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# NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Last updated date: 11th Sep 2024
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## NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements - FREE PDF Download

NCERT for Chapter 1 Units and Measurements Class 11 Solutions by Vedantu, forms the basis for all experimental and theoretical studies in physics. Understanding units and measurements is important as it ensures consistency and accuracy in scientific observations and calculations. This chapter emphasizes the importance of standardized units and accurate measurements in ensuring consistency and precision in scientific observations and calculations. It provides students with the essential tools and techniques to measure physical quantities correctly, which is critical for validating scientific theories and conducting experiments. By practising with Class 11 Physics NCERT Solutions, students can build confidence in their understanding and excel in their studies.

Table of Content
1. NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements - FREE PDF Download
2. Glance on Physics Chapter 1 Class 11 - Units and Measurements
3. Access NCERT Solutions for Class 11 Physics Chapter 1 – Units and Measurements
4. Dimensional Formulae of Physical Quantities
5. Overview of Deleted Syllabus for CBSE Class 11 Physics Units and Measurements
6. Other Study Material for CBSE Class 11 Physics Chapter 1
7. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

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## Glance on Physics Chapter 1 Class 11 - Units and Measurements

• The chapter introduces the concept of base units for fundamental physical quantities and derived units for quantities that are combinations of base quantities.

• Various techniques and instruments for measuring physical quantities accurately are discussed. This includes the use of vernier calipers, micrometers, and other precise tools.

• The chapter also covers different types of errors—systematic and random—and methods to minimize and account for these errors in measurements.

• The chapter discusses the applications of dimensional analysis in checking the correctness of equations, converting units from one system to another, and deriving relationships between physical quantities. Also, it includes significant figures and rules for determining the number of significant figures in a given measurement.

• This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 1 - Units And Measurements, which you can download as PDFs.

• There are 17 fully solved questions in class 11th physics chapter 1 Units And Measurements Exercise.

### System of Units:

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Physics Chapter 1 – Units and Measurements

1. Fill in the Blanks.

a) The Volume of a Cube of 1cm is Equal To ……………… ${{m}^{3}}$.

Ans:

We know that,

$1cm=\frac{1}{100}m$

Volume of a cube of side 1cm would be,

$V=1cm\times 1cm\times 1cm=1c{{m}^{3}}$

On converting it into unit of ${{m}^{3}}$, we get,

$1c{{m}^{3}}={{\left( \frac{1}{100}m \right)}^{3}}={{\left( {{10}^{-2}}m \right)}^{3}}$

$\therefore 1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$

Therefore, the volume of a cube of side 1cm is equal to  ${{10}^{-6}}{{m}^{3}}$.

b) The Surface Area of a Solid Cylinder of Radius 2.0cm and Height 10.0cm is Equal To ……………….${{\left( mm \right)}^{2}}$

Ans:

We know the formula for the total surface area of cylinder of radius r and height h to be,

$S=2\pi r\left( r+h \right)$

We are given:

$r=2cm=20mm$

$h=10cm=100mm$

On substituting the given values into the above expression, we get,

$S=2\pi \times 20\left( 20+100 \right)=15072m{{m}^{2}}=1.5\times {{10}^{4}}m{{m}^{{{2}^{{}}}}}$

Therefore, the surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to $1.5\times {{10}^{4}}{{\left( mm \right)}^{2}}$.

c) A Vehicle Moving With a Speed of $18km{{h}^{-1}}$Covers…………………. m in 1s.

Ans:

We know the following conversion:

$1km/h=\frac{5}{18}m/s$

$\Rightarrow 18km/h=18\times \frac{5}{18}=5m/s$

Now we have the relation:

$\text{Distance = speed }\times \text{ time }$

Substituting the given values, $\text{Distance = 5}\times \text{1 =5m}$

Therefore, a vehicle moving with a speed of $18km{{h}^{-1}}$covers 5m in 1s.

d) The Relative Density of Lead is 11.3. Its Density Is ………………. $gc{{m}^{-3}}$or………………… $kg{{m}^{-3}}$.

