Section 2.2 – Instantaneous Velocity and Speed

Section 2.2 – Instantaneous Velocity and Speed from Class 11 Physics (NCERT Chapter 2: Kinematics):

Section 2.2 – Instantaneous Velocity and Speed

1. Concept of Instantaneous Velocity

• Average velocity gives us an idea of how fast an object moves over a finite time interval.

• However, it does not tell us how fast the object is moving at a particular instant of time during that interval.

To address this, we define:

Instantaneous velocity: The velocity of an object at a specific moment in time.

Mathematically, it is the limit of the average velocity as the time interval becomes infinitesimally small:

$$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$

This is the derivative of position $x$ with respect to time $t$**—also known as the rate of change of position at that instant.

2. Graphical Representation

• Instantaneous velocity can be visualized using the slope of a position-time graph.

• For example, consider a graph representing the motion of a car:

  • To find velocity at $t = 4$ s:

    • Start by calculating average velocity over small time intervals centered at 4 s (like from 3 s to 5 s, 3.5 s to 4.5 s, etc.).

    • As the interval $\Delta t$ decreases, the secant line between two points approaches a tangent line at the point $t = 4$ s.

    • The slope of this tangent line gives the instantaneous velocity at that instant.

This approach, while useful for visual understanding, is often not practical in real-life calculations because:

• It requires precise graph plotting.

• It involves manually calculating slopes of secant lines repeatedly for smaller intervals.

Position-Time Diagram (Responsive)

Position-Time Graph: x = 0.08t³

Velocity-Time Graph: v = 0.24t²

3. Numerical Illustration

To better understand the limiting process, we can use a numerical example.

Suppose the position of a car is given by:

$$x = 0.08t^3$$

We calculate average velocities using different small values of $\Delta t$, centered at $t = 4.0 \, \text{s}$, by computing:

$$\Delta x = x(t + \frac{\Delta t}{2}) - x(t - \frac{\Delta t}{2})$$ $$\text{Average velocity} = \frac{\Delta x}{\Delta t}$$

Here’s how the process looks:

∆t (s)	t₁ (s)	t₂ (s)	x(t₁) (m)	x(t₂) (m)	∆x (m)	∆x/∆t (m/s)
2.0	3.0	5.0	2.16	10.00	7.84	3.92
1.0	3.5	4.5	3.43	8.49	5.06	5.06
0.5	3.75	4.25	4.22	7.64	3.42	6.84
0.1	3.95	4.05	5.03	6.59	1.56	15.6
0.01	3.995	4.005	5.91	5.99	0.08	8.0

(Values are illustrative for explanation purposes; actual values should follow from the exact expression.)

As $\Delta t \to 0$, the average velocity approaches 3.84 m/s, which is the instantaneous velocity at $t = 4.0 \, \text{s}$.

4. Analytical (Calculus) Method

When the position function $x(t)$ is known, the instantaneous velocity is more conveniently found using differential calculus:

$$x = 0.08 t^3 \Rightarrow \frac{dx}{dt} = 0.24 t^2$$

So at $t = 4.0 \, \text{s}$:

$$v = 0.24 \times 16 = 3.84 \, \text{m/s}$$

This confirms the result obtained from the limiting process.

5. Speed vs. Velocity

• Instantaneous speed is the magnitude of instantaneous velocity.

• It is always positive, whereas velocity can be positive or negative depending on direction.

6. Summary

• Instantaneous velocity gives a more accurate description of how fast an object is moving at a particular instant.

• It is defined as the derivative of position with respect to time.

• It can be calculated:

  • Graphically (as slope of tangent to the position-time curve).

  • Numerically (by reducing ∆t in average velocity).

  • Analytically (using calculus, if position-time relation is known).

• Instantaneous speed is the absolute value of instantaneous velocity.

Chapter 2.1 – Introduction to Motion

Class 11 Physics (NCERT) Chapter 2 – Kinematics:

Chapter 2.1 – Introduction to Motion

Understanding Motion

Motion is a fundamental and universal phenomenon. Everything in the universe is constantly in motion. Examples include:

• Everyday human activities like walking, running, or riding a bicycle.

• Biological processes such as the movement of air in and out of lungs and the flow of blood in our body.

• Natural processes like leaves falling, water flowing, and celestial movements.

Even massive celestial bodies are in motion:

• The Earth rotates on its axis every 24 hours and revolves around the Sun once every year.

• The Sun moves within the Milky Way galaxy, and the galaxy itself is in motion within its local group of galaxies.

Definition of Motion

Motion is defined as a change in the position of an object with respect to time. Understanding motion involves answering a fundamental question:
How does the position of an object change with time?Scope of This Chapter

This chapter focuses on:

1. Describing motion using mathematical and graphical tools.

2. Key concepts introduced:

   • Displacement: Change in position.

