NCERT Solutions for Class 11 Physics Chapter 4 Laws Of Motion

NCERT Solutions for Class 11 Physics Chapter 4 Laws Of Motion

Vishal kumarUpdated on 16 Oct 2025, 12:40 AM IST

Have you ever experienced an unexpected jolt as soon as a car accelerates and runs backwards? This everyday life experience is a true demonstration of the Laws of Motion by Newton, which is the main concept of Class 11 Physics Chapter 4 - Laws of Motion. Free-body diagrams, inertia, equilibrium, friction, and force analysis are examples of concepts found in Chapter 4 of 11th Grade Physics that are difficult to learn without the help of NCERT Solutions. The NCERT Solutions for Class 11 Physics Chapter 4 - Laws of Motion are based solely on the newest NCERT syllabus and are highly helpful for CBSE board exams, JEE Main, NEET, and Olympiads.

This Story also Contains

  1. Class 11 Physics Chapter 4 - Laws of Motion Question Answers: Download Solution PDF
  2. Laws of Motion NCERT Solutions: Exercise Questions
  3. Class 11 Physics NCERT Chapter 4: Higher Order Thinking Skills (HOTS) Questions
  4. Class 11 Physics Chapter 4 - Laws of Motion: Topics
  5. Laws of Motion class 11 Question Answers: Important Formulae
  6. Approach to Solve Questions of Class 11 Physics Chapter 4 - Laws of Motion
  7. What Extra Should Students Study Beyond NCERT for JEE?
  8. Importance of NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion
  9. NCERT Solutions for Class 11 Physics Chapter-Wise

Students, through practice of the step-by-step NCERT solutions, could be in a position to enhance their pace of accuracy and confidence in solving both numerical and theoretical problems. More importantly, the solutions allow the students to relate theory and application in the real world-such as learning how rockets take off, how objects resist motion, and how forces can interact in their everyday life. Therefore, the NCERT Solutions for Class 11 Physics Chapter 4 - Laws of Motion are an important study material in the construction of solid basics and success in both board and competitive examinations.

Class 11 Physics Chapter 4 - Laws of Motion Question Answers: Download Solution PDF

The Laws of Motion class 11 question answers can be downloaded in a well-organised PDF format that is free and easy to download by students. This solution PDF is structured to give step-by-step responses to every exercise question that would enable the learners to practice effectively and enhance their ability to solve problems. This makes revision easier as the solutions can be downloaded into one file to do homework, assignments, class tests, or to prepare for final exams.

Download PDF

Laws of Motion NCERT Solutions: Exercise Questions

NCERT Solutions for Class 11 Physics Chapter 4 - Laws of Motion provide answers to each exercise of the NCERT textbook in a clear stepwise manner. This chapter presents some underlying concepts of mechanics, such as the concept of force, inertia, laws of Newton and friction. These remedies not only allow the students to firm up their ideas, but also to perform efficiently in the school exams and the competitive exams, such as the JEE and the NEET

Q4.1 (a) Give the magnitude and direction of the net force acting on a drop of rain falling down at a constant speed

Answer:

Since the drop is falling with constant speed, the net acceleration of the drop is zero. This means the net force on the drop is zero.

Q4.1 (b) Give the magnitude and direction of the net force acting on a cork of mass 10 g floating on water,

Answer:

Since the cork is floating on the water, that implies the gravitational force of the cork is balanced by the buoyant force. Thus net force acting on the cork is zero.

Q4.1 (c) Give the magnitude and direction of the net force acting on a kite skillfully held stationary in the sky.

Answer:

Since the kite is stationary in the sky, according to Newton's law, the net force on the kite is zero.

Q4.1 (d) Give the magnitude and direction of the net force acting on a car moving with a constant velocity of 30 km/h on a rough road.

Answer:

Since the car is moving with constant velocity thus the net acceleration of the car is zero. Thus net force acting on the car is zero.

Q4.1 (e) Give the magnitude and direction of the net force acting on a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Answer:

Since the electron is free from the electric and magnetic field, so net force acting on the electron is zero.

Q4.2 (a) A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble during its upward motion

Answer:

During upward motion, the force acting on the pebble is only the gravitational force, which is acting in a downward direction.

The gravitational force :

=mg=0.05(10)N=0.5N

Q4.2 (b) A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble during its downward motion

Answer:

Since the pebble is moving in the downward direction, the net force acting is also in the downward direction.

The net force is the gravitational force.

