HC Verma Solutions: Chapter 1 – Introduction to Physics

HC Verma Solutions: Chapter 1 – Introduction to Physics

This article is about the HC Verma solutions Part 1 : Chapter 1 – Introduction to Physics. I really hope you find them helpful. If you really like them then help us by sharing the article or you can leave a comment in the comments section.

Question 1: Find the dimensions of

(a) Linear momentum

(b) Frequency

(c) Pressure

Solution:

(a) Linear momentum = $mv$

Dimensions of linear momentum =$ML{T^{ – 1}}$

(b) Frequency =$\frac{1}{T} = [{M^0}{L^0}{T^{ – 1}}]$

(c) Pressure: $\frac{{Force}}{{Area}} = \frac{{[ML{T^{ – 2}}]}}{{[{L^2}]}} = [M{L^{ – 1}}{T^{ – 2}}]$

Question 2: Find the dimensions of

(a) Angular speed $\omega $

(b) Angular acceleration $\alpha $

(c) Torque $\tau $

(d) Moment of inertia $I$

Some of the equations involving these quantities are

$\omega  = \frac{{{\theta _2} – {\theta _1}}}{{{t_2} – {t_1}}}$ , $\alpha  = \frac{{{\omega _2} – {\omega _1}}}{{{t_2} – {t_1}}}$ , $\tau  = F \cdot r$ and $I = m{r^2}$

The symbols have standard meaning.

Solution:

(a) Angular speed $\omega  = \frac{\theta }{t}$

Where, $\theta $ is the angular displacement and $t$ is time.

Angular displacement is defined as the ratio of length and radius.

Angular displacement = $\frac{{length}}{{radius}}$

Now we know
Dimensional Formula of Length = $[{M^0}{L^1}{T^0}]$
Dimensional Formula of Radius  = $[{M^0}{L^1}{T^0}]$

So
Angular displacement= $\frac{{[{M^0}{L^1}{T^0}]}}{{[{M^0}{L^1}{T^0}]}}$

 Hence Dimensional Formula of Angular displacement =$[{M^0}{L^0}{T^0}]$

Hence dimensional formula for angular speed = $\frac{{{M^0}{L^0}{T^0}}}{T} = {M^0}{L^0}{T^{ – 1}}$

(b) Angular acceleration = $\alpha  = \frac{\omega }{t}$

Dimensional formula for angular speed = ${M^0}{L^0}{T^{ – 1}}$

So, dimensional formula for angular acceleration is

$\alpha  = \frac{\omega }{t} = \frac{{{M^0}{L^0}{T^{ – 1}}}}{T} = {M^0}{L^0}{T^{ – 2}}$

(c) Torque, $\tau  = Fr = [ML{T^{ – 2}}][L] = [M{L^2}{T^{ – 2}}]$

(d) Moment of inertia, $I = M{r^2} = [M][{L^2}] = [M{L^2}{T^0}]$

Question 3: Find the dimensions of

(a) Electric field

(b) Magnetic field

(c) Magnetic permeability

The relevant equations are

$F = qE$ , $F = qvB$ , $B = \frac{{{\mu _0}I}}{{2\pi a}}$

Where,$F$ is force, $q$ is charge, $v$ is speed, $I$ is current and $a$ is distance.

Solution:

(a) Electric field $E = \frac{F}{q} = \frac{{ML{T^{ – 2}}}}{{IT}} = [ML{T^{ – 3}}{I^{ – 1}}]$

(b) Magnetic field $B = \frac{F}{{qv}} = \frac{{ML{T^{ – 2}}}}{{[IT][L{T^{ – 1}}}} = [M{T^{ – 2}}{I^{ – 1}}]$

(c) Magnetic permeability ${\mu _0} = \frac{{B \times 2\pi a}}{I} = \frac{{[M{T^{ – 2}}{I^{ – 1}}] \times [L]}}{{[I]}} = [ML{T^{ – 2}}{I^{ – 2}}]$

Question 4: Find the dimensions of



(a) Electric dipole moment $p$

(b) Magnetic dipole moment $M$

The defining equations are $p = q \cdot d$ and $M = IA$ where $A$ is area, $q$ is charge and $I$ is current.