   • Velocity: Rate of change of displacement.

   • Acceleration: Rate of change of velocity.

3. Type of motion considered:

   • Rectilinear Motion – motion along a straight line.

   • Uniform Acceleration – when acceleration remains constant.

Rectilinear Motion

This is the simplest type of motion to study and is limited to one dimension (a straight line). Despite this simplicity, it is a powerful model and lays the foundation for understanding more complex motions.

For rectilinear motion with uniform acceleration, a set of three kinematic equations are introduced:

• $v = u + at$

• $s = ut + \frac{1}{2}at^2$

• $v^2 = u^2 + 2as$

Where:

• $u$ = initial velocity

• $v$ = final velocity

• $a$ = acceleration

• $t$ = time

• $s$ = displacement

Point Object Approximation

To simplify the study of motion, we often treat objects as point objects. This approximation holds true when:

• The size of the object is much smaller than the distance it travels.

• The dimensions of the object do not significantly affect its motion.

This allows us to ignore the complexities of shape and size and focus purely on the motion of the object’s position.

Relative Motion

Another important idea introduced in this chapter is the relative nature of motion. Motion is always relative to a reference point or observer. This leads to the concept of relative velocity, which helps us understand how motion appears differently to different observers.

Kinematics vs. Dynamics

• Kinematics (covered in this chapter) deals with describing motion — it does not concern itself with why motion occurs.

• The causes of motion, such as forces and interactions, are studied in Dynamics, which begins from Chapter 4.

Conclusion

This chapter lays the groundwork for understanding mechanics by introducing the key tools and concepts to describe motion. It uses simplified assumptions like rectilinear motion and point object approximation to develop a deep understanding of how objects move and how this motion can be quantified and analyzed.

Chapter 5 – Introduction to Euclid’s Geometry:

NCERT Class 9 Mathematics Chapter 5 – Introduction to Euclid’s Geometry:

✅ Example 1:

Statement:
If A, B and C are three points on a line, and B lies between A and C, prove that
AB + BC = AC, using Euclid's reasoning.

Reasoning using Euclid’s Axioms:

• Given: A, B, and C are points on the same line, with B between A and C.

• This means the line segment AC is made up of AB and BC.

Using Euclid’s Axiom 2:

If equals are added to equals, the wholes are equal.

• Segment AB and segment BC are "parts" of AC, so:

  $$AB + BC = AC$$

Hence, proved using Euclid’s second axiom.

✅ Example 2:

Statement:
Prove that an equilateral triangle can be constructed on any given line segment.

Let the given line segment be AB.

Construction:

1. With A as the center and AB as the radius, draw a circle.

2. With B as the center and AB as the radius, draw another circle.

3. Let the circles intersect at point C.

4. Join AC and BC.

Reasoning:

• All radii of the same circle are equal (Postulate 3).

• So, AC = AB and BC = AB.

• Hence, AB = BC = CA.

Conclusion:
Triangle ABC is equilateral.

✅ Theorem 5.1:

Statement:
Two distinct lines cannot have more than one point in common.

Proof using Euclid’s Geometry:

• Let’s assume two distinct lines l₁ and l₂ intersect at two points A and B.

• Then both points A and B lie on both lines.

• From Euclid’s Axiom 5.1:

  Given two distinct points, there is a unique line that passes through them.

• But if A and B lie on both lines, then l₁ and l₂ must be the same line, not distinct.

Contradiction arises, hence our assumption is wrong.

Conclusion:
Two distinct lines can intersect at most at one point.

✅ Exercise 5.1 – Q1: True or False with Reasons

(i) Only one line can pass through a single point.
❌ False

• Reason: Infinite number of lines can pass through a single point.

(ii) There are an infinite number of lines which pass through two distinct points.
❌ False

• Reason: Euclid’s Axiom 5.1 states: "Given two distinct points, there is a unique line that passes through them."

(iii) A terminated line can be produced indefinitely on both the sides.
✅ True

• Reason: Euclid’s Postulate 2 states: A terminated line (line segment) can be produced indefinitely.

(iv) If two circles are equal, then their radii are equal.
✅ True

• Reason: By definition, equal circles have equal radii.

(v) If AB = PQ and PQ = XY, then AB = XY.
✅ True

• Reason: Euclid’s Axiom 1: Things which are equal to the same thing are equal to one another.

Exercise 5.1 of Chapter 5: Introduction to Euclid’s Geometry (Class 9 Mathematics):

2. Definitions and Required Pre-definitions

(i) Parallel Lines:
Two lines in a plane that never meet, no matter how far they are extended, are called parallel lines.
✅ Undefined terms required: Line, plane, point, distance.
To define parallel lines, one must understand what a line is and what it means for lines not to intersect.

