NIOS Class 12 Physics Chapter 1 Units, Dimensions and Vectors

NIOS Class 12 Physics Chapter 1 Units, Dimensions and Vectors Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 1 Units, Dimensions and Vectors Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.

NIOS Class 12 Physics Chapter 1 Units, Dimensions and Vectors

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.

Units, Dimensions and Vectors

Chapter: 1

Module – I: Motion, Force and Energy

INTEXT QUESTIONS 1.1

1. Discuss the nature of laws of physics.

Ans: The laws of physics are universal, precise, and consistent across space and time. They describe the behavior of the natural world, such as motion, gravity, and energy, and are based on experimental observations and reasoning. These laws apply to everything in the universe, from subatomic particles to galaxies.

2. How has the application of the laws of physics led to better quality of life?

Ans: Applications of physical laws have enabled the development of technologies like electricity, communication, transportation, and medicine. For example, understanding electromagnetism has led to electric power, and Newton’s laws enabled the design of cars and airplanes — all contributing to better comfort, productivity, and health.

3. What is meant by significant figures in measurement?

Ans: Significant figures in a measurement are the digits that carry meaningful information about the precision of the quantity. They include all the certain digits plus the first uncertain digit. They help reflect the accuracy of measurement.

4. Find the number of significant figures in the following quantities, quoting the relevant laws:

(i) 426.69 

Ans: 5 significant figures.

(ii) 4200304.002

Ans: 10 significant figures. 

(iii) 0.3040 

Ans: 4 significant figures. 

(iv) 4050 m 

Ans: 3 significant. 

(v) 5000

Ans: 1 significant figure.

5. The length of an object is 3.486 m. If it is expressed in centimetre (i.e., 348.6 cm), will there be any change in number of significant figures in the two cases?

Ans: No, there will be no change in the number of significant figures. Both 3.486 m and 348.6 cm have four significant figures. Changing the unit does not affect the number of significant digits.

6. What are the four applications of the principles of dimensions? On what principle are the above based?

Ans: The four applications of dimensional analysis are:

(i) To check the correctness of physical equations.

(ii) To convert units from one system to another.

(iii) To derive relations between physical quantities.

(iv) To identify the dimension of unknown quantities.

These are all based on the principle of dimensional homogeneity, which states that all terms in a physical equation must have the same dimensional formula.

7. The mass of the sun is 2 × 10³⁰ kg. The mass of a proton is 2 × 10⁻²⁷ kg. If the sun was made only of protons, calculate the number of protons in the sun?

Ans: Given:

Mass of the sun = 2 × 10³⁰ kg

Mass of one proton = 2 × 10⁻²⁷ kg

Number of protons = (mass of the sun) / (mass of one proton)

= (2 × 10³⁰) / (2 × 10⁻²⁷)

= 10⁵⁷

10⁵⁷ protons

8. Earlier the wavelength of light was expressed in angstroms. One angstrom equals 10⁻⁸ cm. Now the wavelength is expressed in nanometers. How many angstroms make one nanometre?

Ans: Given:

1 angstrom = 10⁻⁸ cm

1 nanometre = 10⁻⁷ cm

Number of angstroms in 1 nm = (10⁻⁷ cm) / (10⁻⁸ cm)

= 10

= 10 angstroms make one nanometre.

9. A radio station operates at a frequency of 1370 kHz. Express this frequency in GHz.

Ans: Given:

1 GHz = 10⁹ Hz

1370 kHz = 1370 × 10³ Hz 

= 1.37 × 10⁶ Hz

Freuency in GHz = (1.37 × 10⁶) / (10⁹) 

= 1.37 × 10⁻³ GHz

1.37 × 10⁻³ GHz

10. How many decimetres are there in a decametre? How many MW are there in one GW?

Ans: Decimetres in one decametre:

1 decametre = 10 metres

1 metre = 10 decimetres

So, 1 decametre = 10 × 10 

= 100 decimetres

100 decimetres

Megawatts in one gigawatt:

1 gigawatt = 10⁹ watts

1 megawatt = 10⁶ watts

Number of MW in 1 GW = 10⁹ / 10⁶ = 

1000

: 1000 megawatts