Ans:

We know that the relative density of substance could be given by,

$\text{Relative density = }\frac{density\text{ of substance}}{density\text{ of water}}$

$density\text{ of water = 1kg/}{{\text{m}}^{3}}$

$\text{density of lead = Relative density of lead }\times \text{ density of water = 11}\text{.3}\times \text{1= 11}\text{.3g/c}{{\text{m}}^{3}}$

But we know,

$1g={{10}^{-3}}kg$

$1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$

$\Rightarrow 1g/c{{m}^{3}}=\frac{{{10}^{-3}}}{{{10}^{-6}}}kg/{{m}^{3}}={{10}^{3}}kg/{{m}^{3}}$

$\therefore 11.3g/c{{m}^{3}}=11.3\times {{10}^{3}}kg/{{m}^{3}}$

Therefore, the relative density of lead is 11.3. Its density is $11.3gc{{m}^{-3}}$or$11.3\times {{10}^{3}}kg{{m}^{-3}}$.

2. Fill ups.

a) $1kg{{m}^{2}}{{s}^{-2}}=..................gc{{m}^{2}}{{s}^{-2}}$

Ans:

We know that:

$1kg={{10}^{3}}g$

$1{{m}^{2}}={{10}^{4}}c{{m}^{2}}$

$1kg{{m}^{2}}{{s}^{-2}}={{10}^{3}}g\times {{10}^{4}}c{{m}^{2}}\times 1{{s}^{-2}}={{10}^{7}}gc{{m}^{2}}{{s}^{-2}}$

Therefore, $1kg{{m}^{2}}{{s}^{-2}}={{10}^{7}}gc{{m}^{2}}{{s}^{-2}}$

b) $1m=.................ly$

Ans:

We know that light year is the total distance covered by light in one year.

$1ly=\text{Speed of light }\times \text{ one year}$

$\Rightarrow 1ly=\left( 3\times {{10}^{8}}m/s \right)\times \left( 365\times 24\times 60\times 60s \right)=9.46\times {{10}^{15}}m$

$\therefore 1m=\frac{1}{9.46\times {{10}^{15}}}=1.057\times {{10}^{-16}}ly$

Therefore, $1m=1.057\times {{10}^{-16}}ly$

c) $3.0m/{{s}^{2}}=.................km/h{{r}^{2}}$

Ans:

$3.0m/{{s}^{2}}=$ ………….$km/h{{r}^{2}}$

We have, $1m={{10}^{-3}}km$

$1hr=3600s$

$\Rightarrow 1{{s}^{2}}={{\left( \frac{1}{3600} \right)}^{2}}h{{r}^{2}}$

Then,

$3.0m/{{s}^{2}}=\frac{3\times {{10}^{-3}}}{{{\left( \frac{1}{3600}h \right)}^{2}}}km/h{{r}^{2}}$

$\therefore 3.0m/{{s}^{2}}=3.9\times {{10}^{4}}km/h{{r}^{2}}$

d)$6.67\times{{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=.............{{g}^{-1}}c{{m}^{3}}{{s}^{-2}}$

Ans:

We have,

$1N=1kgm{{s}^{-2}}$

$1kg={{10}^{-3}}g$

$1{{m}^{3}}={{10}^{6}}c{{m}^{3}}$

$\Rightarrow 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}=6.67\times {{10}^{-11}}\times \left( 1kgm{{s}^{-2}} \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)$

$=6.67\times {{10}^{-11}}\times \left( 1kg\times 1{{m}^{3}}\times 1{{s}^{-2}} \right)$

$=6.67\times {{10}^{-11}}\times \left( {{10}^{-3}}{{g}^{-1}} \right)\left( {{10}^{6}}c{{m}^{3}} \right)\left( 1{{s}^{-2}} \right)$

$\therefore 6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=6.67\times {{10}^{-8}}c{{m}^{3}}{{s}^{-2}}{{g}^{-1}}$

3. A Calorie is a Unit of Heat or Energy and Is Equivalent to 4.2 J Where $1J=1kg{{m}^{2}}{{s}^{-2}}$. Suppose We Employ a System of Units in Which the Unit of Mass Equals $\alpha \text{ kg}$, the Unit of Length Equals $\beta$ m, the Unit of Time is $\gamma \text{ s}$ . Show That a Calorie Has a Magnitude $4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}$ In Terms of the New Unit.

