{"id":3773,"date":"2016-07-27T13:46:08","date_gmt":"2016-07-27T08:16:08","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=3773"},"modified":"2022-11-04T12:06:31","modified_gmt":"2022-11-04T06:36:31","slug":"hc-verma-solutions-chapter-1","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/hc-verma-solutions-chapter-1\/","title":{"rendered":"HC Verma Solutions: Chapter 1 \u2013 Introduction to Physics"},"content":{"rendered":"

\"HC<\/a><\/p>\n

This article is about the HC Verma solutions Part 1\u00a0: Chapter 1 \u2013 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.<\/div>\n

Question 1: <\/strong>Find the dimensions of<\/p>\n

(a)<\/strong> Linear momentum<\/p>\n

(b)<\/strong> Frequency<\/p>\n

(c)<\/strong> Pressure<\/p>\n

Solution: <\/strong><\/p>\n

(a)<\/strong> Linear momentum = $mv$<\/p>\n

Dimensions of linear momentum =$ML{T^{ – 1}}$<\/p>\n

(b)<\/strong> Frequency =$\\frac{1}{T} = [{M^0}{L^0}{T^{ – 1}}]$<\/p>\n

(c)<\/strong> Pressure: $\\frac{{Force}}{{Area}} = \\frac{{[ML{T^{ – 2}}]}}{{[{L^2}]}} = [M{L^{ – 1}}{T^{ – 2}}]$<\/p>\n

Question 2: <\/strong>Find the dimensions of<\/p>\n

(a)<\/strong> Angular speed $\\omega $<\/p>\n

(b)<\/strong> Angular acceleration $\\alpha $<\/p>\n

(c)<\/strong> Torque $\\tau $<\/p>\n

(d)<\/strong> Moment of inertia $I$<\/p>\n

Some of the equations involving these quantities are<\/p>\n

$\\omega\u00a0 = \\frac{{{\\theta _2} – {\\theta _1}}}{{{t_2} – {t_1}}}$ , $\\alpha\u00a0 = \\frac{{{\\omega _2} – {\\omega _1}}}{{{t_2} – {t_1}}}$ , $\\tau\u00a0 = F \\cdot r$ and $I = m{r^2}$<\/p>\n

The symbols have standard meaning.<\/p>\n

Solution: <\/strong><\/p>\n

(a)<\/strong> Angular speed $\\omega\u00a0 = \\frac{\\theta }{t}$<\/p>\n

Where, $\\theta $ is the angular displacement and $t$ is time.<\/p>\n

Angular displacement is defined as the ratio of length and radius.<\/p>\n

Angular displacement = $\\frac{{length}}{{radius}}$<\/p>\n

Now we know
\nDimensional Formula of Length = $[{M^0}{L^1}{T^0}]$
\nDimensional Formula of Radius\u00a0 = $[{M^0}{L^1}{T^0}]$<\/p>\n

So
\nAngular displacement= $\\frac{{[{M^0}{L^1}{T^0}]}}{{[{M^0}{L^1}{T^0}]}}$<\/p>\n

\u00a0<\/strong>Hence Dimensional Formula of Angular displacement =$[{M^0}{L^0}{T^0}]$<\/p>\n

Hence dimensional formula for angular speed = $\\frac{{{M^0}{L^0}{T^0}}}{T} = {M^0}{L^0}{T^{ – 1}}$<\/p>\n

(b)<\/strong> Angular acceleration = $\\alpha\u00a0 = \\frac{\\omega }{t}$<\/p>\n

Dimensional formula for angular speed = ${M^0}{L^0}{T^{ – 1}}$<\/p>\n

So, dimensional formula for angular acceleration is<\/p>\n

$\\alpha\u00a0 = \\frac{\\omega }{t} = \\frac{{{M^0}{L^0}{T^{ – 1}}}}{T} = {M^0}{L^0}{T^{ – 2}}$<\/p>\n

(c)<\/strong> Torque, $\\tau\u00a0 = Fr = [ML{T^{ – 2}}][L] = [M{L^2}{T^{ – 2}}]$<\/p>\n

(d)<\/strong> Moment of inertia, $I = M{r^2} = [M][{L^2}] = [M{L^2}{T^0}]$<\/p>\n

Question 3:<\/strong> Find the dimensions of<\/p>\n

(a)<\/strong> Electric field<\/p>\n

(b)<\/strong> Magnetic field<\/p>\n

(c)<\/strong> Magnetic permeability<\/p>\n

The relevant equations are<\/p>\n

$F = qE$ , $F = qvB$ , $B = \\frac{{{\\mu _0}I}}{{2\\pi a}}$<\/p>\n

Where,$F$ is force, $q$ is charge, $v$ is speed, $I$ is current and $a$ is distance.<\/p>\n

