Skip to main content

Class 11 Physics Units and Measurements Solutions | NCERT Exercise 1.1 & 1.2

- Dr.Sanjaykumar Pawar 

NCERT Class 11 Physics Chapter 1 Questions and Answers PDF


Student studying Class 11 Physics Units and Measurements chapter with SI units, measurements, formulas, and solved NCERT exercises.

Class 11 Physics Chapter 1 Units and Measurements with NCERT solutions and practice questions.

 INTERNAL LINKS

  1. Class 11 Physics Chapter 2 Motion in a Straight Line Notes

  2. Class 11 Physics Chapter 3 Motion in a Plane Solutions

  3. Class 11 Physics Important Formulas PDF

  4. CBSE Class 11 Physics Sample Questions

  5. Significant Figures Explained with Examples

  6. SI Units and Derived Units Guide

  7. Class 11 Physics MCQs with Answers

  8. NCERT Class 11 Physics Complete Solutions

  9. Measurement and Error Analysis Notes

  10. CBSE Class 11 Study Material Hub



Class 11 Physics - Exercise 1.1 and 1.2

NCERT Class 11 Physics

Chapter 1: Units and Measurements

Exercise 1.1

Q1(a). The volume of a cube of side 1 cm is equal to ____ m³.
Answer:
Volume = (1 cm)³ = 1 cm³
1 cm = 10⁻² m
1 cm³ = (10⁻²)³ m³
= 10⁻⁶ m³

Final Answer: 1 × 10⁻⁶ m³
Q1(b). The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ____ (mm)².
Answer:
Surface Area = 2πr(r+h)
= 2 × 3.14 × 2.0 × (2.0 + 10.0)
= 150.72 cm²
1 cm² = 100 mm²
150.72 × 100 = 15072 mm²

Final Answer: 1.5 × 10⁴ mm²
Q1(c). A vehicle moving with a speed of 18 km h⁻¹ covers ____ m in 1 s.
Answer:
18 km h⁻¹ = 18 × (1000/3600)
= 5 m s⁻¹
Distance = Speed × Time
= 5 × 1
= 5 m

Final Answer: 5 m
Q1(d). The relative density of lead is 11.3. Its density is ____ g cm⁻³ or ____ kg m⁻³.
Answer:
Density = Relative Density × Density of Water
= 11.3 × 1
= 11.3 g cm⁻³

1 g cm⁻³ = 1000 kg m⁻³
11.3 × 1000 = 1.13 × 10⁴ kg m⁻³

Final Answer:
11.3 g cm⁻³
1.13 × 10⁴ kg m⁻³

Exercise 1.2

Q2(a). 1 kg m² s⁻² = ____ g cm² s⁻²
Answer:
1 kg = 10³ g
1 m² = 10⁴ cm²

1 kg m² s⁻²
= 10³ × 10⁴ g cm² s⁻²
= 10⁷ g cm² s⁻²

Final Answer: 10⁷ g cm² s⁻²
Q2(b). 1 m = ____ ly
Answer:
1 light year = 9.46 × 10¹⁵ m

1 m = 1 / (9.46 × 10¹⁵)
= 1.06 × 10⁻¹⁶ ly

Final Answer: 1.06 × 10⁻¹⁶ ly
Q2(c). 3.0 m s⁻² = ____ km h⁻²
Answer:
1 m = 10⁻³ km
1 s = 1/3600 h

3.0 × 10⁻³ × (3600)²
= 38880 km h⁻²

Final Answer: 3.9 × 10⁴ km h⁻²
Q2(d). G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = ____ cm³ s⁻² g⁻¹
Answer:
1 N = kg m s⁻²

G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

1 m³ = 10⁶ cm³
1 kg⁻¹ = 10⁻³ g⁻¹

G = 6.67 × 10⁻¹¹ × 10⁶ × 10⁻³
= 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹

Final Answer: 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹
Class 11 Physics Practice Questions

