{"id":2187,"date":"2026-11-12T22:34:29","date_gmt":"2026-11-12T17:04:29","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2187"},"modified":"2026-11-12T22:34:30","modified_gmt":"2026-11-12T17:04:30","slug":"icse-class-10-physics-2020-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2020-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Physics 2020 Question Paper with Solutions"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_68_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title \" >Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2020-question-paper-with-solutions\/#SECTION_I\" title=\"SECTION I\">SECTION I<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2020-question-paper-with-solutions\/#SECTION_II\" title=\"SECTION II\">SECTION II<\/a><\/li><\/ul><\/nav><\/div>\n<span class=\"rt-reading-time\" style=\"display: block;\"><span class=\"rt-label rt-prefix\">Reading Time: <\/span> <span class=\"rt-time\">35<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span>\n<div class=\"xc-nd xc-icse-physics-pyq\">\n<div class=\"xc-intro-box\">\n<div class=\"xc-intro-label\">ICSE Class 10 Physics PYQ<\/div>\n<p class=\"xc-intro-text\">Practice the complete <strong>ICSE Class 10 Physics 2020 Question Paper with Solutions<\/strong>, including numericals, diagrams and detailed answers from Sections I and II. Use this solved paper to revise important concepts, practise board-style questions and strengthen your exam preparation.<\/p>\n<\/div>\n<div class=\"xc-info-grid\">\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Year<\/div><div class=\"xc-info-card-value\">2020<\/div><\/div>\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Maximum Marks<\/div><div class=\"xc-info-card-value\">80<\/div><\/div>\n<div class=\"xc-info-card\"><div class=\"xc-info-card-label\">Time Allowed<\/div><div class=\"xc-info-card-value\">Three hours<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">\ud83d\udcc4<\/div><div><div class=\"xc-set-title\">Paper Instructions<\/div><div class=\"xc-set-sub\">Read the examination instructions before attempting the paper.<\/div><\/div><\/div>\n<div class=\"xc-qtext\"><p>Section I is compulsory. Attempt any four questions from Section II.<\/p><p>The intended marks for questions or parts of questions are given in brackets [ ].<\/p><\/div>\n<\/div>\n<div class=\"xc-section-heading\">\n<div class=\"xc-section-heading-icon\">\ud83d\udcdd<\/div>\n<div class=\"xc-section-heading-body\"><div class=\"xc-section-heading-label\">ICSE Physics 2020<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"SECTION_I\"><\/span>SECTION I<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">40 MARKS \u2022 Attempt all questions from this Section.<\/div><\/div>\n<div class=\"xc-section-heading-stat\"><div class=\"xc-section-heading-stat-num\">40<\/div><div class=\"xc-section-heading-stat-label\">Marks<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">1<\/div><div><div class=\"xc-set-title\">Question 1<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q1-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>(i) Define moment of force.<\/p>\n<p>(ii) Write the relationship between the Sl and CGS unit of moment of force. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) The product of magnitude of force and the perpendicular distance of line of action of force from the axis of rotation is known as the moment of force. It is also known as torque.<\/p>\n<p>(ii) Since, the SI unit of moment of force is Newton metre, i.e., Nm and the CGS unit of moment of force is dyne cm.<\/p>\n<p>So, the required relationship between the SI and CGS unit of moment of force will be \\(1 \\mathrm{Nm}=10^{5}\\) dyne \\(\\times 10^{2} \\mathrm{~cm}=10^{7}\\) dyne cm<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q1-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Define a kilowatt hour. How is it related to joule ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. One kilowatt hour is defined as the electrical energy which is consumed by the appliance of 1 kW power in duration of 1 hour.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\because 1 \\text { kilowatt hour } &amp; =1 \\text { kilowatt } \\times 1 \\text { hour } \\\\\n&amp; =1000 \\text { watts } \\times 1 \\text { hour } \\\\\n&amp; \\{\\because 1 \\text { kilowatt }=1000 \\text { watts }\\} \\\\\n&amp; =1000 \\text { watts } \\times 3600 \\mathrm{~s} \\\\\n&amp; \\quad\\{\\because 1 \\text { hour }=3600 \\mathrm{~s}\\} \\\\\n&amp; \n&amp; \\text { and } \\quad\\left\\{\\because 1 \\text { watt }=\\frac{1 \\mathrm{~J}}{\\mathrm{~s}}\\right\\} \\\\\n&amp; =1000 \\times \\frac{\\mathrm{J}}{\\mathrm{~s}} \\times 3600 \\mathrm{~s}\n \\\\\n&amp; =3600000 \\mathrm{~J}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n1 \\text { kilowatt hour }=3.6 \\times 10^{6} \\mathrm{~J}\n\\]<\/div>\n<p>So, this is the required relation between kilowatt hour and Joule.