assignment 2.1 physics class 11

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Class 11 Physics (India)

Unit 1: physical world, unit 2: units and measurement, unit 3: basic math concepts for physics (prerequisite), unit 4: differentiation for physics (prerequisite), unit 5: integration for physics (prerequisite), unit 6: motion in a straight line, unit 7: vectors (prerequisite), unit 8: motion in a plane, unit 9: laws of motion, unit 10: work, energy and power, unit 11: system of particles and rotational motion, unit 12: gravitation, unit 13: mechanical properties of solids, unit 14: mechanical properties of fluid, unit 15: thermal properties of matter, unit 16: thermodynamics, unit 17: kinetic theory, unit 18: oscillations, unit 19: waves.

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• NCERT Solutions for Class 11 Physics Chapter 2 - Units And Measurement
• NCERT Solutions

NCERT Class 11 Physics Chapter 2: Complete Resource for Units and Measurement

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List of Important Topics Covered Under Class 11 Physics Chapter 2 Units and Measurement 

Revision notes for class 11 physics chapter 2 units and measurement .

While preparing for the board examination or competitive examination, it can be difficult for you to go through the entire chapter. To simplify and save your precious time, Vedantu is providing revision notes of Class 11 Physics Chapter 2 Units and Measurement. Make use of it and improve your score. 

Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.

Units and Measuements & Basic Mathematics Chapter at a Glance - Class 11 NCERT Solutions

Any quantity which can be measured is called a physical quantity.

Fundamental Unit:

Supplementary Units:

System of Units:

Dimensions of a physical quantity are the powers to which the fundamental units must be raised in order to get the unit of derived quantity.

Dimensional analysis is a tool to find or check relations among physical quantities by using their dimensions.

By using dimensional analysis, we can

1. Convert a physical quantity from one system of unit to another.

2. Check the dimensional consistency of equations

3. Deduce relation among physical quantities.

Limitations of Dimensional Analysis

In some cases, the constant of proportionality also possesses dimensions. In such cases, we cannot use this system.

If one side of the equation contains addition or subtraction of physical quantities, we cannot use this method to derive the expression.

Systematic Errors

Systematic error is a consistent, repeatable error associated with faulty equipment or a flawed experiment design. These errors are usually caused by measuring instruments that are incorrectly calibrated.

These errors cause readings to be shifted one way (or the other) from the true reading.

Causes of Systematic Errors

1. Zero Error

There is not any weight, and the weighing machines are not showing zero.

2. Faulty Instrument

If a ruler is wrongly calibrated, or if it expands, then all the readings will be too low (or all too high).

3. Personal Error

If someone have a habit of taking measurements always from above the reading, then due to parallax you will get a systematic error and all the readings will be too high.

Now, let’s learn about some common terms used during, measurements and error analysis.

Accuracy and Precision

Accuracy is an indication of how close a measurement is to the accepted value.

An accurate experiment has a low systematic error.

Precision is an indication of the agreement among a number of measurements.  

A precise experiment has a low random error

Quadratic Equation

A quadratic equation is an equation of second degree, meaning it contains at least one term that is squared.

The standard form of quadratic equation is $ax^{2}+bx+c=0$ where $a\neq 0$

Discriminant of a Quadratic Equation:

Discriminant of a quadratic $ax^{2}+bx+c=0$ equation is represented by D. 

$D=b^{2}+4ac$

The roots are given by $x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Binomial Expansion

A binomial is a polynomial with two terms.

There are a few similarities between the sine and cosine graphs. They are:

Both have the same curve which is shifted along the x-axis.

Both have an amplitude of 1.

Have a period of 360 o or 2𝜋 radians.

Scalar and Vector

Representation and Properties of Vectors

Types of Vectors

Negative Vector:

A negative vector is a vector that has the opposite direction to the reference positive direction.

(i) Zero Vector

(ii) Unit Vector

(iii) Position Vector

(iv) Co-initial Vector

(v) Like and Unlike Vectors

(vi) Coplanar Vector

(vii) Collinear Vector

(viii) Displacement Vector

A unit vector is a vector that has a magnitude of 1.

Any vector can become a unit vector on dividing it by the vector's magnitude.

A vector representing the straight-line distance and the direction of any point or object with respect to the origin, is called position vector.

Polygon Law: It states that if number of vectors acting on a particle at a time are represented in magnitude and direction by the various sides of an open polygon taken in same order, their resultant vector R is represented in magnitude and direction by the closing side of polygon taken in opposite order.

Addition of Vectors Components:

To get the magnitude R of the resultant, use the Pythagorean theorem:

$R=\sqrt{R_{x}^{2}+R_{y}^{2}}$

To get the direction of the resultant.

$\theta =tan^{-1}\left ( \dfrac{r_{y}}{r_{x}} \right )$

Addition of vectors: Law of Parallelogram of vector addition. Thus, the magnitude of $\vec{a}+\vec{b}$ is $\sqrt{a^{2}+b^{2}+2ab\;cos\theta }$

Its angle with $\vec{a}$ is 𝛼 where $tan\alpha =\dfrac{DE}{AE}=\dfrac{b\;sin\;\theta }{a+b\;cos\;\theta }$

Vector Subtraction:

$\vec{a}=a\hat{}_{x}i+a\hat{}_{y}j$

$\vec{b}=b\hat{}_{x}i+b\hat{}_{y}j$

$\vec{a}-\vec{b}=\vec{a}+\left ( -\vec{b} \right )=a\hat{_{x}}i+a\hat{_{y}}j+\left ( -b\hat{_{x}i-b\hat{_{y}j}} \right )$

$=(a_{x}-b_{x})\hat{}i+(a_{y}-b_{y})\hat{}j$

Scalar Product or Dot Product

$\vec{a}\cdot \vec{b}=\left | \vec{a} \right |\left | \vec{b} \right |cos\theta$

$0\leq \theta \leq \pi$

Dot product gives us a scalar quantity.

Angle between vectors, $cos\;\theta =\dfrac{\vec{a}\cdot\vec{b}}{\left | \vec{a} \right|\left |  \vec{b}\right |}$

Dot product is commutative.

$\vec{b}\cdot \vec{a}=\vec{a}\cdot \vec{b}$

Dot product is distributive over addition or subtraction.

$\vec{a}\cdot \left ( \vec{b} \pm \vec{c}\right )=\vec{a}\cdot \vec{b\pm }\vec{a}\cdot \vec{c}$

Cross Product:

$\vec{a}\times \vec{b}=\left | \vec{a} \right |\left | \vec{b} \right |sin\;\theta \hat{n}$

$0^{\circ}\leq \theta \leq 180^{\circ}$

$\hat{n}$ is the unit vector in direction normal to the a and b

Properties of Cross Product:

Vector product is anti - commutative i.e., $\vec{A}\times \vec{B}=-\vec{B\times \vec{A}}$

Vector product is distributive over addition i.e., $\vec{A}\times \left ( \vec{B}+\vec{C} \right )=\vec{A}\times \vec{B}+\vec{A}\times \vec{C}$

Vector of two parallel or antiparallel vectors is a null vector. Thus $\vec{A}\times \vec{B}=ABsin\left ( 0^{\circ}\;or\;180^{\circ} \right )\hat{n}=\vec{0}$

Vector product of a vector with itself is a null vector.