Gravitational force:-

=mg=0.(05)10=0.5N

Q4.2 (c) A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

Answer:

At the highest point, velocity becomes zero for a moment. At this point, the net force acting is the gravitational force, which acts in the downward direction.

Gravitational force:-

$= mg= (0.05)10= 0.5 N$

When a pebble is thrown at 45 ° with the horizontal direction, then the net force will also be the same gravitational force.

Q4.3 (a) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, just after it is dropped from the window of a stationary train. Neglect air resistance throughout.

Answer:

Since the train is stationary so the net force acting on the stone is the gravitational force, which acts in the downward direction.

Gravitational force :

$F\ =\ ma\ =\ mg$

or $F=\ (0.1)10$

or $F=\ 1\ N$

Q4.3 (b) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg just after it is dropped from the window of a train running at a constant velocity of 36 km/h, Neglect air resistance throughout.

Answer:

Since the train is travelling at a constant speed so the acceleration of the train is zero. Thus, there is no force on the stone due to the train.

The net force acting on the stone will be the gravitational force, which acts in a downward direction.

Gravitational force:-

$F\ =\ mg$

or $F=\ (0.1)10\ N$

or $F=\ 1\ N$

Q4.3 (c) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg just after it is dropped from the window of a train accelerating with $1 ms^{-2}$. Neglect air resistance throughout.

Answer:

Just after the stone is dropped, the stone is free from the acceleration of the train. Thus, the force acting on the stone will be just the gravitational force, which acts in the downward direction.

The gravitational force is given by:-

$F\ =\ mg$

or $F=\ (0.1)10$

or $F=\ 1\ N$

Q4.3 (d) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout.

Answer:

As the stone is in contact with the train, the acceleration of the stone is the same as that of the train, i.e. 1 m/s 2.

Thus force acting on the stone is given by :

$F\ =\ ma$

or $F=\ 0.1\times 1\ =\ 0.1\ N$

This force is acting in the horizontal direction.

Q4.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is :

(i) $T$

(ii) $T - \frac{mv^2}{l}$

(iii) $T + \frac{mv^2}{l}$

(iv) 0

$T$ is the tension in the string. [Choose the correct alternative].

Answer:

When the particle is moving in a circular path, the centripetal force will be :

$F_c\ =\ \frac{mv^2}{r}$

This centripetal force will be balanced by the tension in the string.

So, the net force acting is :

$F\ =\ T\ =\ \frac{mv^2}{l}$

Q4.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of $15 ms^{-1}$. How long does the body take to stop?

Answer:

We are given the retarding force. So we can find the deceleration this force is causing.

By Newton's second law of motion, we get :

$F\ =\ ma$

or $-50\ =\ (20)a$

or $a\ =\ \frac{50}{20}\ =\ - 2.5\ m/s^2$

Now we will use the first equation of motion,

$v\ =\ u\ +\ at$

The final velocity, in this case, will be zero (Since the vehicle stops).

$0\ =\ 15\ +\ (-2.5)t$

Thus $t\ =\ \frac{15}{2.5}\ =\ 6\ s$

Thus, the time taken to stop the vehicle is 6 seconds.

Q4.6 A constant force acting on a body of mass 3.0 kg changes its speed from $2.0 ms^{-1}$ to $3.5 ms^{-1}$ in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Answer:

Since the velocity of the body is increased by applying the force. This is possible only when the force is applied in the direction of the motion.

To find the magnitude of the force, we need to calculate the acceleration.

By using the first equation of the motion,

$v\ =\ u\ +\ at$

or $3.5\ =\ 2\ +\ a(25)$

or $a\ =\ \frac{3.5-2}{25}\ =\ 0.06\ m/s^2$

Thus, force can be written as :

$F\ =\ ma$

$F=\ 3\times 0.06$

$F=\ 0.18\ N$

Q4.7 A body of mass 5 kg is acted upon by two perpendicular forces of 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Answer:

The magnitude of the resulting force can be found by :

$R\ =\ \sqrt{F_1^2\ +\ F_2^2 }$

or $R=\ \sqrt{8^2\ +\ 6^2 }$

or $R=\ 10\ N$

Now force direction,

$\tan \theta \ =\ \frac{6}{8}$

or $\theta \ =\ 37^{\circ}$

The acceleration of the body is given by :

$a\ =\ \frac{F}{m}$

or $a\ =\ \frac{10}{5}\ =\ 2\ m/s^2$

Hence acceleration of the body is $2\ m/s^2$ and its direction is the same as that of the resultant force.

Q4.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg, and the mass of the driver is 65 kg.