Solution:<\/strong><\/p>\n

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

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

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

Question 4:<\/strong> Find the dimensions of<\/p>\n

(a)<\/strong> Electric dipole moment $p$<\/p>\n

(b)<\/strong> Magnetic dipole moment $M$<\/p>\n

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

Solution:<\/strong><\/p>\n

(a)<\/strong> Electric dipole moment $p = qd = [IT] \\times [L] = [LTI]$<\/p>\n

(b)<\/strong> Magnetic dipole moment $M = IA = [I][{L^2}] = [{L^2}I]$<\/p>\n

Question 5: <\/strong>Find the dimension of Plank\u2019s constant $h$ from the equation $E = h\\nu $ where $E$ is the energy is and $\\nu $ is the frequency.<\/p>\n

Solution:<\/strong> Given equation is $E = h\\nu $where $E$ is the energy and $\\nu$ is the frequency.<\/p>\n

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

Question 6:<\/strong> Find the dimensions of<\/p>\n

(a)<\/strong> Specific heat capacity $c$<\/p>\n

(b)<\/strong> The coefficient of linear expansion $\\alpha $<\/p>\n

(c)<\/strong> The gas constant $R$<\/p>\n

Some of the equations involving these quantities are<\/p>\n

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

Solution:<\/strong><\/p>\n

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

(b)<\/strong> The coefficient of linear expansion $\\alpha\u00a0 = \\frac{{\\left( {{l_t} – {l_0}} \\right)}}{{{l_0}({T_2} – {T_1})}} = \\frac{{[L]}}{{[LK]}} = [{K^{ – 1}}]$<\/p>\n

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

Question 7: <\/strong>Taking force, length and time to be the fundamental quantities, find the dimensions of<\/p>\n

(a)<\/strong> \u00a0Density<\/p>\n

(b)<\/strong> Pressure<\/p>\n

(c)<\/strong> Momentum<\/p>\n

(d)<\/strong> Energy<\/p>\n

Solution:<\/strong> 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<\/p>\n

${{M}_{N}}=\\frac{force}{acceleration}=\\frac{F}{L{{T}^{-2}}}=F{{L}^{-1}}{{T}^{2}}$<\/p>\n

(a)<\/strong> \u00a0Density = $\\frac{mass}{volume}=\\frac{F{{L}^{-1}}{{T}^{2}}}{{{L}^{3}}}=F{{L}^{-4}}{{T}^{2}}$<\/p>\n

(b)<\/strong> Pressure = $\\frac{force}{area}=\\frac{F}{{{L}^{2}}}=F{{L}^{-2}}$<\/p>\n

(c)<\/strong> Momentum =$mv=[F{{L}^{-1}}{{T}^{2}}]\\times [L{{T}^{-1}}]=[FT]$<\/p>\n

(d)<\/strong> Energy = $\\frac{1}{2}m{{v}^{2}}=[F{{L}^{-1}}{{T}^{2}}]\\times [{{L}^{2}}{{T}^{-2}}]=[FL]$<\/p>\n

Question 8:<\/strong> Suppose the acceleration due to gravity at a place is 10 m\/s2<\/sup>. Find its value in cm\/(minute)2<\/sup>.<\/p>\n

Solution:<\/strong> \u00a0Acceleration due to gravity g=10 m\/s2<\/sup>.<\/p>\n

We have to convert it into cm\/min2<\/sup>.<\/p>\n

1 m = 100 cm<\/p>\n

1 sec = 0.0166667<\/p>\n

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

Question 9:<\/strong> 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.<\/p>\n

 <\/p>\n

Solution: <\/strong>The average speed of a snail is 0.02 mile\/hr<\/p>\n

We know that 1 mile = 1.6 km = 1600 m<\/p>\n

1 hour = 3600 second<\/p>\n

Converting to S.I. units, $\\frac{0.02\\times 1.6\\times 1000}{3600}$ m\/sec = 0.0089 ms\u20131<\/sup><\/p>\n

The average speed of leopard = 70 miles\/hr<\/p>\n

In SI units = 70 miles\/hour = $\\frac{70\\times 1.6\\times 1000}{3600}$ = 31 m\/s<\/p>\n","protected":false},"excerpt":{"rendered":"

This article is about the HC Verma solutions Part 1\u00a0: Chapter 1 \u2013 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) […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[423],"tags":[],"class_list":["post-3773","post","type-post","status-publish","format-standard","hentry","category-concepts-of-physics-solutions"],"yoast_head":"\nHC Verma Solutions: Chapter 1- Introduction to Physics by physicscatalyst<\/title>\n<meta name=\"description\" content=\"This article is about the HC Verma Solutions Part 1\u00a0: Chapter 1 \u2013 Introduction to Physics. 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