$\vec{A}\times \vec{A}sin0\hat{^{\circ}}n=\vec{0}$

Dimensional Formulae of Physical Quantities

Mastering Class 11 Physics Chapter 2: Units and Measurement - MCQs, Question and Answers, and Tips for Success

1. Fill in the Blanks.

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

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 ……………….${{\left( mm \right)}^{2}}$

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. 

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………………… $kg{{m}^{-3}}$. 

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

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

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

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

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

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

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

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

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

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

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

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

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.

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:

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.

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.

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.

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.

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?

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?

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?

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?

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. 

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

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

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

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

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

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

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?

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?       

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 Physical Quantity P Is Related to Four Observables $a,b,c$ and $d$ as Follows:

$P=\frac{{{a}^{3}}{{b}^{2}}}{\left( \sqrt{cd} \right)}$ 

The Percentage Errors of Measurement in $a$, $b$, $c$ and $d$ are $1%$, $3%$, $4%$ and $2%$ Respectively. What Is the Percentage Error in the Quantity P? If the Value of P Calculated Using the Above Relation Turns Out to Be 3.763, to What Value Should You Round Off the Result?

We are given the relation, 

 $P=\frac{{{a}^{3}}{{b}^{2}}}{\left( \sqrt{cd} \right)}$

and the percentage of error in $a$, $b$, $c$ and $d$ are $1%$, $3%$, $4%$ and $2%$ respectively.

The error could be calculated using the following expression, 

$\frac{\Delta P}{P}=\frac{3\Delta a}{a}+\frac{2\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{\Delta d}{d}$

$\left( \frac{\Delta P}{P}\times 100 \right)%=\left( 3\times \frac{\Delta a}{a}\times 100+2\times \frac{\Delta b}{b}\times 100+\frac{1}{2}\times \frac{\Delta c}{c}\times 100+\frac{\Delta d}{d}\times 100 \right)%$

$=3\times 1+2\times 3+\frac{1}{2}\times 4+2$

$=3+6+2+2=13%$ 

Therefore the percentage error in $P=13%$. 

Value of P is given as $3.763$.

Thus, by rounding off the given value to the first decimal place, we get $P=3.8$.

14. A Book With Many Printing Errors Contains Four Different Formulas for the Displacement $y$ Of a Particle Undergoing a Certain Periodic Motion:

(\[a=\] Maximum Displacement of the Particle, $v=$  Speed of the Particle. $T=$  Time Period of Motion). Rule Out the Wrong Formulas on Dimensional Grounds.

a) $y=a\sin \left( \frac{2\pi t}{T} \right)$ 

It is correct.

Given: $y=a\sin \left( \frac{2\pi t}{T} \right)$

Dimensions of $y={{M}^{0}}{{L}^{1}}{{T}^{0}}$ 

Dimensions of $a={{M}^{0}}{{L}^{1}}{{T}^{0}}$ 

Dimensions of $\sin \left( \frac{2\pi t}{T} \right)={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Since the dimension on the RHS is equal to that on the LHS, the given formula is dimensionally correct.

b) $y=a\sin vt$ 

It is incorrect.

Given: $y=a\sin vt$

Dimensions of $vt={{M}^{0}}{{L}^{1}}{{T}^{-1}}\times {{M}^{0}}{{L}^{0}}{{T}^{1}}={{M}^{0}}{{L}^{1}}{{T}^{0}}$

Since the dimension on the RHS is not equal to that on the LHS, the given formula is dimensionally incorrect. 

c) $y=\left( \frac{a}{T} \right)\sin \frac{t}{a}$ 

Given: $y=\left( \frac{a}{T} \right)\sin \frac{t}{a}$  

Dimensions of $\frac{a}{T}={{M}^{0}}{{L}^{1}}{{T}^{-1}}$ 

Dimensions of $\frac{t}{a}={{M}^{0}}{{L}^{1}}{{T}^{1}}$

d) $y=\left( a\sqrt{2} \right)\left( \sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T} \right)$ 

Given: $y=\left( a\sqrt{2} \right)\left( \sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T} \right)$  

Dimensions of \[a={{M}^{0}}{{L}^{1}}{{T}^{0}}\] 

Dimensions of $\frac{t}{T}={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Since the dimension on the RHS is equal to that of the LHS, the given formula is dimensionally correct.

15. 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:

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

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,

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

16. 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?

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 $3.16\times {{10}^{-7}}{{m}^{3}}$.

17.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?

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. 

18. 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.

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 $\approx 3\times {{10}^{11}}m$. 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?

We know the diameter of Earth’s orbit $=3\times {{10}^{11}}m$.

So, its radius would be, $r=1.5\times {{10}^{11}}m$.

The distance parallax angle, $1''=4.847\times {{10}^{-6}}rad$.

Let D be the distance of the star and we could define parsec as the distance at which the average radius of the earth’s orbit subtends an angle of 1”. 

So, we have, 

$\theta =\frac{r}{D}$

Substituting the given values, we get, 

$D=\frac{r}{\theta }=\frac{1.5\times {{10}^{11}}}{4.847\times {{10}^{-6}}}=0.309\times {{10}^{-6}}$

$\Rightarrow D\approx 3.09\times {{10}^{16}}m$

Therefore, we found the value of 1 parsec to be, 

$1\text{ }par\text{sec }\approx \text{ 3}\text{.09}\times {{10}^{16}}m$

20. The Nearest Star to Our Solar System Is 4.29 Light Years Away. How Much is This Distance in Terms of Parsecs? How Much Parallax Would This Star (named Alpha Centauri) Show When Viewed from Two Locations of the Earth Six Months Apart in Its Orbit Around the Sun?         

Distance of the star from the solar system $=4.29ly$ 

1 light year is the distance travelled by light in one year.

1 light year $=\text{speed of light }\times \text{ 1 year}$ 

$1ly=3\times {{10}^{8}}\times 365\times 24\times 60\times 60=94608\times {{10}^{11}}m$ 

$\Rightarrow 4.29ly=405868.32\times {{10}^{11}}m$ 

But we have, 

$1\text{ parsec}=3.08\times {{10}^{16}}m$ 

$\therefore 4.29ly=\frac{405868.32\times {{10}^{11}}}{3.08\times {{10}^{6}}}=1.32\text{ parsec}$ 

We have another relation, 

$\theta =\frac{d}{D}$ 

Diameter of Earth’s orbit, $d=3\times {{10}^{11}}m$ 

Distance of star from the Earth, \[D=405868\times {{10}^{11}}m\] 

Substituting these values,

$\theta =\frac{3\times {{10}^{11}}}{405868.32\times {{10}^{11}}}=7.39\times {{10}^{-6}}rad$ 

But $1\sec =4.85\times {{10}^{-6}}rad$ 

$\therefore 7.39\times {{10}^{-6}}rad=\frac{7.39\times {{10}^{-6}}}{4.85\times {{10}^{-6}}}=1.52''$

Therefore, 4.2ly is equal to 1.32 parsec, and a star (named Alpha Centauri) would show at 1.52’’ when it is viewed from two locations of the Earth six months apart in its orbit around the Sun.

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 \[\mathbf{II}\]. 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.