(ii) Perpendicular Lines:
Two lines that intersect to form a right angle (90°) are called perpendicular lines.
✅ Undefined terms required: Line, angle, right angle.
A definition of angle and specifically right angle is needed.

(iii) Line Segment:
A part of a line that has two endpoints is called a line segment.
✅ Pre-definition needed: Line, point.

(iv) Radius of a Circle:
The distance from the centre of a circle to any point on the circle is called the radius.
✅ Required terms: Circle, distance, centre.

(v) Square:
A quadrilateral with four equal sides and four right angles is called a square.
✅ Required terms: Line segment, angle, right angle, equality of sides.

3. Postulates Analysis

(i) Postulate: Given any two distinct points A and B, there exists a third point C which lies between A and B.

(ii) Postulate: There exist at least three points that are not on the same line.

(a) Undefined terms involved:

• Point, line, between, and distinct — these are foundational and often taken as primitive terms in geometry.

(b) Are they consistent?

Yes, the two postulates are consistent because:

• They do not contradict each other.

• They can both be true in the same geometrical space.

(c) Do they follow from Euclid’s postulates?

• (i) does not directly follow from Euclid’s postulates but can be accepted as an additional postulate to define betweenness.

• (ii) does not contradict Euclid’s postulates and helps describe configurations beyond collinearity, which is required for planar geometry.

4. If C lies between A and B such that AC = BC, then prove AC = ½ AB

🖊️ Proof:

Let points A, C, and B lie on a straight line with C between A and B.

Given:

• AC = BC

• AB = AC + CB
  But AC = BC ⇒ AB = AC + AC = 2AC
  Therefore, AC = (1/2) AB

🔍 Figure:

A -------- C -------- B  
      <-->    <-->  
      AC      CB (equal lengths)

5. Prove: Every line segment has one and only one midpoint

🖊️ Proof:

Let AB be a line segment.
By definition, a point M is the midpoint of AB if:

• AM = MB

• M lies between A and B

✅ Existence:
We can always find such a point M because a line segment has a measurable length. Halving that gives us a point M such that AM = MB.

✅ Uniqueness:
Suppose there are two midpoints M and N of AB.
Then, AM = MB and AN = NB ⇒ AM = AN
This implies M and N coincide.
Hence, there can be only one such point.

6. If AC = BD, prove that AB = CD (Referring to a straight line A → B → C → D)

🖊️ Given:

Points A, B, C, D lie on a line such that:

• A → B → C → D

• AC = BD

We want to prove AB = CD.

Let’s assume the line is as follows:

A ---- B ---- C ---- D

Then:

• AC = AB + BC

• BD = BC + CD

Given: AC = BD
So, AB + BC = BC + CD
Subtracting BC from both sides:
⇒ AB = CD

Hence proved.

7. Why is Axiom 5 (The whole is greater than the part) a universal truth?

🔍 Explanation:

• Axiom 5: The whole is greater than the part.

This is considered a universal truth because:

• It applies not just to geometry, but to all mathematical and physical quantities — length, area, volume, number, etc.

• It is intuitively obvious and doesn’t need proof.

• It holds true across all fields of math and science, not just in Euclidean geometry.

For example:

• If a rope is cut into two parts, clearly the entire rope is longer than either of the parts.

Hence, it's called a universal axiom — not dependent on any specific geometric figure or construction.

Euclid’s Five Postulates and Related Concepts

GRADE 9 MATHEMATICS – CHAPTER 5: INTRODUCTION TO EUCLID’S GEOMETRY

Comprehensive Note on Euclid’s Five Postulates and Related Concepts

Introduction to Postulates

In his foundational work Elements, Euclid laid out five postulates, or assumptions, upon which the entire structure of Euclidean Geometry is built. These postulates are accepted as self-evident truths without proof, and they apply specifically to the study of geometry. Using these postulates and logical reasoning, Euclid was able to derive hundreds of geometric theorems.

Understanding Euclid’s Five Postulates

Postulate 1

"A straight line may be drawn from any one point to any other point."

• This means at least one straight line can be drawn between any two distinct points.

• Modern Interpretation: There exists exactly one straight line joining any two distinct points.

• This is formalized as Axiom 5.1: Given two distinct points, there is a unique line that passes through them.

  • For example, the line joining points A and B is unique. No other straight line can pass through both.

Postulate 2

"A terminated line can be produced indefinitely."

• A "terminated line" refers to what we now call a line segment.

• This postulate means that a line segment can be extended endlessly in both directions to form a complete straight line.

• It introduces the concept of the infinite length of a line.

Postulate 3

"A circle can be drawn with any centre and any radius."

• Given any point as a center and any length as a radius, a circle can be constructed.