Answer:

Total mass of the system = 400 + 65 = 425 kg

Using the first law of motion, we get

$v\ =\ u\ +\ at$

Since the car comes to rest, the final velocity will be zero.

$0\ =\ 10\ +\ a(4)$

or

$a\ =\ \frac{-10}{4}\ =\ -2.5\ m/s^2$

So the force required :

$F\ =\ ma$

or $F=\ 465\times (-2.5)$

or $F\ =\ -1162.5\ N$

So the magnitude of the force is 1162.5 N, and it is retarding force.

Q4.9 A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of $5.0 ms^{-2}$. Calculate the initial thrust (force) of the blast.

Answer:

Let the initial thrust be F Newton.

Using Newton's second law of motion, we get

$F\ -\ mg\ =\ ma$

or $F\ =\ ma\ +\ mg$

$F\ =\ m(a+g)\ N$

Substituting values in this equation, we get :

$F\ =\ 20000(5+10)$

or $F\ =\ 3\times 10^5\ N$

Q4.10 A body of mass 0.40 kg moving initially with a constant speed of $10ms^{-1}$ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be $t = 0$, the position of the body at that time to be $x = 0$, and predict its position at $t = -5s, 25s, 100s$.

Answer:

The acceleration of force is given by :

$a\ =\ \frac{F}{m}$

$a\ =\ \frac{-\ 8}{0.4}\ =-\ 20\ m/s^2$

At t = - 5 s :

There is no force acting, so acceleration is zero and u = 10 m/s

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or $s=\ 10(-5)\ +\ \frac{1}{2}.0.t^2$

or $s=\ - \ 50\ m$

At t = 25 s :

Acceleration is - 20 m/s 2 and u = 10 m/s

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or $s=\ 10(25)\ +\ \frac{1}{2}(-20)25^2$

or $s=\ -\ 6000\ m$

At t = 100 s

We have acceleration for the first 30 seconds, and then it will move with constant speed.

So for 0 < t < 30 :

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or $s=\ 10(30)\ +\ \frac{1}{2}(-20)30^2$

or $s=\ -\ 8700\ m$

Now for t > 30 s :

We need to calculate the velocity at t = 30 sec, which will be used as the initial velocity for 30 < t < 100.

$v\ =\ u\ +\ at$

or $v=\ 10\ +\ (-20)30\ =\ -\ 590\ m/s$

Now $s\ =\ vt\ +\ \frac{1}{2}at^2$

or $s=\ (-590)70\ +\ \frac{1}{2}.0.t^2$

or $s=\ -\ 41300\ m$

Hence total displacement is : - 8700 + (- 41300 ) = - 50000 m.

Q4.11 (a) A truck starts from rest and accelerates uniformly at $2.0ms^{-2}$. At $t = 10s$, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the velocity of the stone at $t = 11s$? (Neglect air resistance.)

Answer:

The initial velocity of the truck is given as zero.

We need to find the final velocity (at t = 10 s), so we will use the equation of motion :

$v\ =\ u\ +\ at$

or $v=\ 0\ +\ (2)10\ =\ 20\ m/s$

This is the velocity imparted to the stone by the truck; that's why it is a horizontal component of velocity.

The stone is dropped at t = 10 sec. So it has travelled 1 sec in the air (11 - 10 = 1 s). We need to find the final vertical velocity.

$v\ =\ u\ +\ at$

$v=\ 0\ +\ 10(1)\ =\ 10\ m/s$

Thus resultant of both the components is the required velocity.

$R\ =\ \sqrt{v_h^2\ +\ v_v^2}$

or $R=\ \sqrt{20^2\ +\ 10^2}$

or $R=\ 22.36\ m/s$

Direction :

$\tan \theta \ =\ \frac{v_v}{v_h}\ =\ \frac{10}{20}$

or $\theta \ =\ 26.57^{\circ}$

Q4.11 (b) A truck starts from rest and accelerates uniformly at $2.0ms^{-2}$. At $t = 10 s$, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the acceleration of the stone at $t = 11s$? (Neglect air resistance.)

Answer:

When the stone is dropped, the stone comes only in effect of gravity.

So the acceleration of the stone is 10 m/s 2, and it acts in the downward direction.

Answer:

At the extreme positions, the velocity of the bob will become zero for a moment. So if we cut the string at this time, then Bob will fall vertically downward due to gravity.

Q4.12 (b) A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is $1ms^{-1}$. What is the trajectory of the bob if the string is cut when the bob is at its mean position?

Answer:

At the mean position, the bob will have velocity tangential to the circular path (it will be completely horizontal). If the bob is cut at this place, then it will follow a parabolic path having only horizontal velocity.