It is indeed very true that precise measurements of physical quantities are essential for the development of science. Some examples are:

a) Ultrashort laser pulses (time interval $\sim {{10}^{-15}}s$) are used to measure time intervals in several physical and chemical processes.

b) X-ray spectroscopy is used to determine the interatomic separation or inter-planar spacing. 

c) The development of a mass spectrometer makes it possible to measure the mass of atoms precisely.

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): 

A) The Total Mass of Rain-Bearing Clouds Over India During the Monsoon. 

For estimating the total mass of rain-bearing clouds over India during the Monsoon:

During monsoons, a meteorologist records about 215 cm of rainfall in India i.e., the height of water column, \[h=215\text{ }cm=2.15\text{ }m\] 

We have the following information, 

Area of country, \[A=3.3\times {{10}^{12}}{{m}^{2}}\] 

Hence, volume of rainwater, \[V=A\times h=7.09\times {{10}^{12}}{{m}^{3}}\]

Density of water, \[\rho =1\times {{10}^{3}}kg\text{ }{{m}^{-3}}\] 

We could find the mass from the given value of density and volume as,

\[M=\rho \times V=7.09\times {{10}^{15}}kg\] 

Hence, the total mass of rain-bearing clouds over India is approximately found to be  \[7.09\times {{10}^{15}}kg\].

b) The Mass of an Elephant.

For estimating the mass of an elephant: 

Consider a ship floating in the sea whose base area is known. 

Measure its depth in sea (say ${{d}_{1}}$).

Volume of water displaced by the ship would be, ${{V}_{b}}=A{{d}_{1}}$ 

Now one could move an elephant on the ship and then measure the depth of the ship $\left( {{d}_{2}} \right)$.

Let the volume of water displaced by the ship with the elephant on board be ${{V}_{be}}=A{{d}_{2}}$ 

Then the volume of water displaced by the elephant $=A{{d}_{2}}-A{{d}_{1}}$ 

If the density of water $=D$ 

Mass of elephant would be, $M=AD\left( {{d}_{2}}-{{d}_{1}} \right)$ 

c) The Wind Speed During a Storm.

Estimation of wind speed during a storm: 

Wind speed during a storm can be measured by using an anemometer. 

As the wind blows, it rotates and the number of rotations in one second gives the value of the wind speed.

d) The Number of Strands of Hair on Your Head.

Estimation of the number of strands of hair on your head:

Let the area of the head surface carrying hair be \[A\]. 

Let the radius of a hair be \[r\] and it can be determined with the help of a screw gauge.

$\therefore $ Area of one hair $=\pi {{r}^{2}}$

Number of strands of hair $\approx \frac{\text{Total surface area}}{\text{Area of one hair}}=\frac{A}{\pi {{r}^{2}}}$ 

e) the Number of Air Molecules in Your Classroom.         

Estimation of the number of air molecules in your classroom:

Let the volume of the room be \[V\].

One mole of air at NTP occupies \[22.4\text{ }l\] i.e., $22.4\times {{10}^{-3}}{{m}^{3}}$ volume.

Number of molecules in one mole ${{N}_{A}}=6.023\times {{10}^{23}}$ (Avogadro number)

$\therefore $ Number of molecules in room of volume(V) could be found as, 

$n=\frac{6.023\times {{10}^{23}}}{22.4\times {{10}^{-3}}}\times V$  

$\Rightarrow n=134.915\times {{10}^{26}}V$

$\therefore n=1.35\times {{10}^{28}}V$

Therefore, the total number of air molecules in the classroom is $1.35\times {{10}^{28}}V$.

23. 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$=7.0\times {{10}^{8}}m$ .

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. 

24. When the Planet Jupiter Is at a Distance of 824.7 Million Kilometres from the Earth, Its Angular Diameter is Measured to be $35.72''$ of arc. Calculate the Diameter of Jupiter.

Distance of Jupiter from the earth, $D=824.7\times {{10}^{6}}km=824.7\times {{10}^{9}}m$ 

Angular diameter $=35.72''=35.72\times 4.874\times {{10}^{-6}}rad$

Diameter of Jupiter $=d$ 

Using the relation,

\[\begin{align}& \theta =\frac{d}{D} \\ & \Rightarrow d=\theta D=824.7\times {{10}^{9}}\times 35.72\times 4.874\times {{10}^{-6}}=143520.76\times {{10}^{3}} \\ & \therefore d=1.435\times {{10}^{5}}km \\ \end{align}\]

Therefore, the diameter of the jupiter is $1.435\times {{10}^{5}}km$

25. A Man Walking Briskly in Rain With Speed v Must Slant His Umbrella Forward Making An angle $\theta $ with the Vertical. a Student Derives the Following Relation Between $\tan \theta =v$ and Checks That the Relation Has a Correct Limit: as $v\to 0$,$\theta \to 0$as Expected. (we Are Assuming There Is No Strong Wind and That the Rain Falls Vertically for a Stationary Man). Do You Think This Relation Can Be Correct? If Not, Guess the Correct Relation. 

The given relation is incorrect on the basis of dimensional ground. 

The relation is, 

$\tan \theta =v$

Dimension on LHS$={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Dimension on RHS$={{M}^{0}}{{L}^{1}}{{T}^{-1}}$

Clearly, the given relation is dimensionally wrong as the dimensions are different on both sides. 

In order to make it dimensionally right, we could divide RHS by speed of rainfall $v'$. 

So, dimensionally correct relation would be, 

$\tan \theta =\frac{v}{v'}$

26. It Is Claimed That Two Caesium Clocks, If Allowed to Run for 100 Years, Free from Any Disturbance, May Differ by Only About 0.02s. What Does This Imply for the Accuracy of the Standard Caesium Clock in Measuring a Time-Interval of 1s?

The difference in time of caesium clocks is 0.02s and the time required for this difference is 100years $=100\times 365\times 24\times 60\times 60=3.15\times {{10}^{9}}s$.

The time difference for 1s would be $\frac{0.02}{3.15\times {{10}^{9}}}s$.

Therefore, the accuracy of the standard caesium clock in measuring 1s time interval is,  $\frac{3.15\times {{10}^{9}}}{0.02}=157.5\times {{10}^{9}}s\approx 1.5\times {{10}^{11}}s$.

27. Estimate the Average Mass Density of a Sodium Atom Assuming Its Size to Be About $2.5\overset{{}^\circ }{\mathop{\text{A}}}\,$. (use the Known Values of Avogadro's Number and the Atomic Mass of Sodium). Compare it With the Density of Sodium in Its Crystalline Phase: $970kg\text{ }{{m}^{-3}}$. Are the Two Densities of the Same Order of Magnitude? If So, Why?

Diameter of sodium atom = Size of sodium atom $=2.5\overset{{}^\circ }{\mathop{\text{A}}}\,$

Radius of sodium atom, \[r=\frac{1}{2}\times 2.5\overset{{}^\circ }{\mathop{\text{A}}}\,=1.25\overset{{}^\circ }{\mathop{\text{A}}}\,=1.25\times {{10}^{-10}}m\] 

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

$\Rightarrow V=\frac{4}{3}\times 3.14\times {{\left( 1.25\times {{10}^{-10}} \right)}^{3}}$

According to the Avogadro hypothesis, one mole of sodium contains $6.023\times {{10}^{23}}$ atoms and has a mass of \[23\text{ }g\] or $23\times {{10}^{-3}}kg$.