Ans:

We are given that,

$1calorie=4.2\left( 1kg \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)$

Let the new unit of mass $=\alpha \text{ kg}$.

So, one kilogram in terms of the new unit, $1\text{ kg}=\frac{1}{\alpha }={{\alpha }^{-1}}$.

One meter in terms of the new unit of length can be written as, $\text{1m}=\frac{1}{\beta }={{\beta }^{-1}}$  or $\text{1}{{\text{m}}^{2}}={{\beta }^{-2}}$.

And, one second in terms of the new unit of time,

$1\text{ s}=\frac{1}{\gamma }={{\gamma }^{-1}}$

$1\text{ }{{\text{s}}^{2}}={{\gamma }^{-2}}$

$1\text{ }{{\text{s}}^{-2}}={{\gamma }^{2}}$

$\therefore 1calorie=4.2\left( 1{{\alpha }^{-1}} \right)\left( 1{{\beta }^{-2}} \right)\left( 1{{\gamma }^{2}} \right)=4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}$

Therefore, the value equivalent to one calorie in the mentioned new unit system is $4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}$.

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:

Ans:

The given statement is true because a dimensionless quantity may be large or small, but there should be some standard reference to compare that.

For example, the coefficient of friction is dimensionless but we could say that the coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

a) Atoms Are Very Small Objects.

Ans: An atom is very small compared to a soccer ball.

b) A Jet Plane Moves With Great Speed.

Ans: A jet plane moves with a speed greater than that of a bicycle.

c)The Mass of Jupiter is Very Large.

Ans: Mass of Jupiter is very large compared to the mass of a cricket ball.

d) The Air Inside This Room Contains a Large Number of Molecules.

Ans: The air inside this room contains a large number of molecules as compared to that contained by a geometry box.

e) A Proton is Much More Massive than an Electron.

Ans: A proton is more massive than an electron.

f) The Speed of Sound is Much Smaller than the Speed of Light.

Ans:  Speed of sound is less than the speed of light.

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:

Distance between the Sun and the Earth:

$\text{x= Speed of light}\times \text{Time taken by light to cover the distance}$

It is given that in the new system of units, the speed of light $c=1\text{ }unit$.

Time taken, $t=8\text{ }min\text{ }20\text{ }s=500\text{ }s$

Thus, the distance between the Sun and the Earth in this system of units is given by$x'=c\times t=1\times 500=500\text{ }units$

6. Which of the Following is the Most Precise Device for Measuring Length?

Ans: A device which has the minimum least count is considered to be the most precise device to measure length.

a) A Vernier Caliper With 20 Divisions on the Sliding Scale.

Ans:

Least count of vernier calipers is given by

$LC=1\text{ }standard\text{ }division\left( SD \right)-1\text{ }vernier\text{ }division\left( VD \right)$

$\Rightarrow LC=1-\frac{19}{20}=\frac{1}{20}=0.05cm$

b) A Screw Gauge of Pitch 1 Mm and 100 Divisions on the Circular Scale.

Ans:

Least count of screw gauge $=\frac{\text{Pitch}}{\text{No of divisions}}$

$\Rightarrow LC=\frac{1 mm}{100}=\frac{0.1 cm}{100}$

$\Rightarrow LC=\frac{1}{1000}=0.001cm$

c) An Optical Instrument that Can Measure Length to Within a Wavelength of Light.

Ans:

Least count of an optical device $=\text{Wavelength of light}\sim \text{1}{{\text{0}}^{-5}}cm$

$\Rightarrow LC=0.00001cm$

Hence, it can be inferred that an optical instrument with the minimum least count among the given three options is the most suitable device to measure length.

7. A Student Measures the Thickness of a Human Hair Using 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. Estimate the Thickness of Hair.