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q1-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A satellite revolves around a planet in a circular orbit. What is the work done by the satellite at any instant? Give a reason. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. It is given that a satellite is revolving around the Earth, So, let us represent the given situation through the following diagram, at particular instant.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 1(c) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-1_348_554_1055_1180.jpg\"\/><\/div>\n<p>Note: When the direction of acting force on the body is perpendicular to the direction of displacement, then work done remains zero. So, in this given situation, the work done by the satellite at any instant will be zero. It is because the centripetal force \\(\\left(\\overrightarrow{\\mathrm{F}}_{c}\\right)\\) is perpendicular to the direction of the displacement \/motion.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q1-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>(i) Identify the class of the lever shown in the diagram below :<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 1(d) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-1_123_334_1810_1248.jpg\"\/><\/div>\n<p>(ii) How is it possible to increase the M.A. of the above lever without increasing its length ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) Consider the following diagram of lever,<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 1(d) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-1_124_333_2099_1250.jpg\"\/><\/div>\n<p>The given diagram is of class III lever, because Effort(E) is lying between Fulcrum(F) and Load(L).<\/p>\n<p>(ii) As Mechanical Advantage \\(=\\frac{\\text { Effort arm }}{\\text { load arm }}\\).<\/p>\n<p>So, if the distance of the effort from the fulcrum is increased, then the mechanical advantage of the above lever will also be increased without increasing its length.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q1-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>Give one example of each when :<\/p>\n<p>(i) Chemical energy changes into electrical energy.<\/p>\n<p>(ii) Electrical energy changes into sound energy. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) In case of primary cell, chemical energy changes into electrical energy as it is used to convert stored chemical energy into electrical energy.<\/p>\n<p>(ii) In case of electric bell, the electrical energy changes into sound energy as the sounds produced by the electrical signals is converted into a form of sound energy.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">2<\/div><div><div class=\"xc-set-title\">Question 2<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q2-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>A crane &#8216;A&#8217; lifts a heavy load in 5 seconds, whereas another crane &#8216;B&#8217; does the same work in 2 seconds. Compare the power of \\(\\boldsymbol{n}\\) crane &#8216;A&#8217; to that of crane &#8216;B&#8217;. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. According to question, crane A lifts a heavy load is 5 seconds and crane B does the same work in 2 seconds. \u2235<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Power }=\\frac{\\text { work done }}{\\text { time taken }}\n\\]<\/div>\n<p>\u2234 For crane A,<\/p>\n<div class=\"mathjax-overflow\">\\[\nP_{\\mathrm{A}}=\\frac{W}{5 \\mathrm{~s}} \\tag{i}\n\\]<\/div>\n<p>\\(\\{\\because\\) For crane \\(\\mathrm{A}, t=5 s\\}\\) and for crane B,<\/p>\n<div class=\"mathjax-overflow\">\\[\nP_{\\mathrm{B}}=\\frac{W}{2 \\mathrm{~s}} \\tag{ii}\n\\]<\/div>\n<p>\\(\\{\\because\\) For crane \\(\\mathrm{B}, t=2 s\\}\\) On dividing eqns. (i) and (ii), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{P_{\\mathrm{A}}}{P_{\\mathrm{B}}}=\\frac{\\frac{W}{5}}{\\frac{W}{2}} \\Rightarrow \\frac{P_{\\mathrm{A}}}{P_{\\mathrm{B}}}=\\frac{2}{5}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\nP_{A}=1.4 P_{B}\n\\]<\/div>\n<p>i.e., the power of crane A is 0.4 times to that of crane B.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q2-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A ray of light falls normally on a rectangular glass slab. 2 Draw a ray diagram showing the path of the ray till it emerges out of the slab. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. When a ray of light falls normally on a rectangular glass slab, then it will be shown as follows :<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 2(b) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-2_472_473_2121_347.jpg\"\/><\/div>\n<p>So, in the given case, Angle of incidence = Angle of refraction = Angle of emergence = 0 i.e.,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\angle i=\\angle r=\\angle e=0\n\\]<\/div>\n<p>Hence, the above ray diagram is the required ray diagram which shows the path of the ray till it emerges out of the slab.