Q4.13 (a) A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s -1. What would be the readings on the scale?

Answer:

Since the lift is moving with a constant speed, or in this case, constant velocity so the acceleration provided to the man by the lift is zero.

The net force on the man will be zero. So a = 0.

Using Newton's law of motion, we can write :

$R\ -\ mg\ =\ ma$

or $R\ -\ mg\ =\ 0$

Thus $R\ =\ mg$

$R=\ 70\times10\ =\ 700\ N$

So, reading on the weighing scale will be :

$=\ \frac{R}{g}\ =\ \frac{700}{10}\ =\ 70\ Kg$

Q4.13 (b) A man of mass 70 kg stands on a weighing scale in a lift which is moving downwards with a uniform acceleration of $5ms^{-2}$. What would be the readings on the scale?

Answer:

Using Newton's law of motion, we have :

$R\ +\ mg\ =\ ma$

or $R\ -\ 70(10)\ =\ 70(-5)$ (Since we took downward direction as negative and upward as positive).

or $R\ =\ 700\ -\ 350\ =\ 350\ N$

Thus, the reading on the weighing scale will be :

$=\ \frac{R}{g}\ =\ \frac{350}{10}\ =\ 35\ Kg$

Q4.13 (c) A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $5 ms^{-2}$ . What would be the readings on the scale?

Answer:

The acceleration of lift is given to be 5 m/s 2. Let us assume the upward direction to be positive.

Using Newton's law of motion, we can write :

$R\ -\ mg\ =\ ma$

or $R\ -\ 70(10)\ =\ 70(5)$

or $R\ =\ 350 +\ 700\ =\ 1050\ N$

Thus, the reading of the weighing scale will be :

$=\ \frac{R}{g}\ =\ \frac{1050}{10}\ =\ 105\ Kg$

Q44.13 (d) A man of mass 70 kg stands on a weighing scale in a lift which is moving. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:

If the lift falls freely, then the acceleration of the lift will be the acceleration due to gravity.

Using Newton's law of motion, we can write :

$R\ -\ mg\ =\ ma$

or $R\ -\ mg\ =\ m(-g)$

or $R\ =\ m(-g)\ +\ mg\ =\ 0\ N$

Thus, the reading of the weighing scale will also be zero since there is no normal force.

This state is called the state of weightlessness.

Q4.14 (a) Figure shows the position-time graph of a particle of mass 4 kg. What is the force on the particle for $t < 0, t > 4 s, 0 < t < 4 s$ ?

Answer:

(i) For t < 0 :

In this range, the position of the particle coincides with time, which implies no motion takes place. Hence net force on the particle is zero.

(ii) For t > 4 :

In this range displacement of the particle is not changed, so the net force is zero.

(iii) For 0 < t < 4 :

In this range, the slope of the position-time graph is constant, which means the particle is moving with constant speed. And hence net force, in this case, is zero.

Q4.14 (b) Figure shows the position-time graph of a particle of mass 4 kg. What is the impulse at t = 0 and t = 4s? (Consider one-dimensional motion only).

Answer:

Impulse is defined by :

$J\ =\ mv\ -\ mu$ (Change in momentum).

At t = 0 s :

$u\ =\ 0\ and\ v\ =\ \frac{3}{4}\ m/s$

or $J\ =\ 4(\frac{3}{4})\ -\ 0$

or $J\ =\ 3\ Kg\ m/s$

At t = 4 s :

$u\ =\ \frac{3}{4}\ m/s$ and $\ v\ =\ 0$

Thus $J\ =\ 0\ -\ 4\left ( \frac{3}{4} \right )$

or $J\ =\ -3\ Kg\ m/s$

Q4.15 (i) Two bodies of masses 10 kg and 20 kg respectively, kept on a smooth, horizontal surface, are tied to the ends of a light string. A horizontal force F = 600 N is applied to A along the direction of the string. What is the tension in the string?

Answer:

We will consider A and B in a system. So total mass in the system is = 10 + 20 = 30 Kg.

Thus acceleration of the system is given by :

$a\ =\ \frac{F}{m}$

or $a=\ \frac{600}{30}\ =\ 20\ m/s^2$

When force is applied at block A :

Using Newton's law of motion :

$F\ -\ T\ =\ m_1a$

or $600\ -\ T\ =\ 10\times20$

Thus $T\ =\ 400\ N$

Q4.15 (ii) Two bodies of masses 10 kg and 20 kg respectively, kept on a smooth, horizontal surface, are tied to the ends of a light string. A horizontal force F = 600 N is applied to B along the direction of the string. What is the tension in the string?

