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Have you ever thought how scientists determine the extremely accurate distance between the Earth and the Moon, or how engineers construct machines in the range of one millimetre precision? All this becomes achievable due to the precise units and measurements, and they are considered the foundation of any scientific observations. NCERT Exemplar Class 11 Physics Chapter 2 explores the methodologies of making precise measurements of the physical quantities using standard practices. You will learn about concepts such as SI units, dimensional analysis, significant figures, error types and the rules of checking the physical equations with the help of dimension. Such knowledge is critical to the accurate and clear solution of numerical problems in physics.
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NCERT Exemplar Class 11 Physics Solutions Chapter 2 answer is in form of steps with a number of questions of different types being covered such as multiple choice questions (MCQs) and the short answer type, the long answer and the very long answer. These solutions are made by subject professionals which make even the difficult subjects easy to understand. With these guideline solutions, the students will be able to improve their problem-solving abilities, reduce the number of calculation mistakes, and enhance their background in terms of CBSE exams, JEE and NEET, and other competitive exams.
Question:1
The number of significant figures in
Answer:
The answer is option (b) 4Question:2
The sun of the numbers
Answer:
The answer is the optionQuestion:3
Answer:
The answer is the option
Explanation: When we are performing multiplication or division, we should keep in mind to retain as many significant figures in the original no. with the least significant figures.
We know that Density = mass/ volume
As rounding off the number upto 2 significant figures, we get density
Question:4
The numbers
a)
b)
c)
d)
Answer:
The answer is the option
Explanation:
Here the fourth digit is five and is preceded by, viz., an even no. Hence, it becomes 2.74.
Here the fourth digit is 5, but the preceding digit is 3, viz., an odd no. Hence, it is increased by 1, and we get 2.74.
Question:5
Answer:
The answer is the option (i)Question:6
Which of the following pairs of physical quantities does not have same dimensional formula?
a) work and torque
b) angular momentum and Planck’s constant
c) tension and surface tension
d) impulse and linear momentum
Answer:
The answer is option (c): Tension & surface tensionQuestion:7
Measure of two quantities along with the precision of respective measuring instrument is
The value of AB will be
Answer:
The answer is the option (a)Question:8
You measure two quantities as
a)
b)
c)
d)
Answer:
The answer is the option (d)
There are two significant figures in 1.0 & 2.0
After rounding up 1.0 &2.0, we got two significant features, i.e.
Now,
Question:9
Which of the following measurements is most precise?
a)
b)
c)
d)
Answer:
The answer is option (a),Question:10
The mean length of an object is
a)
b)
c)
d)
Answer:
The answer is option (a),
Explanation :
Thus,
Question:11
Young’s modulus of steel is
a)
b)
c)
d)
Answer:
The answer is option (c)
Explanation: Young's modulus (Y)
Converting into dyne/cm2-
Question:12
Answer:
The answer is the option d)Given, fundamental quantities are momentum (p), area (A) and time (T). We can write energy
where
Dimensions of
Putting all the dimensions, we get
By principle of homogeneity of dimensions,
Question:13
Answer:
The answer are the options (b)
Now, by principle of homogeneity of dimensions LHS and RHS of (a) and (d) will be same and is
For (c)
Hence, (c) is not correct option.
In option (b) dimension of angle is [vt]i.e.,
So, option (b) is also not correct.
Question:14
Answer:
The answer are the optionsQuestion:15
Photon is quantum of radiation with energy E = h
a) linear impulse
b) angular impulse
c) linear momentum
d) angular momentum
Answer:
The answer are the options (b) Angular Impulse and (d) Angular momentumThus, the dimensional formulae of Planck’s constant, Angular Impulse and Angular Momentum are the same.