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q2-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Complete the path of the monochromatic light ray AB incident on the surface PQ of the equilateral glass prism PQR till it emerges out of the prism due to refraction. <strong>[2]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 2(c) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-2_383_452_678_1191.jpg\"\/><\/div><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 2(c) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-2_274_464_1109_1155.jpg\"\/><\/div>\n<p>In the given ray diagram,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; \\angle i=\\text { Angle of incidence } \\\\\n&amp; \\angle e=\\text { Angle of emergence }\n\\end{aligned}\n\\]<\/div>\n<p>\\(\\angle r_{1}\\) and \\(\\angle r_{2}\\) are the angle of refractions.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; A B=\\text { Incident ray } \\\\\n&amp; B C=\\text { Refracted ray } \\\\\n&amp; C D=\\text { Emergent ray. }\n\\end{aligned}\n\\]<\/div>\n<p>So, this is the required path of the monochromatic light ray AB incident on the surface PQ of equilateral glass prism PQR till it emerges out of the prism due to refraction.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q2-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>Where should an object be placed in front of a convex lens in order to get :<\/p>\n<p>(i) an enlarged real image<\/p>\n<p>(ii) enlarged virtual image ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans.<\/p>\n<p>(i) To get an enlarged real image, the object should be placed (in front of a convex lens) between F and 2F.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 2(d) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-2_310_687_2176_1130.jpg\"\/><\/div>\n<p>(ii) To get an enlarged virtual image, the object should be placed (in front of a convex lens) between F and O.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 2(d) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-3_316_567_270_382.jpg\"\/><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q2-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>A pond appears to be 2.7 m deep. If the refractive index of water is \\(\\frac{4}{3}\\), find the actual depth of the pond. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Actual depth of pond = ? Apparent depth of pond \\(=2.7 \\mathrm{~m}\\).<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Refractive index of water }=\\frac{4}{3}\n\\]<\/div>\n<p>\u2234 Refractive index of medium<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { Real depth or Actual depth }}{\\text { Apparent depth }}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{4}{3}=\\frac{\\text { Actual depth }}{2.7 \\mathrm{~m}}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Actual depth } &amp; =\\frac{4}{3} \\times 2.7 \\mathrm{~m} \\\\\n&amp; =4 \\times 0.9 \\mathrm{~m}\n\\end{aligned}\n\\]<\/div>\n<p>\\(\\Rightarrow\\) Actual depth \\(=3.6 \\mathrm{~m}\\)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">3<\/div><div><div class=\"xc-set-title\">Question 3<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q3-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The wave lengths for the light of red and blue colours are nearly \\(7.8 \\times 10^{-7} \\mathrm{~m}\\) and \\(4.8 \\times 10^{-7}\\) m respectively.<\/p>\n<p>(i) Which colour has the greater speed in a vacuum ?<\/p>\n<p>(ii) Which colour has a greater speed in glass ? <strong>[2]<\/strong><\/p>\n<\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) Since, speed does not change in vacuum, therefore, all colours will move with the same speed, i.e., with the speed of light ( \\(3 \\times 10^{\\wedge 8}\\) m\/s).<\/p>\n<p>(ii) It is given that wavelength of red light, \\(\\lambda_{\\text {Red }}=\\) \\(7.8 \\times 10^{-7} \\mathrm{~m}\\) and<\/p>\n<p>wavelength of blue light \\(\\lambda_{\\text {blue }}=4.8 \\times 10^{-7} \\mathrm{~cm}\\) \u2235 Velocity = wavelength \u00d7 frequency<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\Rightarrow \\quad V=\\lambda f \\tag{i}\n\\]<\/div>\n<p>As we know that frequency of light does not change when it enters into a medium, i.e., \\(f\\) is constant. \u2234 eqn. (i) becomes<\/p>\n<div class=\"mathjax-overflow\">\\[\nV \\alpha \\lambda \\tag{i}\n\\]<\/div>\n<p>As given<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\lambda_{\\text {red }}&gt;\\lambda_{\\text {blue }} \\tag{iii}\n\\]<\/div>\n<p>from eqns (ii) and (iii), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\nV_{\\text {red }}&gt;V_{\\text {blue }}\n\\]<\/div>\n<p>It means that the red colour light will have greater speed in glass.