**NCERT Solutions Class 11 Physics Chapter 2 Units And Measurement** – Here are all the NCERT solutions for Class 11 Physics Chapter 2. This solution contains questions, answers, images, explanations of the complete chapter 2 titled Of Units And Measurement taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 2 Units And Measurement After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 2 Units And Measurement in one place.

## NCERT Solutions Class 11 Physics Chapter 2 Units And Measurement

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For a better understanding of this chapter, you should also see summary of Chapter 2 Units And Measurement , Physics, Class 11.

Class | 11 |

Subject | Physics |

Book | Physics Part I |

Chapter Number | 2 |

Chapter Name |
Units And Measurement |

### NCERT Solutions Class 11 Physics chapter 2 Units And Measurement

Class 11, Physics chapter 2, Units And Measurement solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Units And Measurement

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### Question & Answer

Q.1:Fill in the blanks (a) The volume of a cube of side 1 cm is equal to ______m³ (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _____ (mm)² (c) A vehicle moving with a speed of 18 km h⁻ⁱ covers _______m in 1 s (d) The relative density of ead is 11 B. Its density is ______g a-n 30r ______kg m⁻⁵.

Ans :(a)\(1 \mathrm{cm}=\frac{1}{100} \mathrm{m}\) Volume of cube = 1 cm³ \(\mathrm{But}, 1 \mathrm{cm} 3=1 \mathrm{cm} \times 1 \mathrm{cm} \times 1 \mathrm{cm}=\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m}\) \(\begin{array}{l}{ 1 \mathrm{cm} 3=10-6 \mathrm{m} 3} \\ {\text { Hence, the volume of a cube of side } 1 \mathrm{cm} \text { is equal to } 10-6 \mathrm{m} 3 \text { . }}\end{array}\) \(\begin{array}{l}{\text { (b) The total surface area of a cylinder of radius } r \text { and height } h \text { is }} \\ {S=2 \pi r(r+h)} \\ {\text { Given that, }} \\ {r=2 \mathrm{cm}=2 \times 1 \mathrm{cm}=2 \times 10 \mathrm{mm}=20 \mathrm{mm}} \\ {h=10 \mathrm{cm}=10 \times 10 \mathrm{mm}=100 \mathrm{mm}} \\ {\therefore S=2 \times 3.14 \times 20 \times(20+100)=15072=1.5 \times 104 \mathrm{mm} 2}\end{array}\) \(\begin{array}{l}{\text { (c) Using the conversion, }} \\ {1 \mathrm{km} / \mathrm{h}=\frac{5}{18} \mathrm{m} / \mathrm{s}} \\ {18 \mathrm{km} \mathrm{h}=18 \times \frac{5}{18}=5 \mathrm{m} / \mathrm{s}}\end{array}\) \(\begin{array}{l}{\text { Therefore, distance can be obtained using the relation: }} \\ {\text { Distance }=\text { Speed } \times \text { Time }=5 \times 1=5 \mathrm{m}} \\ {\text { Hence, the vehicle covers } 5 \mathrm{m} \text { in } 1 \mathrm{s} \text { . }}\end{array}\) (d) Relative density of a substance is given by the relation, Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\) Density of water = 1 g/cm³ \(\begin{aligned} \text { Density of lead } &=\text { Relative density of lead } \times \text { Density of water } \\ &=11.3 \times 1=11.3 \mathrm{g} / \mathrm{cm}^{3} \end{aligned}\) \(1 \mathrm{g}=\frac{1}{1000} \mathrm{kg}\) 1 cm³ = 10⁻⁶ m³ \(1\mathrm{g} / \mathrm{cm}^ 3=\frac{10^{-3}}{10^{-6}} \mathrm{kg} / \mathrm{m}^{3}=10^{3} \mathrm{kg} / \mathrm{m}^{3}\) 11.3 g/cm³ = 11.3 x 10. kg/m³

Q.2:Fill in the blanks by suitable conversion of units: (a) 1 kg m2² s⁻² = _____g cm2 s⁼² (b) 3.0 m s⁻²⁰= ______ km h⁻² (c) G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = ______(cm)³ s⁻² g⁻¹ .