Answer:

We will consider A and B in a system. So total mass in the system is = 10 + 20 = 30 Kg.

Thus acceleration of the system is given by :

$a\ =\ \frac{F}{m}$

or $a=\ \frac{600}{30}\ =\ 20\ m/s^2$

When force is applied at block B :

Using Newton's law of motion, we can write

$F\ -\ T\ =\ m_2a$

or $600 -\ T\ =\ 20(20)$

Thus $T\ =\ 200\ N$

Q4.16 Two masses 8 kg and 12 kg are connected at the two ends of a light, inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer:

Since both the masses are connected with a string so they will have the same acceleration, let say 'a'.

block pully System

We will apply Newton's law for each block individually.

For smaller block (8 Kg) :

$T\ -\ m_sg\ =\ m_sa$ ...................................................(i)

For a larger block (12 Kg) :

The equation of motion is given by :

$m_lg\ -\ T\ =\ m_la$ ......................................................(ii)

Adding both the equations, we get :

The acceleration is given by :

$a\ =\ \left ( \frac{m_l\ -\ m_s}{m_l\ +\ m_s} \right )g$

or $a=\ \left ( \frac{12\ -\ 8}{12\ +\ 8} \right )10$

or $a=\ 2\ m/s^2$

Now, put the value of acceleration in any of the equations to get the value of T.

$T\ =\ \left ( m_l\ -\ \frac{m_l^2\ -\ m_s m_l}{m_s\ +\ m_l} \right )g$

or $T=\ \left ( \frac{2m_s m_l}{m_s\ +\ m_l} \right )g$

or $T=\ \left ( \frac{2\times 12\times 8}{12\ +\ 8} \right )10$

$T=\ 96\ N$

Thu,s the tension in the string is 96 N.

Q4.17 Anucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.

Answer:

Let the mass of parent nuclei be m.

And the mass of daughter nuclei is m1 and m2.

Initial momentum is zero since the nuclei are at rest.

But after dissociation, the momentum becomes :

$M\ =\ m_1\ v_1\ +\ m_2 v_2$

Using conservation of momentum,

$0\ =\ m_1\ v_1\ +\ m_2 v_2$

or $v_1\ =\ \frac{-m_2 v_2}{m_1}$

Thus, both velocities have opposite directions.

Q4.18 Two billiard balls, each of mass 0.05 kg moving in opposite directions with speed $6ms^{-1}$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Answer:

The impulse imparted can be calculated by knowing the change in momentum.

The initial momentum of each ball is :

$p_i\ =\ mu\ =\ (0.05) 6\ =\ 0.3\ Kg\ m/s$

The final momentum is given by :

$p_f\ =\ mv\ =\ (0.05) (-6)\ =\ -\ 0.3\ Kg\ m/s$

So the impulse is :

$J\ =\ p_f\ -\ p_i$

or $J=\ -0.3\ -\ 0.3$

or $J=\ -0.6\ Kg\ m/s$

Q4.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is $80 ms^{-1}$, what is the recoil speed of the gun?

Answer:

In this question, we will use the conservation of momentum.

Final momentum = Initial momentum

For initial momentum :

Both gun and shell are at rest initially, so momentum is zero.

For final momentum :

The direction of the velocity of the shell is opposite to that of the gun.

So, $p_f\ =\ m_s v_s\ -\ m_g V_g$

The recoil speed of the gun :

$V_g\ =\ \frac{m_s v_s}{m_g}$

or $V_g\ =\ \frac{0.020\times 80}{100\times 1000}\ =\ 0.016\ m/s$

Q4.20 A batsman deflects a ball by an angle of 45° without changing its initial speed, which is equal to 54 km/h. What is the impulse imparted to the ball? (The mass of the ball is 0.15 kg.)

Answer:

The situation is shown below :

The horizontal components of velocity are to be considered for imparting impulse, as vertical components are in the same direction thus impulse in the vertical direction is zero.

The impulse is given by a change in momentum.

