NCERT Solutions for class 11 Physics Chapter 4 : Motion in a plane

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The NCERT Solutions for Class 11 Physics Chapter 4 Motion in a plane are provided in the article below. The motion of a body can be referred to as its change in position depending on its surroundings in a given time interval. The motion of any object which has some mass can be measured in distance, displacement, speed and velocity, acceleration, and time.

Class 11 Physics Chapter 4 Motion in a Plane belongs to Unit 2 Kinematics which has a weightage of 23 marks along with Unit 1 Physical World and Measurement and Unit 3 Laws of Motion. NCERT Solutions for Chapter 4 Physics Class 11 deals with the concept of scalars and vectors, uniform circular motion, and relative velocity in two dimensions.

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NCERT Solutions for Class 11 Physics Chapter 4

Class 11 Physics Chapter 4 – Concepts Covered

Motion in a plane is known as motion in two dimensions. For example, projectile motion, circular motion etc.
Scalar Quantities are the physical quantities that are specified by their magnitude or size alone.

Examples of Scalar Quantities – Length, mass, density, speed, work, etc

Vector Quantities are the physical quantities that are characterised by both magnitude and direction.

Example of Vector Quantities – Velocity, displacement, acceleration, force, momentum, torque etc.

A unit vector is a vector with a unit magnitude that points in a particular direction. It is used to only specify the direction.

It can be represented as $\widehat{A} = { \overrightarrow{A} \over |\overrightarrow{A}|}$

Parallelogram Law of Vector Addition: If $\overrightarrow{A}$and $\overrightarrow{B}$ are the two adjacent sides of a parallelogram, inclined at an angle $\theta$, then the magnitude of resultant vector is determined.

$R = \sqrt{A^2 + B^2 +2ABcos\theta}$

The projectile is an initial inclined velocity which subsequently follows a path determined by the gravitational force that acts on it and by the frictional resistance of the air.
The path followed by a projectile is known as its trajectory.

Class 11 Physics Chapter 4 Related Guides –

Resolution of vectors	Displacement Vector	Centripetal Acceleration
Tangential Acceleration Formula	Resultant of Vector Formula	Relation between torque and speed
Horizontal Motion	Velocity Vectors	Trajectory Formula

Class 11 Physics Study Guides:

NCERT Solutions for Class 11 Physics	Physics Formula	Physics Study Notes
Difference between in Physics	Derivations in Physics	SI Units in Physics
Relation between articles in Physics	Physics Constants	Class 11 Physics Notes

CBSE CLASS XII Related Questions

1.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

$\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})$
$\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})$
$\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})$
$\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})$

View Solution

2.
The resistance of a metal wire at $20^\circ\text{C}$ is $1.05\,\Omega$ and at $100^\circ\text{C}$ is $1.38\,\Omega$. Determine the temperature coefficient of resistivity of this metal.

View Solution

3.
A circular coil of 100 turns and radius $\left(\frac{10}{\sqrt{\pi}}\right)$ cm carrying current of $5.0\,\text{A}$ is suspended vertically in a uniform horizontal magnetic field of $2.0\,\text{T}$. The field makes an angle $30^\circ$ with the normal to the coil. Calculate:
(I) the magnetic dipole moment of the coil, and
(II) the magnitude of the counter torque that must be applied to prevent the coil from turning.

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4.
A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:
(I) the formation of maxima and minima in the diffraction pattern, and
(II) why the maxima go on becoming weaker and weaker with its increasing number $n$.

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5.
Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

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6.
A current of $4.0\,\text{A}$ flows through a wire of length $1\,\text{m}$ and cross-sectional area $1.0\,\text{mm}^2$, when a potential difference of $2\,\text{V}$ is applied across its ends. Calculate the resistivity of the material of the wire.

View Solution

NCERT Solutions for class 11 Physics Chapter 4: Motion in a plane

NCERT Solutions for Class 11 Physics Chapter 4

Class 11 Physics Chapter 4 – Concepts Covered

CBSE CLASS XII Related Questions

1.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

2.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

5.
Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

6.
A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

Similar Physics Concepts

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NCERT Solutions for class 11 Physics Chapter 4: Motion in a plane

NCERT Solutions for Class 11 Physics Chapter 4

Class 11 Physics Chapter 4 – Concepts Covered

CBSE CLASS XII Related Questions

1.Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

2.The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

5.Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

6.A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

2.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

5.
Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

6.
A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.