Ans:

We are given that:

Magnification of the microscope $=100$

Average width of the hair in the field of view of the microscope $=3.5\text{ }mm$

$\therefore$ Actual thickness of the hair would be, $\frac{3.5}{100}=0.035\text{ }mm.$

8. Answer the Following:

a) YOu Are Given a Thread and a Meter Scale. How Will You Estimate the Diameter of the Thread?

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 that is wounded by the thread using a metre scale.

The diameter of the thread is given by the relation,

Diameter $=\frac{\text{Length of thread}}{\text{Number of turns}}$

B) 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?

Ans:

Increasing the number divisions of the circular scale will increase its accuracy to a negligible extent only.

C) The Mean Diameter of a Thin Brass Rod Is to Be Measured by Vernier Calipers. 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:

A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved will be reduced on increasing the number of measurements.

9. The Photograph of a House Occupies an Area of $1.75c{{m}^{2}}$ On a 35 Mm Slide. the Slide Is Projected Onto a Screen, and the Area of the House on the Screen is $1.55{{m}^{2}}$. What is the Linear Magnification of the Projector-Screen Arrangement?

Ans:

We are given,

The area of the house on the $35mm$ slide (area of the object) is given by,

${{A}_{O}}=1.75c{{m}^{2}}$.

The area of the image of the house that is formed on the screen is given by, ${{A}_{I}}=1.55{{m}^{2}}=1.55\times {{10}^{4}}c{{m}^{2}}$

We know that areal magnification is given by,

${{m}_{a}}=\frac{{{A}_{I}}}{{{A}_{O}}}$

Substituting the given values,

${{m}_{a}}=\frac{1.55\times {{10}^{4}}}{1.75}$

Now, we have the expression for Linear magnification as, ${{m}_{l}}=\sqrt{{{m}_{a}}}$

$\Rightarrow {{m}_{l}}=\sqrt{\frac{1.55}{1.75}\times {{10}^{4}}}$

$\therefore {{m}_{l}}=94.11$

Thus, we found the linear magnification in the given case to be, ${{m}_{l}}=94.11$.

10. State the Number of Significant Figures in the Following:

a) $0.007{{m}^{2}}$

Ans: We know that when the given number is less than one, all zeroes on the right of the decimal point are insignificant and hence for the given value, only 7 is the significant figure. So, the number of significant figures in this case is 1.

b) $2.64\times {{10}^{24}}kg$

Ans: We know that the power of 10 is considered insignificant and hence, 2, 6 and 4 are the significant figures in the given case. So, the number of significant figures here is 3.

c) $0.2370gc{{m}^{-3}}$

Ans: For decimal numbers, the trailing zeroes are taken significantly. 2, 3, 7 and 0 are the significant figures. So, the number of significant figures here is 4.

d) $6.320J$

Ans: All figures present in the given case are significant. So, the number of significant figures here is 4.

e) $6.032N{{m}^{-2}}$

Ans: Since all the zeros between two non-zero digits are significant, the number of significant figures here is 4.

f) $0.0006032{{m}^{2}}$

Ans: For a decimal number less than 1, all the zeroes lying to the left of a non-zero number are insignificant. Hence, the number of significant digits here is 4.

11. The Length, Breadth and Thickness of a Rectangular Sheet of Metal Are 4.234m, 1.005m and 2.01cm Respectively. Give the Area and Volume of the Sheet to Correct Significant Figure.

Ans:

We are given:

Length of sheet, $l=4.234m$; number of significant figures: 4

Breadth of sheet, $b=1.005m$; number of significant figures: 4

Thickness of sheet, $h=2.01cm=0.0201m$; number of significant figures: 3

So, we found that area and volume should have the least significant figure among the given dimensions, i.e., 3.

Surface area, $A=2\left( l\times b+b\times h+h\times l \right)$

Substituting the given values,

$\Rightarrow A=2\left( 4.234\times 1.005+1.005\times 0.0201+0.0201\times 4.234 \right)=2\left( 4.25517+0.02620+0.08510 \right)$

$\therefore A=8.72{{m}^{2}}$

Volume, $V=l\times b\times h$

Substituting the given values,

$\Rightarrow V=4.234\times 1.005\times 0.0201$

$\therefore V=0.0855{{m}^{3}}$

Therefore, we found the area and volume with 3 significant figures to be $A=8.72{{m}^{2}}$

and $V=0.0855{{m}^{3}}$respectively.

