CBSE Class 11 Physics Exemplar NCERT Solutions: Motion in a Plane

Q: What is the main focus of Chapter 3, 'Motion in a Plane', in the Class 11 Physics Exemplar?

Chapter 3 focuses on understanding motion in two dimensions, covering concepts like vectors, their properties, operations (like the dot product), and the physics of projectile motion, including factors affecting its range.

Q: How do these NCERT Solutions help students prepare for exams?

These solutions provide detailed, step-by-step explanations for MCQs, clarifying the underlying principles of vectors and projectile motion. This helps students build confidence and accuracy in solving similar problems.

Q: What is the significance of the dot product in this chapter?

The dot product is crucial for finding the angle between two vectors. It is also used to determine the component of one vector along another, which is fundamental in physics.

Q: How does the angle of projection affect the horizontal range of a projectile?

The horizontal range is maximum when the angle of projection is 45°. For angles less than or greater than 45° (but complementary, like 15° and 75°), the range is the same, assuming the initial speed and launch height are constant.

Q: What are scalar quantities, and what is a key property highlighted in the solutions?

Scalar quantities are those with only magnitude, like mass or temperature. A key property emphasized is that their value is independent of the observer's orientation of axes.

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 11 PDF

CBSE Class 11 Physics Exemplar, Chapter 3: Motion in a Plane, NCERT Solutions offers a thorough exploration of two-dimensional motion. This resource delves into the essential concepts of vectors, explaining their properties, operations like dot products, and how to resolve them into components. It also provides in-depth analysis of projectile motion, examining the key factors that influence the horizontal range of a projectile. The solutions are crafted to demystify challenging topics, presenting clear, step-by-step explanations and logical reasoning. This guide is an excellent tool for students aiming to solidify their understanding of motion in a plane and build a robust foundation in physics, crucial for exam preparation.

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Quick info

Board	CBSE
Class	Class 11
Subject	Physics Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 3

Chapter summary

Chapter 3, 'Motion in a Plane,' NCERT Solutions for Class 11 Physics Exemplar, focuses on vector algebra and projectile motion. The solutions cover multiple-choice questions that test understanding of vector dot products, scalar quantity properties, vector components in different quadrants, and the conditions for maximum range of a projectile. It provides a clear approach to solving problems involving motion in two dimensions.

Learning outcomes

Understand the concept of the angle between two vectors using the dot product.
Differentiate between scalar and vector quantities and their properties.
Determine the signs of vector components based on their orientation in the coordinate plane.
Analyze the factors affecting the horizontal range of a projectile.
Solve problems related to projectile motion and vector analysis.

Topics covered

Paper topics

Vectors
Vector Components
Dot Product
Scalar Quantities
Projectile Motion
Horizontal Range
Angle of Projection
Motion in a Plane

Important topics

Vector Components and Orientation
Dot Product for Angle Calculation
Properties of Scalar Quantities
Horizontal Range of Projectile
Relationship between Range and Angle of Projection

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Questions and Solutions

Multiple Choice Questions (MCQs) - Q. 1

The angle between the vectors $\mathbf{A} = \hat{\mathbf{i}} + \hat{\mathbf{j}}$, and $\mathbf{B} = \hat{\mathbf{i}} - \hat{\mathbf{j}}$, is:

Solution: To find the angle $\theta$ between two vectors $\mathbf{A}$ and $\mathbf{B}$, we can use the dot product formula: $\mathbf{A} \cdot \mathbf{B} = |A| |B| \cos \theta$. Given vectors are $\mathbf{A} = \hat{\mathbf{i}} + \hat{\mathbf{j}}$, and $\mathbf{B} = \hat{\mathbf{i}} - \hat{\mathbf{j}}$. First, calculate the dot product $\mathbf{A} \cdot \mathbf{B}$: $\mathbf{A} \cdot \mathbf{B} = (1)(\hat{\mathbf{i}} \cdot \hat{\mathbf{i}}) + (1)(\hat{\mathbf{j}} \cdot \hat{\mathbf{i}}) + (1)(\hat{\mathbf{i}} \cdot \hat{\mathbf{j}}) + (1)(\hat{\mathbf{j}} \cdot \hat{\mathbf{j}})$ Since $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1$, $\hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = 1$, and $\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{i}} = 0$, the dot product is: $\mathbf{A} \cdot \mathbf{B} = (1)(1) + (1)(0) + (1)(0) + (1)(1) = 1 - 1 = 0$. Next, calculate the magnitudes of $\mathbf{A}$ and $\mathbf{B}$: $|A| = \sqrt{1^2 + 1^2} = \sqrt{2}$ $|B| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$ Now, substitute these values into the dot product formula: $0 = (\sqrt{2})(\sqrt{2}) \cos \theta$ $0 = 2 \cos \theta$ $\cos \theta = 0$ This implies that $\theta = 90^{\circ}$. Therefore, the angle between the vectors $\mathbf{A}$ and $\mathbf{B}$ is 90 degrees.

Multiple Choice Questions - Q. 2

Which one of the following statements is true regarding scalar quantities?

A scalar quantity is the one that can never take negative values.
A scalar quantity is the one that is conserved in a process.
A scalar quantity is the one that does not vary from one point to another in space.
A scalar quantity has the same value for observers with different orientation of the axes.

Solution: Let's analyze each statement:

A scalar quantity can take negative values. For example, temperature can be negative (e.g., -10°C), and displacement in one dimension can be negative. So, this statement is false.
While some scalar quantities like energy are conserved in certain processes, not all scalar quantities are conserved. For instance, speed is a scalar quantity but is not necessarily conserved. So, this statement is not universally true.
Scalar quantities can vary from one point to another. For example, gravitational potential varies with position. So, this statement is false.
A scalar quantity is defined solely by its magnitude. Its value does not depend on the coordinate system used by an observer. Therefore, it has the same value regardless of the orientation of the axes. This statement is true.

Thus, the correct statement is that a scalar quantity has the same value for observers with different orientations of the axes.

Multiple Choice Questions - Q. 3

Figure shows the orientation of two vectors $\boldsymbol{\mathsf{u}}$, and $\boldsymbol{\mathsf{v}}$, in the xy-plane. If $\mathbf{u} = a\hat{\mathbf{i}} + b\hat{\mathbf{j}}$, and $\mathbf{v} = p\hat{\mathbf{i}} + q\hat{\mathbf{j}}$, which of the following is correct regarding the components a, b, p, and q?

a and p are positive while b and q are negative
a, p and b are positive while q is negative
a, q and b are positive while p is negative
a, b, p and q are all positive

Solution: We need to determine the signs of the components of vectors $\mathbf{u}$ and $\mathbf{v}$ based on their orientation in the xy-plane as shown in the figure. For vector $\mathbf{u} = a\hat{\mathbf{i}} + b\hat{\mathbf{j}}$: The vector $\mathbf{u}$ is shown in the first quadrant of the xy-plane. In the first quadrant, both the x-component and the y-component of a vector are positive. Therefore, $a > 0$ and $b > 0$. For vector $\mathbf{v} = p\hat{\mathbf{i}} + q\hat{\mathbf{j}}$: The vector $\mathbf{v}$ is shown originating from the origin and pointing towards the positive x-axis direction but downwards. This means its x-component is positive, and its y-component is negative. Therefore, $p > 0$ and $q < 0$. Combining these observations, we have $a > 0$, $b > 0$, $p > 0$, and $q < 0$. Comparing this with the given options, option (b) states that a, p, and b are positive while q is negative, which matches our findings.