$\therefore $ Mass of one atom $=\frac{23\times {{10}^{-3}}}{6.023\times {{10}^{23}}}kg$

Density of sodium atom, 

$\rho =\frac{\frac{23\times {{10}^{-3}}}{6.023\times {{10}^{23}}}}{\frac{4}{3}\times 3.14\times {{\left( 1.25\times {{10}^{-10}} \right)}^{3}}}$ 

$\therefore \rho =4.67\times {{10}^{-5}}kg\text{ }{{\text{m}}^{-3}}$

It is given that the density of sodium in crystalline phase is $970kg\text{ }{{m}^{-3}}$ .

Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. 

This is because in solid phase, atoms are closely packed and hence the interatomic separation is very small in the crystalline phase.

28. The Unit of Length Convenient on the Nuclear Scale Is a Fermi: \[\mathbf{1f}={{10}^{-15}}\mathbf{m}\]. Nuclear Sizes Obey Roughly the Following Empirical Relation: $r={{r}_{0}}{{A}^{\frac{1}{3}}}$ , where \[\mathbf{r}\] is the Radius of the Nucleus, \[\mathbf{A}\] Its Mass Number, and ${{r}_{0}}$ Is a Constant Equal to About, \[\mathbf{1}.\mathbf{2}\text{ }\mathbf{f}\]. Show That the Rule Implies That Nuclear Mass Density Is Nearly Constant for Different Nuclei. Estimate the Mass Density of Sodium Nucleus and Compare it With the Average Mass Density of a Sodium Atom Obtained in Exercise. 2.27.

We know that the radius of the nucleus $r$ is given by the relation,

$r={{r}_{0}}{{A}^{\frac{1}{3}}}$

${{r}_{0}}=1.2f=1.2\times {{10}^{-15}}m$

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

\[V=\frac{4}{3}\pi {{\left( {{r}_{0}}{{A}^{\frac{1}{3}}} \right)}^{3}}=\frac{4}{3}\pi {{r}_{0}}^{3}A\] …….. (1)

Now, the mass of the nuclei M is equal to its mass number i.e.,

$M=A\text{ amu}=A\times 1.66\times {{10}^{-27}}kg$ 

Density of nucleus,

$\rho =\frac{\text{Mass of nucleus}}{\text{Volume of nucleus}}$ 

\[\Rightarrow \rho =\frac{A\times 1.66\times {{10}^{-27}}}{\frac{4}{3}\pi {{r}_{0}}^{3}A}=\frac{3\times 1.66\times {{10}^{-27}}}{4\pi {{r}_{0}}^{3}}kg/{{\text{m}}^{3}}\]

This relation shows that nuclear mass depends only on constant ${{r}_{0}}$. Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus could be given by,

${{\rho }_{sodium}}=\frac{3\times 1.66\times {{10}^{-27}}}{4\times 3.14\times {{\left( 1.2\times {{10}^{-15}} \right)}^{3}}}$ 

$\Rightarrow \rho =\frac{4.98}{21.71}\times {{10}^{18}}$

$\therefore \rho =2.29\times {{10}^{17}}kg{{\text{m}}^{-3}}$

29. A Laser Is a Source of a Very Intense, Monochromatic, and Unidirectional Beam of Light. These Properties of a Laser Light Can Be Exploited to Measure Long Distances. The Distance of the Moon from the Earth Has Been Already Determined Very Precisely Using a Laser as a Source of Light. A Laser Light Beamed at the Moon Takes 2.56 S to Return After Reflection at the Moon’s Surface. How Much Is the Radius of the Lunar Orbit Around the Earth?

The time taken by the laser beam to return to Earth after being reflected is $2.56s$.

Speed of light is known to be $3\times {{10}^{8}}m/s$.

Time taken by the beam to reach the moon is $\frac{1}{2}\times 2.56=1.28s$.

We know that the radius of the lunar orbit would be the distance between earth and moon: 

$R=c\times t=3\times {{10}^{8}}\times 1.28=3.84\times {{10}^{8}}m$

$\therefore R=3.84\times {{10}^{5}}km$

30. a Sonar (sound Navigation and Ranging) Uses Ultrasonic Waves to Detect and Locate Objects Underwater. in a Submarine Equipped With a Sonar the Time Delay Between Generation of a Probe Wave and the Reception of Its Echo After Reflection from an Enemy Submarine Is Found to Be \[\mathbf{77}.\mathbf{0}\text{ }\mathbf{s}\] . What Is the Distance of the Enemy Submarine? (speed of Sound in Water \[=\mathbf{1450}m{{s}^{-1}}\]).

Let the distance between the ship and the enemy submarine be $'S'$.

We are given,

Speed of sound in water \[=1450\text{ }m/s\] 

Time lag between transmission and reception of Sonar waves \[=77\text{ }s\] 

In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine \[\left( 2S \right)\] .

So, the time taken for the sound to reach the submarine $=\frac{1}{2}\times 77=38.5s$ 

Thus, the distance between the ship and the submarine is given by

\[S=1450\times 38.5=55825\text{ }m=55.8\text{ }km\]

31. The Farthest Objects in Our Universe Discovered by Modern Astronomers Are So Distant That Light Emitted by Them Takes Billions of Years to Reach the Earth. These Objects (known as Quasars) Have Many Puzzling Features, Which Have Not Yet Been Satisfactorily Explained. What Is the Distance (in Km) of a Quasar from Which Light Takes 3.0 Billion Years to Reach Us?

We are given, time taken by quasar light to reach Earth, \[t=3\text{ }billion\text{ }years\]. 

$t=3\times {{10}^{9}}years$ 

$\Rightarrow t=3\times {{10}^{9}}\times 365\times 24\times 60\times 60s$

We know that,

Speed of light, $c=3\times {{10}^{8}}m/s$ 

Distance between the Earth and quasar,

$x=c\times t$

$\Rightarrow x=\left( 3\times {{10}^{8}} \right)\times \left( 3\times {{10}^{9}}\times 365\times 24\times 60\times 60 \right)$

$\Rightarrow x=283824\times {{10}^{20}}m$

$\therefore x=2.8\times {{10}^{22}}km$

Therefore, we found the distance (in km) of a quasar from which light takes 3.0 billion years to reach us to be $x=2.8\times {{10}^{22}}km$. 

32. It is a Well-Known Fact That During a Total Solar Eclipse the Disc of the Moon Almost Completely Covers the Disc of the Sun. from This Fact and from the Information You Can Gather from Examples 2.3 and 2.4, Determine the Approximate Diameter of the Moon.

(Image will be Uploaded Soon)

The position of the Sun, Moon, and Earth during a lunar eclipse would be as shown in the given figure.

Distance of the Moon from the Earth $=3.84\times {{10}^{8}}m$ 

Distance of the sun from the Earth $=1.496\times {{10}^{11}}m$

Diameter of the sun $=1.39\times {{10}^{9}}m$ 

We could see that $\Delta TRS$ and $\Delta TPQ$ are similar. So, 

$\frac{1.39\times {{10}^{9}}}{RS}=\frac{1.496\times {{10}^{11}}}{3.84\times {{10}^{8}}}$

$\Rightarrow RS=\frac{1.39\times 3.84}{1.496}\times {{10}^{6}}$

$\therefore RS=3.57\times {{10}^{6}}m$

Hence, the diameter of the Moon is found to be $3.57\times {{10}^{6}}m$. 

33. P.A.M. Dirac, a Great Physicist of This Century Loved Playing With Numerical Values of Fundamental Constants of Nature. This Led Him to an Interesting Observation That from the Basic Constants of Atomic Physics (\[c,e\], Mass of Electron, Mass of Proton) and the Gravitational Constant $G$ , One Could Arrive at a Number With the Dimension of Time. Further, it Was a Very Large Number Whose Magnitude Was Close to the Present Estimate on the Age of the Universe (\[\sim\mathbf{15}\] Billion Years). from the Table of Fundamental Constants in This Book, Try to See If You Too Can Construct This Number (or Any Other Interesting Number You Can Think Of). If It's Coincidence With the Age of the Universe Were Significant, What Would This Imply for the Constancy of Fundamental Constants?

We have a relation that is consisting of some fundamental constants to give the age of the Universe given by:

$t=\left( \frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right)\times \frac{1}{{{m}_{p}}{{m}_{e}}^{2}{{c}^{3}}G}$ 

$t=$ Age of universe

\[e=\] Charge of electrons \[=1.6\times {{10}^{-19}}C\]

${{\varepsilon }_{0}}=$ Absolute permittivity

${{m}_{p}}=$ Mass of protons $=1.67\times {{10}^{-27}}kg$

${{m}_{e}}=$ Mass of electrons $=9.1\times {{10}^{-31}}kg$

$c=$ Speed of light $=3\times {{10}^{8}}m/s$ 

$G=$ Universal gravitational constant $=6.67\times {{10}^{11}}N{{m}^{2}}k{{g}^{-2}}$

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}$

Substituting all these values in the above equation, we get,

$t=\frac{{{\left( 1.6\times {{10}^{-19}} \right)}^{4}}\times {{\left( 9\times {{10}^{9}} \right)}^{2}}}{{{\left( 9.1\times {{10}^{-31}} \right)}^{2}}\times 1.67\times {{10}^{-27}}\times {{\left( 3\times {{10}^{8}} \right)}^{3}}\times 6.67\times {{10}^{11}}}$

$\Rightarrow t=\frac{{{\left( 1.6 \right)}^{4}}\times 81}{9.1\times 1.67\times 27\times 6.67\times 365\times 24\times 3600}\times {{10}^{-76+18+62+27-24+11}}years$

$\Rightarrow t\approx 6\times {{10}^{-9}}\times {{10}^{18}}years$

$\therefore t=6\text{ billion years}$.

NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurement

NCERT Grade 11 Physics Chapter 2, Units and Measurement, will strengthen your basic concepts about what is a unit, base units, derived units, system of units and others. It will help you to understand the what is the need of an international system of units and how are they decided. In this chapter, complexities of units and measurement shall be dealt with in a comprehensive manner. Units are used to measure different dimensions such as length, breadth, mass, etc. In Chapter 2, Physics of NCERT Grade 11 you shall learn about parallax method, parallax angle and other methods to measure length. One shall also learn on alternative methods to measure length and estimation of very small distances like size of a molecule. 

Similarly, measurement of time, accuracy, precision of instruments will also be learnt in this chapter. Different instruments are used to measure different dimensions. One can easily learn the operations of a pendulum clock, Vernier calliper, scale and other instruments required to measure the physical quantities on micro and macro level through this chapter, Units and Measurement. NCERT Grade 11 Physics Chapter 2, Units and Measurement, deals with minute and large ranges of length and mass measurement. It will help the learner to understand the identification of errors during measurement, absolute error, zero error, relative error and other concepts when dealing with measurement and units. This chapter is a part of Unit I. Unit I, II and III in combination hold 23 marks of weightage in the final examination of Grade 11 according to the latest pattern

Get 100 percent accurate NCERT Solutions for Class 11 Physics Chapter 2 (Units and Measurement) solved by expert Physics teachers. We provide solutions for questions given in Class 11 Physics text-book as per CBSE Board guidelines from the latest NCERT book for Class 11 Physics. The topics and sub-topics in Chapter 2 Units and Measurement are given below.

Ex 2.1 - Introduction

Ex 2.2 - The international system of units

Ex 2.3 - Measurement of length

Ex 2.4 - Measurement of mass

Ex 2.5 - Measurement of time

Ex 2.6 - Accuracy, precision of instruments and errors in measurement

Ex 2.7 - Significant figures

Ex 2.8 - Dimensions of physical quantities

Ex 2.9 - D imensional formulae and dimensional equations

Ex 2.10 - Dimensional analysis and its applications.

We Cover All Exercises in the Chapter Given Below:-

Chapter 2 - 33 Questions with Solution. Other than given exercises, you should also practice all the solved examples given in the book to clear your concepts on Units and Measurement. Download the free PDF of Chapter 2 Units and Measurement and take the print out to keep it handy for your exam preparation.

NCERT Class 11 Physics Chapter wise Solutions - Free PDF Download

Chapter 1 - Physical World

Chapter 2 - Units and Measurement

Chapter 3 - Motion in a Straight Line

Chapter 4 - Motion in a Plane

Chapter 5 - Law of Motion

Chapter 6 - Work, Energy and Power

Chapter 7 - Systems of Particles and Rotational Motion

Chapter 8 - Gravitation

Chapter 9 - Mechanical Properties of Solids

Chapter 10 - Mechanical Properties of Fluids

Chapter 11 - Thermal Properties of Matter

Chapter 12 - Thermodynamics

Chapter 13 - Kinetic Theory

Chapter 14 - Oscillations

Chapter 15 - Waves

NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurement offered by Vedantu is an excellent resource for students who want to excel in their physics studies. The solutions are designed by subject matter experts according to the CBSE syllabus for Class 11 students and cover all the important topics in the chapter. They provide a comprehensive and detailed explanation of the concepts of units and measurement, including the SI units, dimensions, and errors, making it easier for students to understand and improve their physics skills. Additionally, the solutions offer practice exercises and questions that help students test their understanding of the chapter and prepare for their exams. The solutions are available in both English and Hindi medium, making them accessible to a wider audience. Vedantu also provides interactive live classes and doubt-solving sessions to help students clarify their doubts and improve their understanding of the chapter. Overall, the NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurement offered by Vedantu are an essential resource for students who want to improve their physics skills and score well in their exams.

FAQs on NCERT Solutions for Class 11 Physics Chapter 2 - Units And Measurement

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 Base unit

SI Derived unit

Advantages of learning SI units and CGS units

Measurement of Length.

Measurement of mass.

Measurement of time. 

Accuracy, the precision of instruments and errors in measurement. 

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. 

NCERT Solutions for Class 11 Physics

NCERT Solutions for Class 11 Physics Chapter 2 Free PDF Download

Ncert solutions for class 11 physics chapter 2 – units and measurement.

Class 11 physics is pretty tough and difficult to understand by students. Moreover, there are certain topics that completely bounce off over students head. So to help them know the topic we have prepared NCERT Solutions for Class 11 Physics Chapter 2 that will help them with the topic in which they face difficulty.

Besides, these NCERT Solutions are prepared by our panel of professionals who have developed it after thorough research. In addition, they are in easy language so that students can easily get their meaning.

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CBSE Class 11 Physics Chapter 15 Units and Measurement NCERT Solutions

Unit refers to the international standards that are accepted worldwide. While on the other hand, measurement refers to the use of the unit for measuring the quantity. In this NCERT Solutions for Class 11 Physics Chapter 2, we will discuss all the topic of the chapter in brief and in an easily understandable language so that students can get what the content of the chapter.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 15

2.1 introduction.

This topic discusses what is unit and measurement which we have discussed above.

2.2 The International System of Units

This topic defines that earlier scientist of different countries uses the different unit for measurement. But with time and international usage, they started using SI (Standard Unit).

2.3 Measurement of Length

This topic overview the different unit of measuring the length that is used worldwide.

2.3.1 Measurement of Large Distances- This topic defines the use of parallax method for measuring long distances. Besides, the topic provides various examples of using large distances.

2.3.2 Estimation of very Small Distances: Size of a Molecule- This topic defines the use of the small unit by which we can measure the distance of the molecule.

2.3.3 Range of Lengths- this topic defines the use of such unit that helps to measure the lengths of objects that are spread in this wide universe.

2.4 Measurement of Mass

The weight of any object is its mass and this topic discusses the various unit of measurement of mass.

2.4.1 Range of Masses- It refers to all those objects that are spread over the universe that has a fixed mass and can be measured using these units.

2.5 Measurement of Time

This topic defines the atomic standard of time that is used for measuring the atomic clock and cesium clock.

2.6 Accuracy, Precision of Instruments and Errors in Measurement

This topic firstly defines what are errors and the mistakes we do which causes the errors.

• Systematic Errors
• Instrumental errors
• Imperfection in experimental technique or procedure
• Personal errors
• Random Errors
• Least Count Errors

2.6.1 Absolute Error, Relative Error, and Percentage Error- This topic defines all the three errors and how they affect the results.

2.6.2 Combination of Errors- This topic defines the error that we do while performing several measurements

• An error of a sum or a difference
• The error of a product or a quotient
• Error in case of a measured quantity raised to a power

2.7 Significant Figures

This refers to the first uncertain digit plus the reliable digit. Moreover, this topic several other points related to significant figures.

2.7.1 Rules of Arithmetic Operations with Significant Figures- The topic define the rules related to the arithmetic operations.

2.7.2 Rounding off the Uncertain Digits- This topic defines the rule of rounding off of unclear digits.

2.7.3 Rules of Determining the Uncertainty in the Results of Arithmetic Calculations- This topic defines the rules that help to govern the doubt of the results of mathematics calculations.

2.8 Dimensions of Physical Quantities

This topic defines the scope of physical quantities.

2.9 Dimensional Formulae and Dimensional Equations

This topic describes the dime national equations and formula that we use to equitize a physical quantity.

2.10 Dimensional Analysis and its Applications

This topic defines the various applications by which we can analyze the physical quantities of a dimension.

2.10.1 Checking the Dimensional Consistency of Equations- This topic checks the consistency of dimensional equations.

2.10.2 Deducing Relation among the physical Quantities- This topic defines how we can use the relation of the physical quantities for reasoning.

You can download NCERT Solutions for Class 11 Physics Chapter 2 PDF by clicking on the button below.

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NCERT Solutions for Class 11

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Physics Notes for Class 11 Chapter 2 (PDF Download)

Looking for physics notes for class 11 chapter 2 that you can view online in PDF format or download directly? If yes, then this is the right place for you. 

Here are some notes you may find very useful and helpful, especially if you are from the Gujranwala Board, the Lahore Board, the Multan Board, or the Sargodha Board.

Class 11 Chapter 2 Physics Notes (PDF Download)

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• MCQs Notes  Full Chapter

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Class 11 physics notes fbise & kpk board, 11th class physics chapter 1 measurement.

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11th Class Physics Chapter 2 Vectors & Equilibrium

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99 comments:

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Fastly isn't a word Einstein .....

Please upload full syllabus notes

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PLZ UPLOAD FULL NOTES PLZ PLZ PLZ PLZ PLZ

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Complete 1st year course sir notes and lectures

Plz upload solutions of mcqs

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khud nots bana lo bachy mery pass time nahi hy

beta sab yaad karo

Please upload full notes plz

sir plz upload full syllabus notes of class 11

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SIR NOTES OF COMPREHENSIVE QUESTIONS

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Kindly upload full syllabus notes from chapter 4 to 10

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Chapter 10 Conceptual Questions And Chapter 9 Conceptual Questions Are Full Syllabus Not Reduced

please post full notes as soon as possible and DONT forget to include answers of quizes and and pont to ponder boxes ,, Waiting , JazakAllah

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Chapter 4 numerical

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Physics Book for Class 11

• 1.1: Physics and its Scope ( 2  videos )
• 1.2: System International (SI) ( 2  videos )
• 1.3: Scientific Notation ( 4  videos )
• 1.4: Writing Unit Symbols and Names ( 1  videos )
• 1.5: Errors ( 1  videos )
• Total Sections: 9
• Total Videos: 28
• 2.1: Vectors ( 1  videos )
• 2.2: Cartesian Coordinate System ( 8  videos )
• 2.3: Addition of Vectors ( 5  videos )
• 2.4: Multiplication of a Vector by a Scalar ( 1  videos )
• 2.5: Resolution of a Vectors ( 3  videos )
• Total Sections: 11
• Total Videos: 43
• 3.1: Rest and Motion ( 1  videos )
• 3.2: Displacement ( 1  videos )
• 3.3: Velocity ( 3  videos )
• 3.4: Acceleration ( 4  videos )
• 3.5: Graphical Analysis of Motion ( 4  videos )
• Total Videos: 45
• 4.1: Work ( 4  videos )
• 4.2: Work done in Gravitational Field ( 3  videos )
• 4.3: Power ( 4  videos )
• 4.4: Energy ( 3  videos )
• 4.5: Implication of Energy Losses in Practic ( 3  videos )
• Total Videos: 33
• 5.1: Angular Motion ( 4  videos )
• 5.2: Relation Between Angular and Linear Qua ( 5  videos )
• 5.3: Centripetal Force and Centripetal Accel ( 5  videos )
• 5.4: Torque and Moment of Inertia ( 3  videos )
• 5.5: Angular Momentum and Torque ( 3  videos )
• Total Sections: 13
• 6.1: Viscous Fluids ( 1  videos )
• 6.2: Fluid Friction and Stokes Law ( 2  videos )
• 6.3: Terminal Velocity ( 4  videos )
• 6.4: Fluid Flow ( 2  videos )
• 6.5: Equation of Continuity ( 2  videos )
• Total Sections: 7
• Total Videos: 20
• 7.1: Oscillations ( 1  videos )
• 7.2: Terminology of Oscillatory Motion ( 3  videos )
• 7.3: Simple Harmonic Motion ( 3  videos )
• 7.4: Circular Motion and S.H.M ( 5  videos )
• 7.5: Simple Pendulum ( 5  videos )
• Total Videos: 24
• 8.1: Periodic Waves ( 1  videos )
• 8.2: Progressive Waves ( 3  videos )
• 8.3: Classification of Progressive Waves ( 8  videos )
• 8.4: Speed of Sound ( 6  videos )
• 8.5: Superposition of Waves ( 1  videos )
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• Total Videos: 52
• 9.1: Nature of Light ( 2  videos )
• 9.2: Wave Front ( 1  videos )
• 9.3: Huygen's Principle ( 1  videos )
• 9.4: Coherent sources ( 1  videos )
• 9.5: Interference of Light ( 1  videos )
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• 10.1: Thermal Equilibrium ( 1  videos )
• 10.2: Work ( 1  videos )
• 10.3: Internal Energy ( 1  videos )
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• 10.5: Thermodynamic System ( 2  videos )
• Total Videos: 31

• NCERT Solutions
• NCERT Class 11
• NCERT 11 Physics
• Chapter 1: Physical World

NCERT Solutions for Class 11 Physics Chapter 1 Physical World

Ncert solutions class 11 physics chapter 1 physical world – free pdf download.

*According to the latest update on the CBSE Syllabus 2023-24, this chapter has been removed.

NCERT Solutions Class 11 Physics Chapter 1 Physical World is the most effective tool to practise Physics for the 11 th standard examination. The NCERT Solutions for Class 11 Physics will help students to get familiar with the exam pattern and the syllabus of Chapter 1, Physical World. Before understanding more complex theories in Physics, we will be going through the matters and materials around the world where the laws of physics are being applied by nature.

Physics consists of both concepts and problems as per the latest update of the CBSE Syllabus 2023-24. In order to score well in the exam, students are highly recommended to use the NCERT Solutions while answering the questions from the textbook. The chapter-wise solutions are provided here in a PDF format which can be downloaded by the students free of cost.

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Access Answers of NCERT Solutions for Class 11 Physics Chapter 1 Physical World

1.1 Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said: “The most incomprehensible thing about the world is that it is comprehensible”?

The complex physical world involves different orders of magnitudes in space, time and mass. Despite this, nearly all physical phenomena can be expressed in terms of a few basic laws. By this view, Einstein’s statement, “The most incomprehensible thing about the world is that it is comprehensible”, becomes very clear.

1.2 “Every great physical theory starts as a heresy and ends as a dogma.” Give some examples from the history of the science of the validity of this incisive remark.

Ans: One of the general observations in our daily life is that light travels in a straight line. When Huygens propounded his wave theory, it was a heresy. However, soon it became dogma as interference patterns, refraction, etc., could be successfully explained on the basis of wave theory. It was believed that light was only energy, but when the photoelectric effect was discovered, it was proposed that light had a particle nature, too; this was greatly debated and treated as heresy. However, Einstein finally proved it with his quantum theory of light. Henceforth, it has been treated as a dogma.

1.3 “Politics is the art of the possible.” Similarly, “Science is the art of the soluble.” Explain this beautiful aphorism on the nature and practice of science

Politicians make anything and everything possible to win votes. Science is a systematic study of observation. Scientists and researchers study these observations and then work out certain laws from them. There is a multitude of natural phenomena taking place in this universe, and all of them can be explained in terms of some basic laws. For e.g., F = mg is true for you and me and also for a star. Thus, in science, we see that various phenomena are related, and they are soluble and can be explained with similar or the same law. This goes on to justify that science is the art of the soluble, just as politics is the art of the possible.

1.4 Though India now has a large base in science and technology, which is fast expanding, it is still a long way from realising its potential of becoming a world leader in science. Name some important factors, which in your view, have hindered the advancement of science in India.

Ans: Here are some important factors that have obstructed the growth of science in India:

• Lack of infrastructure and funds for quality research work in science.
• Poor pay scales and other facilities for scientists as compared to administrators.
• Science education is neither properly oriented nor directed. It needs specific directions depending on our requirements. Industrialists are the actual consumers of new technology and research. The industrialists of this country have little confidence in the ability of Indian scientists. There is practically no coordination between the researchers and the industrialists.
• Rural-based science education is nearly non-existent, so the majority of the population is deprived of the benefits of advancements in technology and science.

1.5 No physicist has ever “seen” an electron. Yet, all physicists believe in the existence of electrons. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has ‘seen’ one. How will you refute his argument?

Even though an electron has never been ‘seen’, its effects have been observed, and its practical evidence has been tested and proved, for e.g., electricity. However, regarding spirits and ghosts, even though there are many claims and sightings, standardised scientific reading and evidence have never been observed or successfully tested. Thus, we really cannot state with a cent per cent surety that they exist. 1.6 The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation? (a) A tragic sea accident several centuries ago drowned a young Samurai. As a tribute to his bravery, nature, through its inscrutable ways, immortalised his face by imprinting it on the crab shells in that area.

(b) After the sea tragedy, fishermen in that area, in a gesture of honour to their dead hero, set free any crab shell caught by them that accidentally had a shape resembling the face of a Samurai. Consequently, the particular shape of the crab shell survived longer and therefore, in the course of time, the shape was genetically propagated. This is an example of evolution by artificial selection. [Note: This interesting illustration is taken from Carl Sagan’s ‘The Cosmos’ and highlights the fact that often strange and inexplicable facts that at first sight appear ‘supernatural’ actually turn out to have simple scientific explanations. Try to think of other examples of this kind].

(a) More logical and scientific.

1.7 The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were these advances?

The rise of modern science and the industrial revolution are closely connected to each other. The other profound impact effect that Science had on the development of modern science is that the application of Science to solve industrial problems garnered public support and interest in Science.

Some of the key discoveries and their uses are listed below:

• The application of thermodynamics and heat to form the steam engine are some of the key advances in Science and technology during the period of the industrial revolution.
• The implementation and discovery of electricity helped in the invention of motors and dynamos. Likewise, the study of gravitation led to the study of motion, which further led to the development of cannons and guns. This invention gave power to the hands of western countries, and they ruled over the rest of the world.
• The discovery of explosives not only helped the army but also helped in mineral exploration.

These were a few examples of scientific breakthroughs that helped England and Europe to have their prominent positions in the world.

1.8 It is often said that the world is now witnessing a second industrial revolution, which will transform society as radically as the first. List some key contemporary areas of science and technology which are responsible for this revolution.

Some key contemporary areas of technology and Science, which are chiefly responsible for a new industrial revolution taking place now and likely to take place in the near future, are

• Artificial intelligence
• Design of super-fast computers
• Biotechnology
• Development of superconducting materials at room temperature
• Advancements in the field of

→ Electronics

→ Information technology

→ Nanotechnology

→ Developments in the field of Space Sciences

1.9 Write in about 1000 words a fiction piece based on your speculation on the Science and technology of the twenty-second century.

Ans: In the 22 nd century, humans will be able to create wormholes allowing people to travel to distant places in the universe. We will be in contact with aliens and have established human settlements outside the earth as well. With advancements in quantum Physics, we shall be more aware and understanding of the true nature of our universe and existence. Our technology will not pollute and degrade the earth. Artificial intelligence and humans could have some clashes.

1.10 Attempt to formulate your ‘moral’ views on the practice of Science. Imagine yourself stumbling upon a discovery which has great academic interest but is certain to have nothing but dangerous consequences for human society. How, if at all, will you resolve your dilemma?

Ans: A scientist works for the truth. Every scientific discovery reveals a certain truth about nature. So, any discovery, bad or good for humankind, must be made public. But with that being said, we cannot afford to be blind to the consequences. Before disclosing it, we must ascertain the degree of good or bad consequences it will have. If we know that a certain discovery has nothing but dangerous consequences to offer to the mass, the discovery is best kept limited only to the knowledge of the scientist and researchers working on it. This way, the discovery can help societies in the long run without completely destroying them now.