12. The Mass of a Box Measured by a Grocer's Balance is 2.300 Kg. Two Gold Pieces of Masses 20.15 G and 20.17 G Are Added to the Box. What Is:

a) The Total Mass of the Box?

Ans:

We are given:

Mass of grocer’s box $=2.300kg$

Mass of gold piece $I=20.15g=0.02015kg$

Mass of gold piece $II=20.17g=0.02017kg$

Total mass of the box $=2.3+0.02015+0.02017=2.34032kg$

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 $2.3kg$.

b) The Difference in the Masses of the Pieces to Correct Significant Figures?

Ans:

Difference in masses $=20.17-20.15=0.02g$

While subtracting, the final result should retain as many decimal places as there are in the number with the least decimal places.

13. A Famous Relation in Physics Relates ‘Moving Mass’ M to the ‘Rest Mass’ ${{m}_{0}}$of a Particle in Terms of Its Speed v and Speed of Light c. (This Relation First Arise as a Consequence of Special Relativity Due to Albert Einstein). A Boy Recalls the Relation Almost Correctly but Forgets Where to Put the Constant c. He Writes:

$m=\frac{{{m}_{0}}}{{{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}}$

Ans:

We are given the following relation:

$m=\frac{{{m}_{0}}}{{{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}}$

Dimension of m, ${{M}^{1}}{{L}^{0}}{{T}^{0}}$

Dimension of ${{m}_{0}}$, ${{M}^{1}}{{L}^{0}}{{T}^{0}}$

Dimension of v, ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$

Dimension of ${{v}^{2}}$, ${{M}^{0}}{{L}^{2}}{{T}^{-2}}$

Dimension of c, ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$

For the formula to be dimensionally correct, the dimensions on the LHS should be the same as those on the RHS. In order to satisfy this condition, ${{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}$should be dimensionless and for that we require ${{v}^{2}}$ be divided by ${{c}^{2}}$. So, the dimensionally correct version of the above relation would be,

$m=\frac{{{m}_{0}}}{{{\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)}^{\frac{1}{2}}}}$

14. The Unit of Length Convenient on the Atomic Scale is Known as an Angstrom and is Denoted By $\overset{{}^\circ }{\mathop{A}}\,:1\overset{{}^\circ }{\mathop{A}}\,={{10}^{-10}}m$. The Size of a Hydrogen Atom Is About 0.5a. What is the Total Atomic Volume In ${{m}^{3}}$of a Mole of Hydrogen Atoms?

Ans:

Radius of hydrogen atom is given to be,

$r=0.5\overset{{}^\circ }{\mathop{A}}\,=0.5\times {{10}^{-10}}m$

The expression for the volume is,

$V=\frac{4}{3}\pi {{r}^{3}}$

Now on substituting the given values,

$V=\frac{4}{3}\pi {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

But we know that 1 mole of hydrogen would contain Avogadro number of hydrogen atoms, so volume of 1 mole of hydrogen atoms would be,

$V'={{N}_{A}}V=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$

Therefore, we found the required volume to be $3.16\times {{10}^{-7}}{{m}^{3}}$.

15.One Mole of an Ideal Gas at Standard Temperature and Pressure Occupies $22.4L$ (molar Volume). What is the Ratio of Molar Volume to the Atomic Volume of a Mole of Hydrogen? (Take the Size of a Hydrogen Molecule to Be About $1\overset{{}^\circ }{\mathop{\text{A}}}\,$). Why is This Ratio So Large?

Ans:

Radius of hydrogen atom, $r=0.5\overset{{}^\circ }{\mathop{\text{A}}}\,=0.5\times {{10}^{-10}}m$

Volume of hydrogen atom, $V=\frac{4}{3}\pi {{r}^{3}}$

$\Rightarrow V=\frac{4}{3}\times \frac{22}{7}\times {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

Now, 1 mole of hydrogen contains $6.023\times {{10}^{23}}$ hydrogen atoms.