Multiple Choice Questions - Q. 4

The component of a vector $\mathbf{r}$ along the X-axis will have its maximum value if the vector $\mathbf{r}$ is:

Along the positive Y-axis
Along the positive X-axis
Making an angle of 45° with the X-axis
Along the negative Y-axis

Solution: Let the vector be $\mathbf{r}$ and let it make an angle $\theta$ with the positive X-axis. The component of $\mathbf{r}$ along the X-axis, denoted as $r_x$, is given by the formula:

$r_x = |\mathbf{r}| \cos \theta$

To find the maximum value of $r_x$, we need to find the maximum value of $\cos \theta$, since $|\mathbf{r}|$ is constant for a given vector. The maximum value of $\cos \theta$ is 1, which occurs when $\theta = 0^{\circ}$. When $\theta = 0^{\circ}$, the vector $\mathbf{r}$ is aligned with the positive X-axis. Therefore, the component of $\mathbf{r}$ along the X-axis will have its maximum value when $\mathbf{r}$ is along the positive X-axis.

Multiple Choice Questions - Q. 5

The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be:

60 m
71 m
100 m
141 m

Solution: The horizontal range $R$ of a projectile fired with an initial speed $u$ at an angle $\theta$ with the horizontal is given by the formula:

$R = \frac{u^2 \sin 2 \theta}{g}$

where $g$ is the acceleration due to gravity. We are given that when $\theta = 15^{\circ}$, the range $R = 50 \text{ m}$. Substituting these values into the formula:

$50 = \frac{u^2 \sin (2 \times 15^{\circ})}{g}$

$50 = \frac{u^2 \sin 30^{\circ}}{g}$

Since $\sin 30^{\circ} = \frac{1}{2}$, we have:

$50 = \frac{u^2}{g} \times \frac{1}{2}$

This gives us $\frac{u^2}{g} = 100$ m. Now, we need to find the range $R'$ when the projectile is fired with the same speed $u$ at an angle $\theta' = 45^{\circ}$:

$R' = \frac{u^2 \sin (2 \times 45^{\circ})}{g}$

$R' = \frac{u^2 \sin 90^{\circ}}{g}$

Since $\sin 90^{\circ} = 1$, we get:

$R' = \frac{u^2}{g} \times 1$

We already found that $\frac{u^2}{g} = 100$ m. Substituting this value:

$R' = 100 \text{ m}$

Therefore, if the projectile is fired with the same speed at an angle of 45°, its range will be 100 m.

Common mistakes

Incorrectly applying the dot product formula to find the angle between vectors.
Misinterpreting the properties of scalar quantities.
Errors in determining the signs of vector components from diagrams.
Confusing the conditions for maximum range with other projectile motion parameters.

Revision tips

Review the dot product formula and its application for finding the angle between vectors.
Clearly visualize the quadrants of the xy-plane to determine the signs of vector components.
Understand the relationship between the angle of projection and the horizontal range of a projectile.
Practice solving MCQs to reinforce understanding of key concepts in motion in a plane.

Practice MCQs

Q1. What is the angle between the vectors $ $\mathbf{A}$ = $\hat{\mathbf{i}}$ + $\hat{\mathbf{j}}$ $, and $ $\mathbf{B}$ = $\hat{\mathbf{i}}$ - $\hat{\mathbf{j}}$ $?

Q2. Which of the following statements about a scalar quantity is always true?

Q3. Given vectors $ $\mathbf{u}$ = a $\hat{\mathbf{i}}$ + b $\hat{\mathbf{j}}$ $, and $ $\mathbf{v}$ = p $\hat{\mathbf{i}}$ + q $\hat{\mathbf{j}}$ $, if $ $\mathbf{u}$ $ is in the first quadrant and $ $\mathbf{v}$ $ points in the positive x-direction but downwards, which of the following is correct about the components?

Q4. For a projectile, the component of its position vector $ $\mathbf{r}$ $ along the X-axis will have its maximum value when the projectile is:

Q5. A projectile fired at 15° has a horizontal range of 50 m. If fired with the same speed at 45°, what will be its new range?

Frequently asked questions

What is the main focus of Chapter 3, 'Motion in a Plane', in the Class 11 Physics Exemplar?