NCERT Solutions for Class 11 Physics Chapter 2

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 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?

Ans:

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

Ans:

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) $y=a\sin vt$ 

Ans:

It is incorrect.

Given: $y=a\sin vt$

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

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

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

Ans:

It is incorrect.

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

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

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

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

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

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

Ans:

It is correct.

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

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

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

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

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:

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

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,

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

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?

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

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?

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. 

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?

Ans:

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?         

Ans: 

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, 

$\theta =\frac{d}{D}$ 

Where,

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.

Ans: 

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. 

Ans:

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.

Ans:

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.

Ans:

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.

Ans:

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.         

Ans:

Estimation of the number of air molecules in your classroom:

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

We know that:

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

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. 

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.

Ans:

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. 

Ans:

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?

Ans:

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?

Ans: 

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.

Ans: 

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?

Ans:

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 \[=\mathbf{1450}m{{s}^{-1}}\]).

Ans: 

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?

Ans: 

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

That is,

$t=3\times {{10}^{9}}years$ 

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

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.

Ans:

(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.

We know that, 

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?

Ans: 

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

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

Where,

$t=$ 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 $=6.67\times {{10}^{11}}N{{m}^{2}}k{{g}^{-2}}$

Also,

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

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 - Dimensional 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.