Volume of 1 mole of hydrogen atoms,${{V}_{a}}=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$.

Molar volume of 1 mole of hydrogen atoms at STP, ${{V}_{m}}=22.4L=22.4\times {{10}^{-3}}{{m}^{3}}$

So, the required ratio would be,

$\frac{{{V}_{m}}}{{{V}_{a}}}=\frac{22.4\times {{10}^{-3}}}{3.16\times {{10}^{-7}}}=7.08\times {{10}^{4}}$

Hence, we found that the molar volume is $7.08\times {{10}^{4}}$ times higher than the atomic volume.

For this reason, the interatomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

16. Explain This Common Observation Clearly: If You Look Out of the Window of a Fast-Moving Train, the Nearby Trees, Houses Etc. Seems to Move Rapidly in a Direction Opposite to the Train's Motion, but the Distant Objects (hill Tops, the Moon, the Stars Etc.) Seems 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.

17. The Sun Is a Hot Plasma (ionized Matter) With Its Inner Core at a Temperature Exceeding ${{10}^{7}}K$ and Its Outer Surface at a Temperature of About 6000k. 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$=2.0\times {{10}^{30}}kg$, Radius of the Sun$=7.0\times {{10}^{8}}m$.

Ans:

We are given the following:

Mass of the sun, $M=2.0\times {{10}^{30}}kg$

Radius of the sun, $R=7.0\times {{10}^{8}}m$

Now we find the volume of the sun to be,

$V=\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\pi {{\left( 7.0\times {{10}^{8}} \right)}^{3}}=1437.3\times {{10}^{24}}{{m}^{3}}$

Density of the sun is found to be,

$\rho =\frac{M}{V}=\frac{2.0\times {{10}^{30}}}{1437.3\times {{10}^{24}}}$

$\therefore \rho \sim 1.4\times {{10}^{3}}kg/{{m}^{3}}$

So, we found the density of the sun to lie in the density range of solids and liquids.

Clearly, the high intensity is attributed to the intense gravitational attraction of the inner layers on the outer layer of the sun.

## Dimensional Formulae of Physical Quantities

 S. No Physical quantity Relationship with other physical quantities Dimensions Dimensional formula 1. Area Length × breadth [L2] [M0L2 T0] 2. Volume Length × breadth × height [L3] [M0L3 T0] 3. Mass density Mass/volume [M]/[L3] or [ML−2] [ML−3T0] 4. Frequency 1/time period 1/[T] [M0L0T−1] 5. Velocity, speed Displacement/time [L]/[T] [M0LT−1] 6. Acceleration Velocity/time [LT−1]/[T] [M0LT−1] 7. Force Mass × acceleration [M][LT−1] [MLT−1] 8. Impulse Force × time [MLT−1][T] [MLT−1] 9. Work, Energy Force × distance [MLT1][L] [ML1T1] 10. Power Work/time [ML1T2]/[T] [ML1T1] 11. Momentum Mass × velocity [M][LT−1] [MLT−1] 12. Pressure, stress Force/area [MLT−1]/[L2] [ML−1T1] 13. Strain $\dfrac{change\;in\;dimension}{original\;dimension}$ [L]/[L][L1]/ [L1] [M0L0T0] 14. Modulus of elasticity Stress/strain $\left [ \dfrac{ML^{-1}T^{-2}}{M^{0}L^{0}T^{0}} \right ]$ [ML−1T−2] 15. Surface tension Force/length [MLT−2]/[L] [ML0T−2] 16. Surface energy Energy/area [ML2T2]/[L2] [ML0T−2] 17. Velocity gradient Velocity/distance [LT1]/[L] [M0L0T−1] 18. Pressure gradient Pressure/distance [M1L1T−2]/[L2] [M1L−2T−2] 19. Pressure energy Pressure × volume [ML−1T−2][L3] [ML2T−2] 20. Coefficient of viscosity Force/area × velocity gradient $\dfrac{\left [ MLT^{-2} \right ]}{\left [ L^{2} \right ]\left [ LT^{-1}L \right ]}$ [ML−1T−1] 21. Angle, Angular displacement Arc/radius [L]/[L] [M0L0T0] 22. Trigonometric ratio (sin $\theta$ , cos $\theta$ , tan $\theta$ , etc Length/length [L]/[L] [M0L0T0] 23. Angular velocity Angle/time [L0]/[T] [M0L0T−1]