Initial momentum = $-mv \cos \theta$

Final momentum = $mv \cos \theta$

Thus impulse is, J $=\ mv \cos \theta\ -\ (-\ mv \cos \theta)\ =\ 2mv \cos \theta$

or $J=\ 2\times0.15 \times15\ \cos 22.5 ^{\circ}$

or $J=\ 4.16\ Kg\ m/s$

Q4.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Answer:

We are given, frequency :

$n\ =\ \frac{40}{60}\ =\ \frac{2}{3}$

So, the angular velocity becomes :

$\omega \ =\ 2\pi n$

By Newton's law of motion, we can write :

$T\ =\ F_{centripetal }$

or $T=\ \frac{mv^2}{r}\ =\ mw^2r$

or $T=\ (0.25)\left ( 2\times\Pi \times \frac{2}{3} \right )^2(1.5)$

or $T=\ 6.57\ N$


Now, we are given maximum tension, and we need to find the maximum velocity for that :

$T_{max}\ =\ \frac{mv_{max}^2}{r}$

Thus, $v_{max}\ =\ \sqrt{\frac{T_{max}.r}{m}}$

or $v_{max}=\ \sqrt{\frac{200\times 1.5}{0.25}}$

or $v_{max}=\ 34.64\ m/s$

Q4.22 (a) If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :

(a) The stone moves radially outwards,

Answer:

The stone should move in the direction of its velocity at that instant. Since velocity is tangential at that moment, the stone will not move radially outward.

Q4.22 (b) If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :

(b) The stone flies off tangentially from the instant the string breaks,

Answer:

(b) This statement is correct as the direction of velocity at the instant of the breaking of the string is tangential, thus the stone will move tangentially.

Q4.22 (c) If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :

(c) The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

Answer:

The direction of velocity at the instant of breakage of the string is tangential, so the stone will fly tangentially. Hence given statement is false.

Q4.23 (a) Explain why a horse cannot pull a cart and run in empty space

Answer:

The horse moves forward by pushing the ground backwards. The ground will then give the normal force to the horse (action-reaction pair), which is responsible for the movement of the horse. In an empty space, no such force is present, so the horse cannot get the push to run forward. Thus, a horse cannot pull a cart and run in empty space.

Q4.23 (b) Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly

Answer:

This is because of inertia. When the bus is moving our body has the same speed. But when the bus comes to rest, the inertia of our body opposes stopping and continues its motion. That's why passengers are thrown forward from their seats when a speeding bus stops suddenly.

Q4.23 (c) Explain why it is easier to pull a lawnmower than to push it

Answer:

Because when you pull the lawnmower at some angle, one component of force is in the upward direction and one in the horizontal (to move). The vertical force reduces the effective weight of the motor which makes it easier. But in case of a push, the vertical force is directed downward which makes its effective weight even greater than before. That's why it is said 'pull is easier than push'.

Q4.23 (d) Explain why a cricketer moves his hands backwards while holding a catch.

Answer:

According to Newton's law, we can write :

$F\ =\ ma$

or $F\ =\ m\frac{dv}{dt}$

Thus $F\ \propto \frac{dv}{dt}$

It can be seen from the equation that if we increase the impact time, then the experienced force will be less.

So a cricketer increases the impact time by taking their hands backwards while holding a catch, resulting in less force on their hand.

Class 11 Physics NCERT Chapter 4: Higher Order Thinking Skills (HOTS) Questions

Chapter 4: Laws of Motion Higher Order Thinking Skills (HOTS) questions of Class 11 Physics improvise students apply the conceptual knowledge to life-like situations and difficult problems. These are not simple memorisation questions, but they can help the learners practice analytical thinking and problem-solving in a simple question, which is necessary in competitive examinations such as JEE and NEET.

Q1. Consider a block and trolley system as shown in the figure. If the coefficient of kinetic friction between the trolley and the surface is 0.4, the acceleration of the system $\mathrm{ms}^{-2}$ is: ( Consider that the string is massless and unstretchable and the pulley is also massless and frictionless)

Block Trolly system
Answer:

Block Trolly system

$\begin{aligned} 50-T & =5 a \rightarrow( 1) \\ T-f & =10 a \rightarrow(2)\end{aligned}$

$\begin{aligned} & 10=15 a \\ & a=\frac{2}{3} \mathrm{~m} / \mathrm{s}^2\end{aligned}$

Hence, the answer is 2/3.

