Free NCERT Solutions For Class 11 Physics Chapter 2 : Units and Measurements Free PDF Download

Question 1: Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to…. 3 m
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to …  2 mm
(c) A vehicle moving with a speed of 18 km 1 h  covers….m in 1 s
(d) The relative density of lead is 11.3. Its density is ….g 3 cm or . …kg 3

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

Ans:
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.
An atom is a very small object in comparison to a soccer ball.
A jet plane moves with a speed greater than that of a bicycle.
Mass of Jupiter is very large as compared to the mass of a cricket ball.
The air inside this room contains a large number of molecules as compared to that present
in a geometry box.
A proton is more massive than an electron.
Speed of sound is less than the speed of light.

Question 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 tem of the new unit if light takes 8 min and 20 s to cover this distance?

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

Question 6: Which of the following is the most precise device for measuring length:
a vernier callipers with 20 divisions on the sliding scale
a screw gauge of pitch 1 mm and 100 divisions on the circular scale
an optical instrument that can measure length to within a wavelength of light ?

Ans:
A device with minimum count is the most suitable to measure length.
Least count of vernier callipers
= 1 standard division (SD) – 1 vernier division (VD)
9 1 1 0.01
10 10
    cm
Least count of screw gauge =
pitch
Number of divisions
1
0.001
1000
  cm
Least count of an optical device = Wavelength of light
= 0.00001 cm
Hence, it can be inferred that an optical instrument is the most suitable device to measure
length.

Question 7: A student measures the thickness of a human hair by looking at it through a microscope
of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

Ans:
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
3.4 0.035
100
 mm

Question 8: Answer the following:
You are given a thread and a metre scale How will you estimate the diameter of the
thread?
A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Ans:
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.
Lenght of thread Diameter =
Number of turns
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.
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.

Question 11: The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures

Question 12: The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces
masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b)
the difference in the masses of the pieces to correct significant figures?

Ans:
Mass of grocer’s box = 2.300 kg
Mass of gold piece I = 20.15g = 0.02015 kg
Mass of gold piece II = 20.17 g = 0.02017 kg
Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg
In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.
Difference in masses
   20.17 20.15 0.02g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Question14: A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
(a) y = a sin 2π t/T
(b) y = a sin vt
(c) y = (a/T) sin t/a
(d) y = (a √2) (sin 2πt /T + cos 2πt /T)
(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Ans: (a = ma×imum displacement of the particle, v = speed of the particle. T= timeperiod of motion).
Rule out the wrong formulas on dimensional grounds.
(a) Correct Y=asin2πt/T
Dimension of y=M1
L
1
T
0
Dimension of a= M1
L
1
T
0
Dimension of sin2πt/T = M1
L
0
T
0
Dimension of L.H.S = Dimension of R.H.S
Hence, the given formula is dimensionally correct.
(b) Incorrect
Y = a sin vt
Dimension of y= M0
L
1
T
0
Dimension of a = M0
L
1
T
0
Dimension of vt = M0
L
1
T
-1× M0
L
1
T
1= M0
L
1
T
0
But the argument of the trigonometric function must be dimensionless, which is not so in the
given case. Hence, the given formula is dimensionally incorrect.
(c) Incorrect
Y= (a/T) sin(t/a)
Dimension of y= M0
L
1
T
0
Dimension of a/T = M0
L
1
T
-1
Dimension of =t/a = M0
L
-1
T
1
But the argument of the trigonometric function must be dimensionless, which is not so in the
given case. Hence, the formula is dimensionally incorrect.
(d) Correct
Y=(
a √2 )sin(2π t/T )+ cos(2π t/T)
Dimension of y= M0
L
1
T
0
Dimension of a= M0
L
1
T
0
Dimension of t/T = M0
L
0
T
0
Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimension of y and a are the same.
Hence, the given formula is dimensionally correct

Question 16: The unit of length convenient on the atomic scale is known as an angstrom and is denoted by 10 A A m :1 10  The size of a hydrogen atom is about 0.5A what is the total atomic volume in 3 m of a mole of hydrogen atoms?

Question 18: Explain this common observation clearly : If you look out of the window of a fast moving
train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Ans:
Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

Question 19:
The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth ‘s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit 11  3 10 m
However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters?

Question 21: Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Ans:
It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses ultra-shot laser pulses ( time interval 15 10 s  ) are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.
The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

Question 22: Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations.
Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) the total mass of rain-bearing clouds over India during the Monsoon
the mass of an elephant
the wind speed during a storm
the number of strands of hair on your head
the number of air molecules in your classroom.

Solution 22:
During monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of
water column h = 215 cm = 2.15 m
Area of country,
12 2 A m   3.3 10
Hence, volume of rain water,
12 3 V A h m     7.09 10
Density of water
3 3  1 10 kg m
 
Hence, mass of rain water
15      V kg 7.09 10
Hence, the total mass of rain-bearing clouds over India is approximately
15 7.09 10  kg
Page 15
Consider a ship of known base area floating in the sea. Measure its depth in sea (say
1 d
)
Volume of water displaced by the ship,
V Ad b 
1
Now, move an elephant on the ship and measure the depth of the ship
  2 d
in this case
Volume of water displaced by the ship with the elephant on board
V Ad be 
2
Volume of water displaced by the elephant
Ad Ad 2 1 
Density of water
 D
Mass of elephant
  AD d d   2 1
Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates.
The rotation made by the anemometer in one second gives the value of wind speed
Area of the head surface carrying hair
 A
With the help of a screw gauge, the diameter and hence, the radius of a hair can be
determined. Let it be r.
Area of one hair
2
r
Number of strands of hair
2
Total surfacearea A
Areaof onehari r 
 
Let the volume of the room be
V
One mole of air at NTP occupies 22.4 l i.e.,
3 3 22.4,22.4 10 m


volume.
Number of molecules in one mole
23   6.023 10

Number of molecules in room of volume V
23
26
3
28
6.023 10 134.915 10
22.4 10
1.35 10
V V
V


   

 
Question 23:
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding
7
10 K
and its outer surface at a temperature of about 6000 K At these high temperatures no
substance remains in a solid or liquid phase. In what range do you expect the mass density of
the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is
correct from the following data: mass of the Sun
30   2.0 10
kg, radius of the Sun
8
  7.0 10 m
Solution 23:
Mass of the Sun,
30 M kg   2.0 10
Radius of the Sun,
8 R m   7.0 10
Page 16
Volume of the Sun,
4 3
3
V R  
 3
8
24 24 3
4 22 7.0 10
3 7
88 343 10 1437.3 10
21
m
   
    
Density of the Sun
30
3 5
24
2.0 10 1.4 10 /
1437.3 10
Mass kg m
Volume

  

The density of the Sun is in the density range of solids and liquids. This high density is
attributed to the intense gravitational attraction of the inner layers on the outer layer of the
Sun.

Question 24: When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter