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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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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.

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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


Chapter-Specific NCERT Solutions for Class 11 Physics

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.



CBSE Class 11 Physics Study Materials

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

1. What do the NCERT Solutions for Class 11 Physics Chapter 1 provide for the 2025-26 session?

The NCERT Solutions for Class 11 Physics Chapter 1, Units and Measurements, provide detailed, step-by-step answers for all the exercise questions in the textbook. They are fully updated for the CBSE 2025-26 syllabus and focus on explaining the correct method for solving problems related to dimensional analysis, error calculation, and significant figures.

2. How do you solve problems involving unit conversion using the method in NCERT Solutions?

The correct method, as shown in the NCERT solutions, involves using dimensional analysis. The steps are:

  • Identify the given physical quantity and its current units.
  • Write down the dimensional formula for the quantity (e.g., Force is [MLT⁻²]).
  • Determine the conversion factors between the old and new base units (e.g., 1 m = 100 cm).
  • Substitute the conversion factors into the dimensional formula and calculate the new numerical value.

3. What are the key rules for determining significant figures in a measurement as per the Class 11 Physics Chapter 1 solutions?

The NCERT solutions for Chapter 1 clarify the following rules for significant figures:

  • All non-zero digits are significant.
  • Zeros between two non-zero digits are significant (e.g., 6.032 has 4 significant figures).
  • Leading zeros (zeros before non-zero digits) are not significant (e.g., 0.007 has 1 significant figure).
  • Trailing zeros in a number with a decimal point are significant (e.g., 6.320 has 4 significant figures).

4. How are different types of errors in measurement classified in the NCERT exercises?

In Chapter 1, measurement errors are primarily classified into two types:

  • Systematic errors: These are consistent, directional errors due to instrument flaws (like a zero error) or incorrect experimental procedures. They can often be minimized.
  • Random errors: These are unpredictable, non-directional errors caused by unknown factors or limitations of the observer. They can be reduced by taking multiple readings and averaging them.

5. Why is using the SI system of units recommended in the NCERT Solutions for solving Physics problems?

The SI (Système International) system is a coherent and rationalised system of units. Using SI units, as emphasised in the NCERT Solutions, is crucial because it standardises calculations, prevents conversion mistakes, and ensures consistency across different topics in physics. This is the standard for CBSE exams and scientific communication worldwide, making it easier to derive and check equations.

6. What happens if a physical equation is not dimensionally homogeneous?

If an equation is not dimensionally homogeneous, it means the dimensions on the left-hand side do not match the dimensions on the right-hand side. According to the principle of homogeneity, such an equation is fundamentally incorrect. For example, you cannot add a quantity with the dimension of length [L] to one with the dimension of time [T]. Spotting this indicates an error in the formula's derivation or application.

7. What is the step-by-step approach to checking the dimensional correctness of a formula like Einstein's mass-energy relation?

To check an equation's dimensional correctness, as done in the NCERT solutions, follow these steps:

  • Step 1: Write down the dimensional formula for the quantity on the Left Hand Side (LHS).
  • Step 2: Write down the dimensional formula for all terms on the Right Hand Side (RHS).
  • Step 3: Apply the principle of homogeneity. The dimensions of the LHS must equal the dimensions of the RHS.
  • Step 4: For a relation like m = m₀ / √(1-v²/c²), you ensure that the term v²/c² is dimensionless, as it is subtracted from a dimensionless number (1).

8. How do the NCERT Solutions explain the difference between accuracy and precision in measurement?

The solutions clarify that:

  • Accuracy refers to how close a measured value is to the true or accepted value of the quantity.
  • Precision refers to how close multiple measurements of the same quantity are to each other, indicating the resolution or limit of the measuring instrument.
A measurement can be precise but inaccurate if there is a systematic error.

9. Why can't dimensional analysis determine the value of dimensionless constants in a formula?

Dimensional analysis is a powerful tool for checking consistency and deriving relationships between physical quantities, but it has limitations. It cannot determine dimensionless constants (like 'k', π, or 1/2) because these constants do not have physical dimensions. Their values must be found through experimentation or more advanced theoretical derivation, which is a key concept explained through problems in the NCERT solutions.

10. Why is it meaningless to call a quantity 'large' or 'small' without a standard for comparison, as explained in an NCERT problem?

Describing a physical quantity as 'large' or 'small' is relative and lacks scientific meaning without a reference point. For instance, the mass of Jupiter is 'large' compared to the mass of Earth, but 'small' compared to the mass of the Sun. The NCERT solutions stress that for a description to be meaningful, you must specify what it is being compared to, establishing a clear standard of comparison.

11. How does taking more measurements improve the reliability of an estimate, like measuring the diameter of a rod?

Taking a larger number of measurements helps to minimise the effect of random errors. Individual random errors can be positive or negative. When you take the arithmetic mean of many observations (e.g., 100 measurements instead of 5), these random positive and negative errors tend to cancel each other out. This makes the final average value a more reliable and accurate estimate of the true value, a key practical skill covered in the NCERT exercises.

12. How should a student use these NCERT Solutions to master numerical problems for their Class 11 exams?

For effective exam preparation, students should first attempt to solve the NCERT exercise problems independently. Afterwards, use these solutions to:

  • Verify the final answer.
  • Understand the correct step-by-step method if you are stuck.
  • Identify and learn from mistakes in applying formulas or performing unit conversions.
  • Pay close attention to how significant figures and error propagation are handled in the final answers, as marks are often allocated for these steps in CBSE exams.