Ans :\(\begin{aligned}(a) 1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2} &=\frac{1 \mathrm{kg} \mathrm{m}^{2}}{s^{2}}=\frac{1 \times 1000 \times\left(10^{2}\right)^{2}}{s^{2}} \mathrm{g} \mathrm{cm}^{2} \\ &=10^{7} \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2} \end{aligned}\) \(1 \mathrm{m} \quad=\frac{1}{9.46 \times 10^{5}} \mathrm{ly} \approx \frac{1}{10^{16}} \mathrm{ly}=10^{-16} \mathrm{ly}\) \(\begin{aligned}(b) 3 \mathrm{ms}^{-2} &=\frac{3 \times 10^{-3} \mathrm{km}}{\left(\frac{1}{3600}\right)^{2} \mathrm{h}^{2}}=3 \times 3600 \times 3600 \times 10^{-3} \mathrm{km} \mathrm{h}^{-2} \\ &=3.888 \times 10^{4} \mathrm{km} \mathrm{h}^{-2} \end{aligned}\) \(\begin{aligned}(c) G &=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{kg}^{-2}=6.67 \times 10^{-11} \frac{\mathrm{kg} \mathrm{m}}{\mathrm{s}^{2}} \mathrm{m}^{2} \mathrm{kg}^{-2} \\ &=6.67 \times 10^{-11} \mathrm{kg}^{-1} \mathrm{m}^{3} \mathrm{s}^{-2} \\ &=6.67 \times 10^{-11} \frac{\mathrm{m}^{3}}{\mathrm{kgs}^{2}}=\frac{6.67 \times 10^{-11} \times\left(10^{2}\right)^{3}}{\left(10^{3}\right)^{2}} \\ &=6.67 \times 10^{-8} \mathrm{cm}^{-3} \mathrm{s}^{-2} \mathrm{g}^{-1} \end{aligned}\)

Q.3:A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm² s⁻². Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals j8 m, the. unit of time is ys. Show that a calorie has a magnitude \(4.2^{\alpha-1 \beta-2 \mathrm{y} 2}\)in terms of the new units.

Ans :\(1 \mathrm{cal}=4.2 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}\) Hence in terms of new unit , = \(1 \mathrm{kg}=\frac{1}{\mathrm{a}}=\alpha^{-1}\) In terms of the new unit length, \(\mathrm{I} \mathrm{m}=\frac{1}{\beta}=\beta^{-1} \text { or } \mathrm{Im}^{2}=\beta^{-2}\) And, in terms of the new unit of time, \(\begin{array}{l}{1 \mathrm{s}=\frac{1}{\gamma}=\gamma^{-1}} \\ {1 \mathrm{s}^{2}=\gamma^{-2}} \\ {1 \mathrm{s}^{-2}=\gamma^{2}}\end{array}\) Calorie = \(4.2(1 \mathrm{a}-1)(1 \beta-2)(1 \mathrm{y} 2)=4.2 \mathrm{a}-1 \beta-2 \mathrm{y}^2\)

Q.4:Explain this statement clearly: "To call a dimensional quantity 'large' or is meaningless without specifying a standard for comparison". In view of this, reframe the following statements wherever necessary. (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light.

Ans :Physical quantities are called large or small depending on the unit (standard) of measurement For example, the distance between two cities on earth is measured in kilometres but the distance between stars or intergalactic distances are measured in parsec The later standard parsec is equal to \(3.08 \times 10^{16} \mathrm{m} \text { or } 3.08 \times 10^{12} \) km is certainly larger than metre or kilometre Therefore, the inter-stellar or intergalactic distances are certainly larger than the distances between two cities on earth. (a) The size of an atom is much smaller than even the sharp tip of a pin. (b) A Jet plane moves with a speed greater than that of a super fast train. (c) The mass of Jupiter is very large compared to that of the earth. (d) The air inside this room contains more number of molecules than in one mole of air. (e) This is a correct statement. (f) This is a correct statement.

Q.5:A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Ans :\(\begin{array}{l}{\text { Distance between Sun and Earth }} \\ {=\text { Speed of light in vacuum } x \text { time taken by light to travel from Sim to Earth }=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 8} \\ {\min 20 \mathrm{s}=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 500 \mathrm{s}=500 \times 3 \times 10^{8} \mathrm{m} \text { . }} \\ {\text { In the new system, the speed of light in vacuum is unity. So, the new unit of length is } 3 \times 10^{8} \mathrm{m} \text { . }} \\ {\therefore \text { distance between Sun and Earth }=500 \text { new units. }}\end{array}\)

## NCERT / CBSE Book for Class 11 Physics

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