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NCERT Solution for Class 11 Science Physics Chapter 2 - Units and Measurements

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NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 11 Physics exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 2 - Units and Measurements.

All our solutions for Chapter 2 - Units and Measurements are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 11 Physics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

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NCERT Solution for Class 11 Science Physics Chapter 2 - Units and Measurements Page/Excercise 35

Solution 1

(a) 1 cm =

Volume of the cube = 1 cm3

But, 1 cm3 = 1 cm 1 cm 1 cm =


1 cm3 = 10-6 m3

Hence, the volume of a cube of side 1 cm is equal to 10-6 m3.

(b) The total surface area of a cylinder of radius r and height h is

S = 2 r (r + h).

Given that,

r = 2 cm = 2 1 cm = 2 10 mm = 20 mm

h = 10 cm = 10 10 mm = 100 mm

= 15072 = 1.5 104 mm2

(c) Using the conversion,

1 km/h =


Therefore, distance can be obtained using the relation:

Distance = Speed Time = 5 1 = 5 m

Hence, the vehicle covers 5 m in 1 s.

(d) Relative density of a substance is given by the relation,

Relative density =

Density of water = 1 g/cm3


Again, 1g = kg

1 cm3 = 10-6 m3

1 g/cm3 =

11.3 g/cm3 = 11.3 103 kg/m3

Solution 2

Solution 3

Solution 4

The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

(a) An atom is a very small object in comparison to a soccer ball.

(b) A jet plane moves with a speed greater than that of a bicycle.

(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.

(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.

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

 (f) Speed of sound is less than the speed of light.

Solution 5

Distance between the Sun and the Earth:

= Speed of light Time taken by light to cover the distance

Given that in the new unit, speed of light = 1 unit

Time taken, t = 8 min 20 s = 500 s

Distance between the Sun and the Earth = 1 500 = 500 units

Solution 6

A device with minimum count is the most suitable to measure length.

(a) Least count of vernier callipers

= 1 standard division (SD) - 1 vernier division (VD)

(b) Least count of screw gauge =

(c) Least count of an optical device = Wavelength of light 10-5 cm

= 0.00001 cm

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

Solution 7

Magnification of the microscope = 100

Average width of the hair in the field of view of the microscope = 3.5 mm

Actual thickness of the hair is   = 0.035 mm.

Solution 8

(a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,  

(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

(c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

Solution 9

Area of the house on the slide = 1.75 cm2
Area of the image of the house formed on the screen = 1.55 m2
= 1.55 104 cm2
Arial magnification, ma =
  Linear magnifications, ml =      

Solution 10

NCERT Solution for Class 11 Science Physics Chapter 2 - Units and Measurements Page/Excercise 36

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Solution 19

NCERT Solution for Class 11 Science Physics Chapter 2 - Units and Measurements Page/Excercise 37

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NCERT Solution for Class 11 Science Physics Chapter 2 - Units and Measurements Page/Excercise 38

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TopperLearning provides step-by-step solutions for each question in each chapter in the NCERT textbook. Access Chapter 2 - Units and Measurements here for free.

Our NCERT Solutions for Class 11 Physics are by our subject matter experts. These NCERT Textbook Solutions will help you to revise the whole chapter, and you can increase your knowledge of Physics. If you would like to know more, please get in touch with our counsellor today!

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