• This postulate allows the construction of circles, which are key elements in geometry.

• It also implies that the plane is continuous and unbounded.

Postulate 4

"All right angles are equal to one another."

• A right angle is the angle made when two lines intersect to form equal adjacent angles (each measuring 90°).

• This postulate states that all such angles, regardless of how or where they are formed, are equal in measure.

• This is a foundational idea used to compare angles and to establish angle congruence.

Postulate 5 (The Parallel Postulate)

"If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles."

• This is the most complex and controversial of Euclid's postulates.

• It talks about the intersection of lines based on angle relationships.

• In simpler terms:

  • If a transversal cuts two lines and the sum of interior angles on one side is less than 180°, then those two lines will eventually intersect on that side.

🡪 This postulate led to much mathematical debate and attempts to prove it using the other four postulates. It was ultimately accepted as an independent assumption. In modern geometry, different versions of this postulate give rise to non-Euclidean geometries (e.g., spherical and hyperbolic geometry).

Postulates vs. Axioms

• In Euclid’s time:

  • Axioms (also called common notions) were universal truths used across mathematics.

  • Postulates were assumptions specific to geometry.

• Today, the terms postulate and axiom are used interchangeably.

• A postulate is essentially a statement accepted as true without proof, used as the starting point for further reasoning.

Consistency of Axiomatic Systems

• A system of axioms is said to be consistent if no contradictions can be derived from it.

• Euclid’s geometry is based on a consistent and logical system of postulates and axioms.

Deductive Reasoning and Propositions

• Euclid used deductive logic to build a vast structure of geometry.

• Using definitions, axioms, and postulates, he proved statements called propositions or theorems.

• In total, he logically derived 465 propositions that form the backbone of Euclidean Geometry.

• These include results about lines, angles, triangles, circles, and other geometric figures.

Conclusion

Euclid’s five postulates serve as the foundation of classical geometry. His logical method of building geometry step by step using definitions, axioms, and postulates laid the path for mathematical reasoning for centuries.
While most postulates seem intuitive, Postulate 5 stands out in complexity and importance.
In upcoming chapters, students will use these postulates and axioms to explore geometric figures, prove theorems, and gain a deeper understanding of logical mathematical thinking.

GRADE 9 MATHEMATICS – CHAPTER 5: INTRODUCTION TO EUCLID’S GEOMETRY

 GRADE 9 MATHEMATICS – CHAPTER 5: INTRODUCTION TO EUCLID’S GEOMETRY

Comprehensive Note

---

Overview:

Geometry, as we know it today, was developed as an abstract study of shapes, sizes, and spaces. The foundational concepts of geometry were first organized systematically by the Greek mathematician Euclid in his famous work Elements. His approach to geometry is known as Euclidean Geometry.

Euclid’s Contribution:

Euclid developed geometry based on logical reasoning from a small set of basic assumptions. His work included:

• 23 Definitions

• 5 Postulates (specific to geometry)

• Common Notions or Axioms (general truths used in mathematics)

He began by defining fundamental geometrical ideas such as points, lines, surfaces, and solids, and then used these to build more complex results.

Euclid’s Definitions (Selected):

1. A Point – That which has no part (no length, breadth, or thickness).

2. A Line – Breadthless length (has only length).

3. Ends of a Line – Points.

4. Straight Line – A line which lies evenly with the points on itself.

5. Surface – That which has only length and breadth.

6. Edges of a Surface – Lines.

7. Plane Surface – A surface that lies evenly with the straight lines on itself.

🡪 Concept of Dimensions:

• Solid: 3D (length, breadth, height)

• Surface: 2D (length, breadth)

• Line: 1D (length only)

• Point: 0D (no dimensions)

Undefined Terms in Geometry:

Some basic geometric terms are left undefined because defining them leads to an infinite chain of definitions. These include:

• Point

• Line

• Plane

We understand these intuitively:

• A point is represented as a dot.

• A line is imagined as a straight path with no thickness.

• A plane is like a flat sheet extending infinitely.

Euclid’s Assumptions:

Euclid’s structure of geometry was built on assumptions that were not proven but accepted as true. He classified them into:

Axioms (Common Notions) – Universally accepted truths used across all branches of mathematics.

Some examples:

1. Things equal to the same thing are equal to one another.

2. If equals are added to equals, the wholes are equal.

3. If equals are subtracted from equals, the remainders are equal.

4. Things which coincide with one another are equal.

5. The whole is greater than the part.

6. Things which are double of the same things are equal.

7. Things which are halves of the same things are equal.

🡪 These are logical statements that help in understanding equality, comparison, and arithmetic of geometric magnitudes.

Postulates – Assumptions specific to geometry. These are dealt with in detail in later parts of the chapter.

