NCERT solutions can help students master the entire chapter with ease.

Significance of NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion:

To understand the NCERT solutions for Class 11 Physics chapter 7 System of Particles and Rotational Motion knowledge of previous chapters of Class 11 Physics is a prerequisite.
The NCERT solutions for class 11 help to understand how to apply concepts studied in numerical problems.
For exams like NEET and JEE Mains, two or more questions are expected from the chapter System of Particles and Rotational Motion.

NCERT Physics Exemplar Solutions Class 11 For All The Chapters:

Chapter 1 Physical World

Chapter 2 Units and Measurements

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

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1. What is the weighage for the Class 11 NCERT chapter 7 for NEET exam

For NEET exam around 5% of questions are asked from the chapter system of particles and rotational motion. NEET questions can be practiced refering the previous papers of NEET. For more number of questions refer to NCERT exemplar questions for Rotational Motion.

2. How many questions can be expected from the chapter system of particles and rotational motion for JEE Main exam

2 questions can be expected from System of Particles and Rotational motion for JEE Main exam. Oscillation and Waves is one of the important topics for exams like KVPY, NSEP and JEE Main

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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Option 3) $zero\;$	Option 4) $K/4$

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Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

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Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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NCERT Solutions for Class 11 Physics Chapter 7 System Of Particles And Rotational Motion

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NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

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System of Particles Class 11 NCERT Solutions - Additional Exercis e

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Center of mass

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