## Overview of Deleted Syllabus for CBSE Class 11 Physics Units and Measurements

 Chapter Dropped Topics Units and Measurements 2.3 Measurement of Length 2.4 Measurement of Mass 2.5 Measurement of Time 2.6 Accuracy, Precision of Instruments and Errors in Measurement Exercises 2.13, 2.14 2.19–2.22 2.24–2.33

## Conclusion

NCERT Class 11 Physics Chapter 1 Solutions on Units And Measurements provided by Vedantu serves as the foundation for all scientific study and experimentation. This chapter introduces students to the fundamental concepts of measuring physical quantities, ensuring consistency and accuracy in their scientific observations and calculations. Understanding units and measurements is critical for conducting experiments, interpreting results, and communicating scientific findings effectively. These concepts are essential for mastering the topic and are often tested in exams. From previous year's question papers, typically around 2-3 questions are asked from this chapter. These questions test students' theoretical concepts as well as their problem-solving skills.

## Other Study Material for CBSE Class 11 Physics Chapter 1

 S. No Important Links for Chapter 1 Units and Measurements 1 Class 11 Units and Measurements Important Questions 2 Class 11 Units and Measurements Revision Notes 3 Class 11 Units and Measurements NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

1. What are the topics that are covered in class 11 physics chapter 2?

The concepts that are covered in chapter 2 are:

Introduction to units and measurement.

The international system of the units.

SI unit

SI Base unit

SI Derived unit

Advantages of learning SI units and CGS units

Significant figures.

Applications of significant figures

Exact Number

Dimensions of physical quantities.

Dimensional formulae and dimensional equations.

Dimensional analysis and its applications.

Unit conversion and dimensional analysis

Using dimensional analysis to check the correctness of physical equation.

Homogeneity principle of dimensional analysis.

Applications of dimensional analysis.

Limitations of dimensional analysis.

2. What is dimensional analysis?

We quantify the size and shape of things using Dimensional Analysis. It helps us study the nature of objects mathematically. It involves lengths and angles as well as geometrical properties such as flatness and straightness. The basic concept of dimension is that we can add and subtract only those quantities that have the same dimensions. Similarly, two physical quantities are equal if they have the same dimensions.

The main benefits of the dimensional analysis of a problem have reduced the number of variables in the issue by combining dimensional variables to one form of non-dimensional parameters. By far the easiest and most useful method in the analysis of any fluid problem is that of direct mathematical solution.

3. What are gross errors?

This section basically takes into account human oversight and other mistakes while reading, recording and readings. The most common errors, the human error in the measurement fall which is under this category of errors in measurement. For instance, a person taking the reading from the meter of the instrument and he may read 33 as 38. Gross errors could be avoided by using two suitable measures and they are mentioned below:

• A proper care should always be taken while reading and recording the data. Also, calculation of error should be done accurately.

• By increasing the total number of experimenters we can reduce the gross errors. If each experimenter takes different reading at various points, then by taking an average of more readings we can reduce the gross errors.

4. How Vedantu will help me in exam preparation?

Our NCERT solutions are prepared by our maths experts with various real-life examples. These examples will make you understand the concept quickly and memorise them for a longer time. Solutions provided to the questions are 100% accurate in the exercises which are crisp and concise to the point.

Our solutions are the best study guides, which help you in smart learning and efficient answering of questions. These solutions will also help you in improving a strong conceptual base with all the important concepts in a very easy and understandable language. You will also enjoy learning from our solutions which are really fun and interactive.