Key Ideas from the Axioms:

• Equality: If a triangle has the same area as a rectangle, and that rectangle has the same area as a square, then the triangle and square have equal areas.

• Comparison: Only magnitudes of the same type can be compared (e.g., comparing a line to another line, but not a line to a rectangle).

• Superposition Principle: Based on the 4th axiom, two shapes are equal if they coincide completely when placed on one another.

• Part-Whole Relationship: From the 5th axiom, we understand that any quantity is greater than its part.

Conclusion:

Euclid’s work laid the foundation for classical geometry by starting with intuitive definitions, leaving some basic concepts undefined, and building logical deductions using axioms and postulates. This structured, logical approach allows us to explore and prove deeper properties in geometry. Understanding Euclid’s approach helps students appreciate the rigor and logic behind modern mathematics.

EXERCISE NCERT Class 11 Physics Chapter 1: Units and Measurement

Exercises 1.1 to 1.5 from NCERT Class 11 Physics Chapter 1: Units and Measurement, step by step with explanations and appropriate unit conversions.

1.1 Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to:

$$V = (1\, \text{cm})^3 = 1\, \text{cm}^3 = 1 \times 10^{-6}\, \text{m}^3$$

✅ Answer: $1 \times 10^{-6}\, \text{m}^3$

(b) Surface area of a solid cylinder:

Formula:

$$A = 2\pi r^2 + 2\pi rh$$

Given:
$r = 2.0\, \text{cm} = 20\, \text{mm},\quad h = 10.0\, \text{cm} = 100\, \text{mm}$

$$A = 2\pi r(r + h) = 2\pi (20)(20 + 100) = 2\pi (20)(120) = 4800\pi\, \text{mm}^2 \approx 15080\, \text{mm}^2$$

✅ Answer: $\approx 1.51 \times 10^4\, \text{mm}^2$

(c) Speed = 18 km/h
Convert to m/s:

$$\frac{18 \times 1000}{3600} = 5\, \text{m/s}$$

So, in 1 s, it covers 5 m.

✅ Answer: $5\, \text{m}$

(d) Relative density of lead = 11.3
That means:

$$\text{Density} = 11.3\, \text{g/cm}^3$$

Convert to kg/m³:

$$1\, \text{g/cm}^3 = 1000\, \text{kg/m}^3 \Rightarrow 11.3\, \text{g/cm}^3 = 11300\, \text{kg/m}^3$$

✅ Answer: $11.3\, \text{g/cm}^3$ or $11300\, \text{kg/m}^3$

1.2 Fill in the blanks by suitable conversion of units

(a)

$$1\, \text{kg} = 1000\, \text{g},\quad 1\, \text{m} = 100\, \text{cm}$$ $$1\, \text{kg m}^2 \text{s}^{-2} = 1000\, \text{g} \times (100\, \text{cm})^2 \times \text{s}^{-2} = 1000 \times 10^4\, \text{g cm}^2 \text{s}^{-2} = 10^7\, \text{g cm}^2 \text{s}^{-2}$$

✅ Answer: $10^7\, \text{g cm}^2 \text{s}^{-2}$

(b)
1 ly (light year) = distance light travels in 1 year
Speed of light $c = 3 \times 10^8\, \text{m/s}$

$$1\, \text{year} = 365.25 \times 24 \times 3600 \approx 3.156 \times 10^7\, \text{s}$$ $$1\, \text{ly} = 3 \times 10^8 \times 3.156 \times 10^7 = 9.47 \times 10^{15}\, \text{m}$$

So,

$$1\, \text{m} = \frac{1}{9.47 \times 10^{15}}\, \text{ly}$$

✅ Answer: $\approx 1.06 \times 10^{-16}\, \text{ly}$

(c)
Convert $3.0\, \text{m/s}^2$ to km/h²

$$1\, \text{m/s}^2 = \left(\frac{1}{1000}\, \text{km}\right)/\left(\frac{1}{3600}\, \text{h}\right)^2 = \frac{1}{1000} \times (3600)^2 = \frac{1}{1000} \times 1.296 \times 10^7 = 12960\, \text{km/h}^2$$ $$3.0\, \text{m/s}^2 = 3.0 \times 12960 = 38880\, \text{km/h}^2$$

✅ Answer: $3.89 \times 10^4\, \text{km/h}^2$

(d)
Given:

$$G = 6.67 \times 10^{-11}\, \text{N m}^2 \text{kg}^{-2}$$

We know:

$$1\, \text{N} = 10^5\, \text{dyne},\quad 1\, \text{m} = 100\, \text{cm},\quad 1\, \text{kg} = 1000\, \text{g}$$ $$G = 6.67 \times 10^{-11} \times 10^5\, \text{dyne} \times (100)^2\, \text{cm}^2 / (1000)^2\, \text{g}^2$$ $$= 6.67 \times 10^{-11} \times 10^5 \times 10^4 / 10^6 = 6.67 \times 10^{-8}$$