Q.2 The system shown in the figure is in equilibrium and at rest. The spring and string are massless. Now the string between B and C is cut. Find the acceleration of A, B and C just after the string is cut. $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Spring block system

Answer:

At equilibrium, the free body diagrams are-

Spring block system

For block A,

$\begin{aligned}
T_A & =m g+k x \\
& =3 m g
\end{aligned}$

For block B,

$\begin{aligned}
k x&=T_B+m g \\
& =2 m g
\end{aligned}$

For block C,

$T_B=m g$

When string is cut $T_B$ will be equal to zero. Net force acting on block A is zero

Block A $a_a=0$
Block B $a_B=\frac{(2 m g-m g)}{m}=g$
Block C $a_c=g$


Q.3 The reading of the spring balance (mass-stress) in the figure given below is $\left ( m_{2}>m_{1} \right )$

Spring balance system

Answer:

FBD of the block of mass m1

$T-m g=m_1 a$......(1)

FBD of block mass m2

$m_2 g-T=m_2 a$.......(2)

For massless spring balance reading $=\mathrm{kx}=\mathrm{T}$
From the equations (1) and (2), we get-

$
\begin{aligned}
& m_2 g-m_1 g=\left(m_1+m_2\right) a \\
& \Rightarrow a=\left(\frac{m_2-m_1}{m_1+m_2}\right) g
\end{aligned}
$
$
T=m_1 g+m_1 a=m_1 g+\frac{m_1 m_2 g}{m_1+m_2}-\frac{m_1^2+g}{\left(m_1+m_2\right)}
$
$
T=\frac{2 m_1 m_2 g}{m_1+m_2}
$
Spring balance reading

$
=\frac{2 m_1 m_2 g}{m_1+m_2}
$

Q.4 The arrangement is shown in the given figure. If the coefficient of friction between the 2kg block and the table is 0.2. What would be the maximum value of m (in kg) such that the two blocks do not move?(10=m/s2) ?

Block pully system

Answer:

Given-
mass of block on table $=2 \mathrm{~kg}$,
coefficient of friction, $\mu=0.2$
As we keep on increasing the value of $m$ from zero, the tension as well as the friction force increase to keep the blocks at rest. For the maximum value of $m$, the friction force is maximum and the blocks are just about to move. Therefore, at that instant, the friction force is limiting.

Let the tension in the string be $T$.
F.B.D of a block of mass $m$ -

For equilibrium-

$
\mathrm{T}=\mathrm{mg} \ldots(1)
$

F.B.D of a block of mass 2 kg -

$
N=20
$
Limiting friction-

$
\mathrm{f}_{\mathrm{l}}=\mu \mathrm{N}=4
$
For equilibrium-

$
\mathrm{T}=\mathrm{f}_{\mathrm{l}}=4 \ldots(2)
$
From equations (1) and (2) -

$
\begin{aligned}
& \mathrm{mg}=4 \\
& \Rightarrow \mathrm{~m}=0.4
\end{aligned}
$

Q.5 A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in the figure. The other end of the string is pulled with constant acceleration ' a ' vertically downward.The tension in the string is equal to :

Answer:

As we learned

Newton's 2nd Law -

$F=m a$

Apply the concept of pseudo force as shown in the below figure


Applying Newton's law perpendicular to string $m g \sin \theta=m a \cos \theta$

$
\Rightarrow \tan \theta=\frac{a}{g}
$
Applying Newton's law along string,

$
\begin{aligned}
& T-m g \cos \theta-m a \sin \theta=m a \\
& \Rightarrow T=m \sqrt{g^2+a^2}+m a
\end{aligned}
$

Class 11 Physics Chapter 4 - Laws of Motion: Topics

In Class 11 Physics, the chapter Laws of Motion present the basic laws of motion that govern the movement of the objects. It discusses Newton laws of motion, the idea of force, inertia, momentum and friction to provide a foundation to appreciate both simple phenomena, and complicated physical systems.

4.1 Introduction
4.2 Aristotle’s Fallacy
4.3 The Law Of Inertia
4.4 Newton’s First Law Of Motion
4.5 Newton’s Second Law Of Motion
4.6 Newton’s Third Law Of Motion
4.7 Conservation Of Momentum
4.8 Equilibrium Of A Particle
4.9 Common Forces In Mechanics
4.9.1 Friction
4.10 Circular Motion
4.11 Solving Problems In Mechanics

4.1 Introduction
4.2 Aristotle’s Fallacy
4.3 The Law Of Inertia
4.4 Newton’s First Law Of Motion
4.5 Newton’s Second Law Of Motion
4.6 Newton’s Third Law Of Motion
4.7 Conservation Of Momentum
4.8 Equilibrium Of A Particle
4.9 Common Forces In Mechanics
4.9.1 Friction
4.10 Circular Motion
4.11 Solving Problems In Mechanics

Laws of Motion class 11 Question Answers: Important Formulae

Laws of Motion NCERT Solutions – Important Formulae compile all the key equations related to vectors, projectile motion, and uniform circular motion. These formulas act as quick reference tools, making problem-solving faster and strengthening exam preparation.