5. Are NCERT Solutions important for Class 11 Physics Chapter 2?

Referring to NCERT Solutions is as important as referring to the questions while preparing for your Class 11 Physics Chapter 2. Only reading the questions that are provided in the NCERT is not going to be helpful if students are not able to answer them correctly. NCERT Solutions for Class 11 Physics Chapter 2 available on Vedantu provide students the correct step-by-step solutions for all NCERT questions so that students do not lose any marks in their exams.

6. Do I need to practice all the questions provided in Class 11 Physics Chapter 2 NCERT Solutions?

Questions provided in Class 11 Physics NCERT Solutions for Chapter 2 are to be considered crucial when preparing for your Class 11 Physics exams. The exam question papers always include questions that have been provided in the Physics NCERT for Class 11. Practicing all the questions will only help you increase your understanding of all the concepts taught in Chapter 2 and also the possibility of scoring well in the questions framed from the chapter.

7. How can I understand the Class 11 Physics Chapter 2?

Chapter 2 in Class 11 Physics NCERT is called “Units and Measurement”. This chapter talks in detail about various units used for determining the measurement of different physical quantities, instruments used for such measurements and their accuracy, etc. Students can easily understand this chapter by indulging in regular reading of the chapter and solving the questions provided in the NCERT. For more help, students can also refer to NCERT Solutions for Class 11 Physics Chapter 2 on the official website of Vedantu or download the Vedantu app where these resources are available at free of cost.

8. What is the marks distribution for Class 11 Physics Chapter 2?

Chapter 2 - Units and Measurement in Class 11 Physics is a part of Unit - I along with Chapter 1 - Physical World. According to the marks distribution provided by CBSE for Class 11 Physics, Unit - I consisting of both the chapters, carries a total of 23 marks. Hence, preparation from both chapters should be given equal priority to avoid losing any marks in questions framed from them in the exam.

9. What are the important topics covered in NCERT Solutions for Class 11 Physics Chapter 2?

NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurements includes various topics like the International System of Units, Measurement of length, mass, and time, Application of Significant Figures, etc. Among these, the most important topics from the chapter include SI Units, Absolute Errors, Dimensional Analysis, and Significant Figures. Short-answer and numerical-based questions can also be asked from the topic of evaluating errors during the measurement of quantities.

10. What is the concept of units and measurements in class 11 exercise solutions?

In units and measurements class 11 exercise solutions in Physics Chapter 1 involves quantifying physical quantities in a standardized manner. A unit is a definite magnitude of a quantity, defined and adopted by convention or law, used as a standard for measuring the same kind of quantity. Measurement is the process of determining the size, length, or amount of something, typically using standard units.

11. Which topics are important in units and measurements class 11 NCERT solutions?

Key topics in Physics Class 11 Chapter 1 Exercise Solutions include:

• Physical Quantities and Units: Base and derived quantities, SI units, and other systems of units.

• Measurement Techniques: Direct and indirect measurement methods.

• Significant Figures: Rules for determining significant figures in measurements.

• Errors in Measurement: Types of errors (systematic and random) and methods to minimize them.

• Dimensional Analysis: Checking the dimensional consistency of equations and converting units.

12. What is the direct method of measurement of length class 11 physics chapter 1 NCERT solutions?

In units and measurements class 11 exercise solutions, the direct method of measurement of length involves using instruments that directly provide the measurement of length, such as rulers, measuring tapes, vernier calipers, and micrometer screw gauges. These tools are used to measure the length of objects by comparing them directly against a standard unit of length.

13. What is the full form of SI unit?

According to physics class 11 chapter 1, the full form of SI unit is "Système International 'Unités", which translates to the International System of Units. It is the modern form of the metric system and the most widely used system of measurement.

14. What are the three main units of measurement in Class 11 Physics Ch 1 NCERT Solutions?

The three main base units and measurements class 11 ncert solutions in the SI system are:

• Meter (m) for length

• Kilogram (kg) for mass

• Second (s) for time

15. What is a measurement of time in units and measurements class 11 solutions?

The measurement of time is the process of quantifying the duration or interval between two events. The standard unit of time in the SI system is the second (s). Time can be measured using various devices such as clocks, stopwatches, and atomic clocks.