✅ Answer: $6.67 \times 10^{-8}\, \text{cm}^3 \text{s}^{-2} \text{g}^{-1}$

1.3 Calorie in new units

Given:

• $1\, \text{cal} = 4.2\, \text{J} = 4.2\, \text{kg m}^2 \text{s}^{-2}$

In new units:
$1\, \text{kg} = \alpha,\quad 1\, \text{m} = \beta,\quad 1\, \text{s} = \gamma$

So,

$$1\, \text{J} = \frac{1}{\alpha}\, \text{mass unit} \times \frac{1}{\beta^2}\, \text{length unit}^2 \times \gamma^2\, \text{time unit}^{-2} = \frac{\gamma^2}{\alpha \beta^2}$$

So,

✅ Answer:

$$\text{Magnitude of 1 calorie} = 4.2 \alpha^{-1} \beta^{-2} \gamma^2$$

1.4 Rewriting Statements (Explanation-based)

The term “large” or “small” is relative. Without a reference standard, these terms are meaningless.

Refined statements:

(a) Atoms are very small compared to a grain of sand.

(b) A jet plane moves with great speed compared to a bicycle.

(c) The mass of Jupiter is very large compared to Earth's mass.

(d) The air inside this room contains a large number of molecules compared to the number in an evacuated container.

(e) A proton is about 1836 times more massive than an electron.

(f) The speed of sound (~343 m/s) is much smaller than the speed of light (~3 × 10^8 m/s).

1.5 Sun-Earth Distance in New Units

Let $c = 1$ in new units.

Time light takes = 8 min 20 s = $8 \times 60 + 20 = 500\, \text{s}$

In new units where $c = 1$:

$$\text{Distance} = c \times t = 1 \times 500 = 500\, \text{new units}$$

✅ Answer:
Distance between Sun and Earth = 500 units (in new units where speed of light = 1)

Detailed solutions to questions 1.6 to 1.17 from Chapter 1: Units and Measurement (Class 11 Physics - NCERT):

1.6 Most precise device:

• (a) Vernier Callipers: Least count = 0.01 cm = 0.1 mm

• (b) Screw Gauge: Least count = pitch / number of divisions = 1 mm / 100 = 0.01 mm

• (c) Optical Instrument: Precision ≈ wavelength of light ≈ 0.0005 mm

✅ Answer: (c) An optical instrument is the most precise.

1.7 Estimate thickness of hair:

Given:

• Magnified image width = 3.5 mm

• Magnification = 100

So, actual thickness = $\frac{3.5\ \text{mm}}{100} = 0.035\ \text{mm} = 35\ \mu m$

✅ Answer: Thickness of hair = 35 micrometers

1.8

(a) Estimate thread diameter:
Wrap the thread around a pencil multiple times (say, 50 turns), measure total length using a scale, and divide by number of turns.

$$\text{Diameter} = \frac{\text{Total length}}{\text{Number of turns}}$$

(b) Can accuracy of screw gauge be increased arbitrarily?
No. Mechanical limitations like backlash error and wear & tear limit practical accuracy despite increasing scale divisions.

(c) Why 100 measurements more reliable than 5?
Because more data minimizes random errors through averaging → improves reliability and accuracy.

1.9

Given:

• Slide area = 1.75 cm²

• Screen area = 1.55 m² = 15,500 cm²

$$\text{Area magnification} = \frac{15,500}{1.75} = 8857.14$$ $$\text{Linear magnification} = \sqrt{8857.14} \approx 94.1$$

✅ Answer: Linear magnification ≈ 94.1

1.10 Significant Figures:

(a) 0.007 → 1 sig. fig.
(b) 2.64 × 10²⁴ → 3 sig. figs.
(c) 0.2370 → 4 sig. figs.
(d) 6.320 → 4 sig. figs.
(e) 6.032 → 4 sig. figs.
(f) 0.0006032 → 4 sig. figs.

1.11

• Length = 4.234 m (4 sig. figs.)

• Breadth = 1.005 m (4 sig. figs.)

• Thickness = 2.01 cm = 0.0201 m (3 sig. figs.)