1.11 Science, like any knowledge, can be put to good or bad use, depending on the user. Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorised.

(a) Mass vaccination against smallpox to curb and finally eradicate this disease from the population. (This has already been successfully done in India.) (b) Television for the eradication of illiteracy and mass communication of news and ideas. (c) Prenatal sex determination. (d) Computers for the increase in work efficiency. (e) Putting artificial satellites into orbits around the Earth. (f ) Development of nuclear weapons. (g) Development of new and powerful techniques of chemical and biological warfare. (h) Purification of water for drinking. (i) Plastic surgery (j ) Cloning

(a)  Good. Mass vaccination helped eradicate the dreaded diseases from the Earth.

(b) Good. Television helps in the literacy campaign and is an effective method of mass communication and entertainment.

(c) Bad. Prenatal sex determination is considered bad because it gives rise to the practice of abortion in the case of the female foetus.

(d) Good. Computer increases work efficiency.

 (e)  Good. Artificial satellites help in the worldwide communication process.

(f) Bad. Nuclear weapons, if misused, may cause the mass destruction of humankind.

(g) Bad. These techniques may be misused for destructive purposes.

(h) Good. Purified water improves the health of people.

(i) Neither good nor bad. Plastic surgery is something which can’t be classified as either good or bad because it helps to remove a certain types of deformations in needy persons.

(j) Bad. Cloning has the potential to ruin the normal family life of human society.

1.12 India has had a long and unbroken tradition of great scholarship — in mathematics, astronomy, linguistics, logic and ethics. Yet, in parallel with this, several superstitious and obscurantist attitudes and practices flourished in our society and unfortunately continue even today — among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?

In order to popularise scientific explanations of everyday phenomena, mass media like

→ Newspapers

→ Television

should be used. Knowledge of science should be educated to the masses so that they learn about the real causes of phenomena, allowing their superstitious beliefs to wash away.

1.13 Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice, give them a secondary status and role. Demolish this view using scientific arguments and by quoting examples of great women in science and other spheres, and persuade yourself and others that, given equal opportunity, women are on par with men.

No difference is noticed in the ability of women and men as far as work, intelligence, decision making is concerned. Nature makes little difference in the anatomy and feeling of men and women. The nutrition content of prenatal and postnatal diets contributes a lot towards the development of the human body. If equal opportunities are provided to both women and men, then the female mind and body will be just as efficient as a man’s. The list of great women who have excelled in their respective fields is enormous. Names of

→ Madame Curie

→ Indira Gandhi

→ Florence Nightingale

→ Margaret Thatcher

→ Mother Teresa

→ Sarojini Naidu

→ Kalpana Chawla

been taken from fields varying from Sociology to Science, and they are very well-known for their contribution to the world.  Reflecting on the contribution being made to each and every sphere of life in the country, it can be positively debated that women are no less essential to society than men.

1.14 “It is more important to have beauty in the equations of Physics than to have them agree with experiments”. The great British physicist P. A. M. Dirac held this view. Criticise this statement. Look out for some equations and results in this book which strike you as beautiful.

Dirac’s belief holds true. Equations which represent entire concepts and hold up against experimental results are automatically simple, small and symmetrical, making them truly beautiful. Some examples of beautiful equations are: E = mc 2 , E = hv, F = mg, P.E = mgh

NCERT Solutions for Class 11 Physics Chapter 1

NCERT Solutions provided here present students with answers to the textbook questions, along with extra questions, NCERT Exemplar problems with solutions, worksheets, MCQs and short answer questions. Solving these questions will help them to prepare notes and to understand the concept thoroughly.

We will see here how things work and the reasons behind their workings. As Einstein once said, “Imagination is more important than knowledge,” this chapter will make students aware of the things that surround them, the history of science and its origin. We will also see many aphorisms below and reasons to justify them with real-life Physics concepts. We will be relating ourselves to the theories and observations done by many scientists and deriving conclusions from them.

We will be getting acquainted with the factors responsible for the growth of Science and technology in India. The existence of an electron has been discussed in this chapter, along with the scientific resemblance of a crab species with that of a Japanese samurai. We will be glancing at the history of the industrial revolution in Western Europe and England, which was two centuries ago. Besides, we will be seeing the significant areas of science and technology responsible for the current revolution.

We will also be writing a fictional article about the speculation of Science and technology in the 22nd century. Moreover, we will work on moralising our views on science and the consequences of practising Science. We will ask students to give their opinions about certain applications and processes of Science, whether it was helpful or not for the betterment of human civilisation. We will be focusing on the history of India and the superstitious practices and attitudes followed by Indians. Students can check out NCERT Solutions for Class 11 for more chapter-wise solutions of Physics.

Furthermore, we will be talking about the law behind people having irrational ideas about women’s intelligence, capacity and nature and about giving them a status along with the role of technology in the development of society. We will let students express their judgements on the statement provided by Physicist P.A.M. Dirac, and we will also be knowing some beautiful equations of Physics. We have heard about the adventures of scientists and many more outgoing individuals who have climbed tall mountains to record temperatures and even dive deep to collect data regarding flora and fauna. Below, we will see various questions on the above-mentioned topics.

The NCERT Solutions for Class 11 Physics Chapter 1 are prepared by expert subject teachers in such a way that it covers all the concepts of the topic. Check the NCERT Solutions for Class 11 Physics Chapter 1 PDF.

BYJU’S brings students NCERT study materials, notes, previous years’ question papers, Physics problems and sample papers. These study materials are prepared as per the latest CBSE Syllabus 2023-24. For more information, register with BYJU’S or download BYJU’S – The Learning App for a personalised learning experience.

Important Concepts of NCERT Solutions for Class 11 Physics Chapter 1

Students who are not able to answer the questions from the textbook are advised to refer to the NCERT Solutions for Class 11 Physics from BYJU’S. A highly experienced faculty creates the solutions in accordance with the latest CBSE Syllabus 2023-24 and its guidelines. Some of the important concepts covered in Chapter 1 are

1.1 WHAT IS PHYSICS?

1.2 SCOPE AND EXCITEMENT OF PHYSICS

1.3 PHYSICS, TECHNOLOGY AND SOCIETY

1.4 FUNDAMENTAL FORCES IN NATURE

1.4.1 Gravitational Force

1.4.2 Electromagnetic Force

1.4.3 Strong Nuclear Force

1.4.4 Weak Nuclear Force

1.4.5 Towards Unification of Forces

1.5 NATURE OF PHYSICAL LAWS

Disclaimer – 

Dropped Topics – 

1.1 What Is Physics? 1.2 Scope and Excitement of Physics 1.3 Physics, Technology and Society 1.4 Fundamental Forces in Nature 1.5 Nature of Physical Laws

Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 1

How do ncert solutions of class 11 physics chapter 1 physical world help in the exam preparation, is byju’s providing answers for ncert solutions for class 11 physics chapter 1, list out the important concepts discussed in ncert solutions for class 11 physics chapter 1..

The main concepts covered in NCERT Solutions for Class 11 Physics Chapter 1 are we will be going through the matters and materials around the world where the laws of Physics are being applied by nature.

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