Question:16
If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which of the following can in addition be taken to express length, mass, and time in terms of the three chosen fundamental quantities?
a) mass of electron (me)
b) universal gravitational constant (G)
c) charge of electron (e)
d) mass of proton (mp)
Answer:
The answer are the optsion (a) Mass of electron (me) and (b) Universal gravitational constant (G) and (d) Mass of proton (mp)Similarly,
Hence, (a), (b) or (d) any can be used to express
Question:17
Which of the following ratios express pressure?
a) Force/area
b) Energy/volume
c) Energy/area
d) Force/volume
Answer:
The answer is the option (a) Force/Area and (b) Energy/Volume
Explanation:
Let us start with calculating the dimension of pressure,
We know that Pressure(P)
This also verifies the option (a).
Now let us compare it with other options-
Thus, option (b) is verified as well.
This is not equal to the Dimensions of P
This again is not equal to the dimensions of P
Thus, opt (c) & (d) are discarded.
Question:18
Which of the following is not a unit of time?
a) second
b) parsec
c) year
d) light year
Answer:
The answer are the options (b) Parsec and (d) Light yearQuestion:19
Why do we have different units for the same physical quantity?
Answer:
(i) Different orders of quantity are measured by the same physical quantities.Question:20
The radius of atom is of the order of
Answer:
Radius of atom
Radius of nucleus
Volume of atom
Volume of nucleus
Question:21
Name the device used for measuring the mass of atoms and molecules.
Answer:
The mass spectrograph is the device used for measuring the mass of atoms and molecules.Question:22
Express unified atomic mass unit in kg.
Answer:
We know that the unified atomic mass unit or amu orQuestion:23
A function
Why is it necessary for
Answer:
(i)Question:24
Why length, mass and time are chosen as base quantities in mechanics?
Answer:
Length, mass & time is chosen as base quantities in mechanics because unlike other physical quantities, they cannot be derived from any other physical quantities.Question:25
(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?
(b) Moon is seen to be of
(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of the sun's to Earth's diameters.
Answer:
(a) Here we know that,
Now, the angle from the moon to the diameter of the earth will be
(b) If the moon is seen from the earth diametrically, the angle will be
& if the earth is seen from the moon, the angle will be
Thus, the size of the moon/ size of the earth =
Therefore, as compared to the earth, the size of the moon is
(c)Let rem = x be the distance between the earth and the moon
Now, the distance between the sun and the earth (rse) = 400x
On a solar eclipse, the sun is completely covered by the moon; also, the angle formed by the diameter of the sun, moon and earth are equal.
Thus,
Now since
Thus,
Thus,
Therefore,
Question:26
Which of the following time-measuring devices is most precise?
(a) A wall clock.
(b) A stopwatch.
(c) A digital watch.
(d) An atomic clock.
Give a reason for your answer.
Answer:
The least counts of the devices are as follows-
Wall clock = 1 sec
Stopwatch =
Digital watch =
Atomic clock =
Hence, the atomic clock is the most precise.
Question:27
Answer:
It is given that the distance travelled by light from the galaxy to earth =Question:28
Answer:
Let ‘n’ be the no. of the parts on the Vernier scale, n = 50Question:29
Answer:
Consider the diagram given above
Let angle made by sun and moon is
Here,
(Here, radius of sun and moon represents their sizes respectively)
Question:30
Answer:
The dimensions of -
Force =
Length =
Time =
Now let us substitute the values of eq. (ii) & (iii) in eq. (i)
We get,
Thus,
Question:31
Give an example of
(a) a physical quantity which has a unit but no dimensions.
(b) a physical quantity which has neither unit nor dimensions.
(c) a constant which has a unit.
(d) a constant which has no unit.
Answer:
(a) A plane angle is an example of a physical quantity which has unit but no dimension since, plane angle = arc/radius in radian & solid angle.Question:32
Answer:
Given :radius = 31 cmQuestion:33
Answer:
We know that solid angleQuestion:34
Answer:
According to the principle of homogeneity, the dimensional formula on L.H.S. & R.H.S. is equal.Question:35
Answer:
Given data :
and
L.C. of the instrument is 0.1 sec, hence, Precision (LC) = 0.1 sec.