Area:

$$A = 4.234 \times 1.005 = 4.25517 \Rightarrow 4.255\ \text{m}^2\ (\text{3 sig. figs.})$$

Volume:

$$V = A \times \text{thickness} = 4.255 \times 0.0201 = 0.08561 \Rightarrow 0.0856\ \text{m}^3\ (\text{3 sig. figs.})$$

1.12

(a) Total mass =

$$2.30\ \text{kg} + 0.02015\ \text{kg} + 0.02017\ \text{kg} = 2.34032\ \text{kg} \Rightarrow 2.34\ \text{kg}$$

(b) Difference =

$$|20.17 - 20.15| = 0.02\ \text{g} \ (\text{2 sig. figs.})$$

✅ Answer:
(a) Total mass = 2.34 kg
(b) Difference = 0.02 g

1.14

1 mole of H = 6.022 × 10²³ atoms
Size of one H atom ≈ 0.5 Å → radius
Volume of one atom ≈ $\frac{4}{3} \pi r^3$

$$r = 0.5 \times 10^{-10} \text{ m},\quad V_{\text{atom}} = \frac{4}{3}\pi (0.5 \times 10^{-10})^3 \approx 5.24 \times 10^{-31} \text{ m}^3$$

Total atomic volume = $5.24 \times 10^{-31} \times 6.022 \times 10^{23} \approx 3.15 \times 10^{-7} \text{ m}^3$

✅ Answer: Total atomic volume ≈ $3.15 \times 10^{-7}\ \text{m}^3$

1.15

Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Atomic volume from 1.14 ≈ $3.15 \times 10^{-7} \text{ m}^3$

$$\text{Ratio} = \frac{22.4 \times 10^{-3}}{3.15 \times 10^{-7}} \approx 7.1 \times 10^{4}$$

✅ Answer:
Ratio ≈ $7.1 \times 10^4$
This is large because atoms are point-like and gases have vast empty space between them.

1.16

This is due to parallax. Nearby objects subtend larger angles and shift more rapidly relative to your line of sight, while distant objects subtend very small angles and appear almost stationary.

1.17

$$\text{Volume of Sun} = \frac{4}{3}\pi R^3 = \frac{4}{3} \pi (7 \times 10^8)^3 \approx 1.43 \times 10^{27} \text{ m}^3$$ $$\text{Density} = \frac{2.0 \times 10^{30}}{1.43 \times 10^{27}} \approx 1400\ \text{kg/m}^3$$

✅ Answer: Density of Sun ≈ 1400 kg/m³ → closer to liquids (e.g., water = 1000 kg/m³), not gases.

.

Grade 11 Physics – Chapter: Units and Measurements EXAMINATION PAPER

full Examination Question Paper based on the entire chapter: Units and Measurements, aligned with the NEP 2020 and competency-based learning approach. This paper includes a mix of Competency-Based, Application-Oriented, and Analytical Thinking questions with various formats.

🧪 Grade 11 Physics – Chapter: Units and Measurements

Max Marks: 50
Time: 1 hour 30 minutes
NEP-Based Examination Pattern
Competency-Focused | Application-Oriented | Conceptual Clarity

Section A: Very Short Answer Questions (1 mark each)

(Answer in one sentence or definition. Attempt all questions.)
[1 × 6 = 6 marks]

1. Define a base quantity with an example.

2. Write the dimensional formula of force.

3. What do you mean by dimensional consistency of an equation?

4. Name two physical quantities having the same dimensional formula.

5. Give one example of a dimensionless quantity.

6. Write the SI unit of Planck’s constant.

Section B: Conceptual Understanding (2 marks each)

(Answer in brief: 2–3 lines)
[2 × 5 = 10 marks]

7. Why can't quantities with different dimensions be added or subtracted?

8. Show by dimensional analysis that $x = x_0 + v_0 t + \frac{1}{2} a t^2$ is dimensionally consistent.

9. Differentiate between fundamental and derived quantities.

10. Find the dimensional formula of pressure and write its SI unit.

11. What is meant by the order of accuracy in measurements?

Section C: Application-Based Questions (3 marks each)

(Answer using calculation or short explanation)
[3 × 4 = 12 marks]

12. Derive an expression for the time period of a simple pendulum using the method of dimensions.

13. Two physical quantities have the same dimensions. Does it mean they represent the same physical entity? Justify with an example.

14. If the unit of length is increased by 2%, how much percentage error will be introduced in the measurement of volume?

15. A formula is given as $T = 2\pi \sqrt{m/k}$. Check whether this formula is dimensionally correct. Assume $T$ is time, $m$ is mass and $k$ is spring constant.

Section D: Higher Order Thinking Skills (HOTS) (4 marks each)

(Analytical and creative questions)
[4 × 3 = 12 marks]

16. A new quantity $Z$ depends on mass $m$, acceleration $a$, and time $t$ as:

$$Z = k \cdot m^x a^y t^z$$

Find the values of $x, y, z$ such that $Z$ has the dimensions of energy.

17. The gravitational force between two masses is given by:

$$F = G \frac{m_1 m_2}{r^2}$$

Using dimensional analysis, find the dimensional formula of the gravitational constant $G$.