Solution:

(a) Electric dipole moment $p = qd = [IT] \times [L] = [LTI]$

(b) Magnetic dipole moment $M = IA = [I][{L^2}] = [{L^2}I]$

Question 5: Find the dimension of Plank’s constant $h$ from the equation $E = h\nu $ where $E$ is the energy is and $\nu $ is the frequency.

Solution: Given equation is $E = h\nu $where $E$ is the energy and $\nu$ is the frequency.

This implies that $h = \frac{E}{\nu } = \frac{{[M{L^2}{T^{ – 2}}]}}{{[{T^{ – 1}}]}} = [M{L^2}{T^{ – 1}}]$

Question 6: Find the dimensions of

(a) Specific heat capacity $c$

(b) The coefficient of linear expansion $\alpha $

(c) The gas constant $R$

Some of the equations involving these quantities are

$Q = mc({T_2} – {T_1})$ , ${l_t} = {l_0}[1 + \alpha ({T_2} – {T_1})]$ and $PV = nRT$

Solution:

(a) Specific heat capacity $c = \frac{Q}{{m\Delta T}} = \frac{{[M{L^2}{T^{ – 2}}]}}{{[M][K]}} = [{L^2}{T^{ – 2}}{K^{ – 1}}]$

(b) The coefficient of linear expansion $\alpha  = \frac{{\left( {{l_t} – {l_0}} \right)}}{{{l_0}({T_2} – {T_1})}} = \frac{{[L]}}{{[LK]}} = [{K^{ – 1}}]$

(c) The gas constant $R = \frac{{PV}}{{nT}} = \frac{{[M{L^{ – 1}}{T^{ – 2}}][{L^3}]}}{{[mol][K]}} = [M{L^2}{T^{ – 2}}{K^{ – 1}}{(mol)^{ – 1}}]$

Question 7: Taking force, length and time to be the fundamental quantities, find the dimensions of

(a)  Density

(b) Pressure

(c) Momentum

(d) Energy

Solution: Here we are taking force $F$ , length $L$ and time $T$ as the fundamental quantities. So in terms of new system of units dimensions of mass would be

${{M}_{N}}=\frac{force}{acceleration}=\frac{F}{L{{T}^{-2}}}=F{{L}^{-1}}{{T}^{2}}$

(a)  Density = $\frac{mass}{volume}=\frac{F{{L}^{-1}}{{T}^{2}}}{{{L}^{3}}}=F{{L}^{-4}}{{T}^{2}}$

(b) Pressure = $\frac{force}{area}=\frac{F}{{{L}^{2}}}=F{{L}^{-2}}$

(c) Momentum =$mv=[F{{L}^{-1}}{{T}^{2}}]\times [L{{T}^{-1}}]=[FT]$

(d) Energy = $\frac{1}{2}m{{v}^{2}}=[F{{L}^{-1}}{{T}^{2}}]\times [{{L}^{2}}{{T}^{-2}}]=[FL]$

Question 8: Suppose the acceleration due to gravity at a place is 10 m/s2. Find its value in cm/(minute)2.

Solution:  Acceleration due to gravity g=10 m/s2.

We have to convert it into cm/min2.

1 m = 100 cm

1 sec = 0.0166667

$g=10m/{{\sec }^{2}}=10\times \frac{100}{{{\left( 0.0166667 \right)}^{2}}}=36\times {{10}^{5}}cm/{{\min }^{2}}$

Question 9: The average speed of a snail is 0.020 miles/hour and that of a leopard is 70 miles/hour. Convert these speeds into SI units.

 

Solution: The average speed of a snail is 0.02 mile/hr

We know that 1 mile = 1.6 km = 1600 m

1 hour = 3600 second

Converting to S.I. units, $\frac{0.02\times 1.6\times 1000}{3600}$ m/sec = 0.0089 ms–1

The average speed of leopard = 70 miles/hr

In SI units = 70 miles/hour = $\frac{70\times 1.6\times 1000}{3600}$ = 31 m/s



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Rupam Tirkey

Thakyou for the solution