Now,
For 20 oscillations, the mean value will be
Now the absolute errors in measurement
Now, Mean absolute error
& Accuracy of measurement
Question:36
Answer:
Joules is a unit of work/energy, so let us first write the dimensions of energy-
Energy
Let
Thus,
Now,
Thus,
Thus, by the new system,
Question:37
Answer:
Let us write dimensions of -Question:38
Answer:
As per the information given in the question, the percentage error in
As per the information given in the question, the Percentage error in
As per the information given in the question, the Percentage error in
As per the information given in the question, the Percentage error in
Percentage error in
Now, let us calculate the mean absolute error
Therefore, we get ‘2.8’ after rounding X i.e.
Question:39
Answer:
The dimensional formula of –
Now let us find out the dimension of P
Hence, it is clear that P is a dimensionless quantity.
Question:40
Answer:
Let us first write down the dimensions of-
(i) Let us consider that Mass m is directly proportional to
Comparing the powers on R.H.S. & L.H.S.
Thus,
Substituting the value of a in eq (i) & (ii)
By adding (iii) & (iv), we get,
Thus,
Substituting these values in eq. (iii)
Thus,
Now, let
Thus,
Equating the powers on both sides
Thus,
Now, since
& eq. (ii)
Subtracting eq (iv) from eq (iii) we get,
Thus,
And eq (iv) will be
Therefore,
Let us consider
Thus,
On adding eq (ii) & (iii) we get,
From eq (i) &(iv)
Thus,
Thus,
Question:41
Answer:
According to Kepler’s 3rd law of planetary motion,Question:42
In an experiment to estimate the size of a molecule of oleic acid, 1mL of oleic acid is dissolved in 19mL of alcohol. Then 1mL of this solution is diluted to 20mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film we can calculate the thickness of the film, which will give us the size of the oleic acid molecule.
Read the passage carefully and answer the following questions:
a) Why do we dissolve oleic acid in alcohol?
b) What is the role of lycopodium powder?
c) What would be the volume of oleic acid in each mL of solution prepared?
d) How will you calculate the volume of n drops of this solution of oleic?
e) What will be the volume of oleic acid in one drop of this solution?
Answer:
(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.Question:43
a) How many astronomical units (AU) make 1 parsec?
b) Consider the sun like a star at a distance of 2 parsec. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2) degree from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
c) Mars has approximately half of the earth’s diameter. When it is closer to the earth it is at about ½ AU from the earth. Calculate at what size it will disappear when seen through the same telescope.
Answer:
(a)1 A.U. long arc subtends the angle of 1s or 1 arc sec at distance of 1 parsec.
Thus, angle 1 sec =
Thus, 1 parsec =
Thus, 1 parsec=
(b)Angle of the sun’s diameter
Now,
If the sunlike star is at 2 parsec, the angle becomes half
Thus, angle
When it is seen with a telescope that has a magnification of 100, the angle formed will be
Hence, it can’t be observed by a telescope.
(c)
Given that
where
From answer 25(e)
we know that,
This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.
Question:44
Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E = mc2, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV where
a) Show that the energy equivalent of 1 u is 931.5 MeV.
b) A student writes the relation as
Answer:
(a) Given :
By the formula
(b) 1 u mass converted into total energy will be released by 931.5 MeV.
However, 1 amu = 931.5 MeV is dimensionally incorrect.
Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement
These solutions from Chapter 2 can help better understand the basic concepts and prepare for your academic exams. Questions of multiple choice type, short answer and long answer types are solved.
These Class 11 physics NCERT exemplar solutions chapter 1 can be downloaded from the website and the solution page directly in PDF format.
This chapter's essential topics are the International System of Units, Measurement of quantities, accuracy and precision, dimensional formulae, and dimensional analysis.
Significant figures show how accurate your measurement is. They help you write your answers correctly and avoid silly mistakes in calculations
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