18. In a lab, a student writes an equation $E = \frac{1}{2} m v$. Another student writes $E = \frac{1}{2} m v^2$.
    (a) Use dimensional analysis to identify which is correct.
    (b) What does this equation represent physically?

Section E: Competency-Based Extended Response (5 marks)

[1 × 5 = 5 marks]

19. You are given three variables:

• mass $m$

• velocity $v$

• radius $r$

You are asked to derive the expression for a physical quantity $Q$, which has the dimensions of angular momentum using dimensional analysis.
(a) Write down the assumed relation.
(b) Derive the values of exponents.
(c) Write the final expression.
(d) Mention one real-life application of this physical quantity.

Here’s the Answer Key for the Grade 11 Physics Chapter: Units and Measurements NEP-Based Examination Paper.

---

🧪 Answer Key – Units and Measurements

Max Marks: 50

---

Section A: Very Short Answer Questions (1 mark each)

[1 × 6 = 6 marks]

1. Base Quantity: A quantity that is independent of other quantities and is defined arbitrarily, e.g., mass.

2. Dimensional formula of force: $[M^1 L^1 T^{-2}]$

3. Dimensional consistency: An equation is dimensionally consistent if all terms have the same dimensions.

4. Examples: Velocity and speed.

5. Dimensionless quantity: Refractive index or angle (θ in radians).

6. SI Unit of Planck’s Constant:

   J·s (joule second)
   or, in base units:
   kg·m²·s⁻¹

   Explanation:

   • 1 Joule (J) = 1 kg·m²/s²

   • So, J·s = (kg·m²/s²)·s = kg·m²·s⁻¹

   $\text{kg·m}^2\text{s}^{-1}$

   
   

Section B: Conceptual Understanding (2 marks each)

[2 × 5 = 10 marks]

7. Different dimensions represent different physical entities; adding them is meaningless (e.g., adding mass and length).

8. LHS: [L]; RHS: $[L] + [LT^{-1}T] + [LT^{-2}T^2] = [L]$. Hence, it is consistent.

9. Fundamental quantities are independent (e.g., mass), while derived quantities depend on them (e.g., velocity).

10. Pressure = Force/Area = $[MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$, SI unit = Pascal (Pa).

11. Order of accuracy refers to the closeness of a measurement to the true value; smaller the error, higher the accuracy.

Section C: Application-Based Questions (3 marks each)

[3 × 4 = 12 marks]

$$T = k l^x g^y m^z \Rightarrow [T] = [L]^x [LT^{-2}]^y [M]^z = L^{x+y} T^{-2y} M^z \Rightarrow x + y = 0, -2y = 1, z = 0 \Rightarrow x = \frac{1}{2}, y = -\frac{1}{2}, z = 0 \Rightarrow T = k\sqrt{l/g}$$

13. No. Velocity and speed have same dimensions but different directions. E.g., velocity is a vector, speed is scalar.

14. Volume ∝ $L^3$
    % Error = 3 × 2% = 6%

$$LHS = [T]; RHS = \sqrt{[M]/[MT^{-2}]} = \sqrt{T^2} = [T] \Rightarrow \text{Yes, dimensionally correct.}$$

Section D: Higher Order Thinking Skills (4 marks each)

[4 × 3 = 12 marks]

Let $Z = m^x a^y t^z$ and Z has dimensions of energy: $[ML^2T^{-2}]$
Acceleration $a = [LT^{-2}]$
So,

$$[M^x][LT^{-2}]^y[T]^z = M^x L^y T^{-2y+z} \Rightarrow x = 1, y = 2, -2y + z = -2 \Rightarrow z = 2 \Rightarrow x = 1, y = 2, z = 2$$

$$F = G \frac{m_1 m_2}{r^2} \Rightarrow G = \frac{F r^2}{m^2} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$$

(a) $E = \frac{1}{2} m v$ → $[MLT^{-1}]$;
$E = \frac{1}{2} m v^2$ → $[ML^2T^{-2}]$ → correct
(b) Represents kinetic energy of the body.

Section E: Competency-Based Extended Response (5 marks)

[1 × 5 = 5 marks]

(a) Assume: $Q = m^a v^b r^c$
(b) Angular momentum $[ML^2T^{-1}]$
$m = [M], v = [LT^{-1}], r = [L]$
So,

$$[M^a][LT^{-1}]^b [L]^c = M^a L^{b+c} T^{-b} \Rightarrow a = 1, b + c = 2, -b = -1 \Rightarrow b = 1, c = 1 \Rightarrow Q = mvr$$

(c) Expression: $L = mvr$
(d) Application: Rotational motion, planetary motion, quantum mechanics (electron orbitals)