Newton’s Laws of Motion

  • First Law (Law of Inertia):
    A body remains at rest or in uniform motion unless acted upon by an external force.
  • Second Law

F = ma
Force is the product of mass and acceleration.

  • Third Law:
    For every action, there is an equal and opposite reaction.

Momentum and Impulse

$
\vec{p}=m \vec{v}
$
Momentum is the product of mass and velocity.

  • Impulse

$
\vec{J}=\vec{F} \Delta t=\Delta \vec{p}
$

Friction

  • Static Friction

$
f_s \leq \mu_s N
$

  • Kinetic Friction

$
f_k=\mu_k N
$

  • Limiting Friction:
    Maximum value of static friction: $f_{\max }=\mu_s N$

Circular Motion

  • Centripetal Force

$
F_c=\frac{m v^2}{r}
$

  • Banking of Roads (without friction):

$
v=\sqrt{r g \tan \theta}
$

Approach to Solve Questions of Class 11 Physics Chapter 4 - Laws of Motion

In answering NCERT questions, it is only important to have a clear approach, besides remembering the formulae. The Laws of Motion NCERT Solutions in a step-by-step manner (Class 11 Physics) will teach the learners how to organize answers, apply concepts, make sense, and prevent errors most of the time. These solutions not only construct accuracy but also enhance speed and confidence in taking tests.

  • Learn Newton’s Laws: Always start with the basics, like what Newton’s three laws mean and how they apply in real-life situations, like pushing or pulling objects.
  • Draw Free Body Diagrams: Always draw FBDs to show all the forces acting on a body. It helps in making equations, especially in tension and friction problems.
  • Learn Friction Concepts Well: Know the main difference between static and kinetic friction. Understand when an object starts moving and how friction changes.
  • Use Newton’s Second Law (F = ma): This formula is the key to most questions. Break forces into components and apply the equation properly in the direction of motion.
  • Practice Pulley and Inclined Plane Problems: These are exam favourites! Learn how to deal with tension and forces on inclined surfaces. Use components of weight (mg sinθ and mg cosθ) smartly.
  • Revise Formulas: Keep a small list of important formulas, short notes like for net force, acceleration, and friction. They save time and help solve questions faster in exams.
  • Solve Mixed and Previous Year Questions: Practice a variety of Questions, especially from PYQs. It boosts confidence and helps you to understand question patterns easily.

What Extra Should Students Study Beyond NCERT for JEE?

Solving a variety of problems that deal with the application of the laws of Newton in real-life situations is, in itself, a great idea to do over and above the NCERT, along with Chapter 4: Laws of Motion of JEE. These are within inclined plane friction, pulley problem, circular motion and problems of constraint. It is always necessary to solve difficult problems that enhance conceptual understanding and improve the ability to solve problems needed to get good scores in competitive exams.

Importance of NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion

To understand the solutions of NCERT Class 11 Physics Chapter 4 Laws of Motion, it is necessary to know how to draw a free-body diagram. To familiarise with free body diagrams practice as many problems as possible. NCERT solutions for Class 11 are important for competitive exams like NEET and JEE Main and final exams of Class 11. NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion helps to do well in these exams.

Also, Check NCERT Books and NCERT Syllabus here

NCERT Solutions for Class 11 Physics Chapter-Wise

The NCERT Solutions to Class 11 Physics are available in chapter-wise format to have inexpensive, step-by-step solutions to each chapter. The links are useful in revising major concepts, doing numerical problems, and preparing a sound base on exams such as CBSE, JEE, and NEET. All solutions are designed by professionals according to the new syllabus.


Subject-wise NCERT solutions

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

Q: What is the need to study laws of motion?
A:

In order to learn how to determine the behaviour of objects in various forces, as well as to resolve real-life problems in mechanics.

Q: Do Newton laws apply under all circumstances?
A:

No, it can only apply in the frames that are inertial and break down at very high velocities or quantum levels.

Q: How would Newton law three actually fit in real life?
A:

Practical examples are rocket propulsion, jumping, swimming and walking.

Q: Can an object in motion possess no net force on it?
A:

Yes, it will move at a constant velocity according to the first law of Newton.

Q: What makes a motion come to rest?
A:

Unbalanced external forces, such as friction or air resistance, tend to bring motion to a halt.

Articles
Upcoming School Exams
Ongoing Dates
Maharashtra SSC Board Application Date

1 Aug'25 - 31 Oct'25 (Online)

Ongoing Dates
Maharashtra HSC Board Application Date

1 Aug'25 - 31 Oct'25 (Online)