Class 11 Physics - Units and Measurements

20 Practice Questions with Answers

Q1. A cube has a side of 2 cm. Find its volume.
Volume = a³
= 2³
= 8 cm³
Answer: 8 cm³
Q2. Convert 500 cm into metres.
500 ÷ 100 = 5 m
Answer: 5 m
Q3. Convert 1 km into metres.
1 km = 1000 m
Answer: 1000 m
Q4. Convert 72 km h⁻¹ into m s⁻¹.
72 × (1000/3600)
= 20 m s⁻¹
Answer: 20 m s⁻¹
Q5. A body moves with a speed of 15 m s⁻¹ for 4 s. Find the distance covered.
Distance = Speed × Time
= 15 × 4
= 60 m
Answer: 60 m
Q6. Convert 1 m² into cm².
1 m = 100 cm
1 m² = (100)² cm²
= 10000 cm²
Answer: 10000 cm²
Q7. Find the area of a square of side 5 m.
Area = side²
= 5²
= 25 m²
Answer: 25 m²
Q8. Convert 1 litre into cm³.
1 litre = 1000 cm³
Answer: 1000 cm³
Q9. A body has mass 200 g and volume 50 cm³. Find density.
Density = Mass / Volume
= 200 / 50
= 4 g cm⁻³
Answer: 4 g cm⁻³
Q10. What is the SI unit of force?
Answer: Newton (N)
Q11. What is the SI unit of energy?
Answer: Joule (J)
Q12. Convert 3 kg into grams.
3 × 1000 = 3000 g
Answer: 3000 g
Q13. Convert 2500 g into kilograms.
2500 ÷ 1000 = 2.5 kg
Answer: 2.5 kg
Q14. Find the total surface area of a cube of side 4 cm.
TSA = 6a²
= 6 × 4²
= 6 × 16
= 96 cm²
Answer: 96 cm²
Q15. What is the SI unit of time?
Answer: Second (s)
Q16. A cyclist covers 120 m in 20 s. Find speed.
Speed = Distance / Time
= 120 / 20
= 6 m s⁻¹
Answer: 6 m s⁻¹
Q17. Convert 36 km h⁻¹ into m s⁻¹.
36 × (1000 / 3600)
= 10 m s⁻¹
Answer: 10 m s⁻¹
Q18. What is the relative density of water?
Answer: 1
Q19. Find the volume of a cylinder of radius 3 cm and height 7 cm.
Volume = πr²h
= (22/7) × 3² × 7
= 198 cm³
Answer: 198 cm³
Q20. Express 1 day in seconds.
1 day = 24 × 60 × 60
= 86400 s
Answer: 86400 s

Quick Revision

  • SI unit of length = metre (m)
  • SI unit of mass = kilogram (kg)
  • SI unit of time = second (s)
  • SI unit of force = newton (N)
  • SI unit of energy = joule (J)
  • 1 km = 1000 m
  • 1 m = 100 cm
  • 1 litre = 1000 cm³
  • Density = Mass / Volume
  • Speed = Distance / Time

Comments

Popular posts from this blog

Block and Trolley System NEET Solution | Acceleration & Tension Explained

NEET Physics Example 4.9: Block and Trolley System (Step-by-Step) - Dr.Sanjaykumar Pawar  INTERNAL LINKS  /neet-physics-newtons-laws /friction-notes-class-11 /tension-in-string-problems /pulley-system-problems-neet /mechanics-important-questions /class-11-physics-motion-in-a-line /neet-important-derivations Physics NEET diagram showing a block and trolley system with forces, tension, friction, and acceleration clearly labeled.         Example 4.9 – Block and Trolley System Example 4.9 – Block and Trolley System (Easy NEET Notes) Given: Mass of hanging block, \( m_1 = 3\,kg \) Mass of trolley, \( m_2 = 20\,kg \) Coefficient of kinetic friction, \( \mu_k = 0.04 \) Acceleration due to gravity, \( g = 10\,m\,s^{-2} \) String is light (massless) and inextensible Pulley is smooth (frictionless) To Find: Acceleration of the system \( a \) Tension in the string \( T \) Step 1: Understand the Motion The 3 kg block hangs vert...

Example 3.4 Solution Explained for Beginners | Velocity and Acceleration

Step-by-step solution of Example 3.4 showing velocity and acceleration in vector form. Dr.Sanjaykumar pawar Internal Links Introduction to Vectors in Physics Difference Between Speed and Velocity Motion in a Straight Line Notes Vector Addition and Subtraction How to Differentiate Position Vectors Magnitude of Vector Formula Explained Direction Cosines in Physics NCERT Kinematics Solutions Class 11 Physics Chapter Motion Notes Solved Problems on Acceleration Example 3.4 Solution Example 3.4 Solution Example 3.4 The position of a particle is given by where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0 s The position of a particle is given by: r(t) = 3.0t î + 2.0t² ĵ + 5.0 k̂ where time t is in seconds and position is in metres. Part (a): Find Velocity and Acceleration Step 1: Write the Position Vector r(t) = 3.0t î + ...

Fundamental Forces in Nature: Strength, Range & Comparison Guide

The four fundamental interactions that govern everything from the smallest atom to the largest galaxy.  Internal Link  * Newton’s Law of Universal Gravitation: (when mentioning mass-dependent attraction).  * Atomic Structure & Nucleus: (when discussing the Strong Nuclear force and Quarks).  * Radioactivity and Half-life: (when explaining the Weak Nuclear force and \beta-decay).  * Coulomb’s Law: ( the Electromagnetic section regarding charges at rest).   -Dr.Sanjaykumar pawar  wed25March Physics Notes: Fundamental Forces in Nature Fundamental Forces in Nature The four basic interactions that govern everything in the universe. 1. Gravitational Force The force of mutual attraction between any two objects by virtue of their masses . Nature: Weakest of all forces but infinite in range. It is always attractive . Scope: Governs large-scale phenomena like the formation of stars, gal...