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q3-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Draw a graph between displacement from mean position and time for a body executing free vibration in a vacuum. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 3(b) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-3_291_707_381_1107.jpg\"\/><\/div>\n<p>This is the required graph between the displacement from mean position and time for a body executing free vibrations in a vacuum.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q3-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A sound wave travelling in water has wavelength 0.4 m. 2 Is this wave audible in air ? (The speed of sound in water \\(\\boldsymbol{=} \\mathbf{1 4 0 0 ~ m s}{ }^{\\boldsymbol{- 1}}\\) ) <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. It is given that a sound wave is travelling in water has wavelength of 0.4 m, i.e., \\(\\lambda=0.4 \\mathrm{~m}\\) Speed of sound in water, \\(V=1400 \\mathrm{~ms}^{-1}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{rlrl}\n\\because &amp; &amp; \\text { Velocity } &amp; =\\text { wavelength \u00d7 Frequency } \\\\\n\\Rightarrow &amp; V &amp; =\\lambda \\times f \\\\\n\\Rightarrow &amp; &amp; f &amp; =\\frac{V}{\\lambda}=\\frac{1400}{0.4}=\\frac{14000}{4}=3500 \\mathrm{~Hz}\n\\end{array}\n\\]<\/div>\n<p>As we know that audible frequency ranges between 20 Hz to 20,000 Hz. And the obtained value of frequency, i.e., 3500 Hz lies between this range, i.e., in between 20 Hz and 20,000 Hz. So, yes this sound wave is audible in water.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q3-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>Why does stone lying in the sun get heated up much more than water lying for the same duration of time ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. The stone lying in the sun get heated much more than water lying for the same duration of time. It is because the specific heat capacity of stone is less than that of water, due to which the temperature of stone is more.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q3-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>Why is it not advisable to use a piece of copper wire as fuse wire in an electric circuit ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. It is not advisable to use a piece of copper wire as fuse wire in an electric circuit. It is because copper has low specific resistance and high melting point, due to which it will not break (fuse) during the overloading in the circuit.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">4<\/div><div><div class=\"xc-set-title\">Question 4<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q4-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Calculate the total resistance across AB : <strong>[2]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 4(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-3_155_690_2101_1125.jpg\"\/><\/div><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Consider the given diagram,<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 4(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-3_153_689_2302_1125.jpg\"\/><\/div>\n<p>Let<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; R_{1}=3 \\Omega \\\\\n&amp; R_{2}=6 \\Omega \\\\\n&amp; R_{3}=5 \\Omega\n\\end{aligned}\n\\]<\/div>\n<p>It can be seen from the figure that \\(\\mathrm{R}_{1}\\) and \\(\\mathrm{R}_{2}\\) are in parallel. So, their equivalent, \\(\\operatorname{Req}_{1}=\\frac{R_{1} R_{2}}{R_{1}+R_{2}}\\) \u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\operatorname{Req}_{1}=\\frac{3 \\times 6}{3+6}=\\frac{18}{9}=2 \\Omega\n\\]<\/div>\n<p>Now, the above given diagram can be shown as follows<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 4(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-4_110_581_615_402.jpg\"\/><\/div>\n<p>\u2235 \\(\\mathrm{Req}_{1}\\) and \\(\\mathrm{R}_{3}\\) are in series. \u2234 equivalent resistance across AB will be \\(R_{\\mathrm{AB}}=\\operatorname{Req}_{1}+R_{3}=(2+5) \\Omega=7 \\Omega\\).<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q4-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Two metallic blocks P and Q having masses in ratio 2:1 are supplied with the same amount of heat. If their temperatures rise by same degree, compare their specific heat capacities. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Consider a metallic block having mass &#8216; \\(m\\) &#8216;, specific heat capacity &#8216;S&#8217;, gains heat energy &#8216;Q&#8217; after rising the temperature &#8216; \\(\\Delta \\mathrm{T}\\) &#8216;, then the relation among them is given by, \u2235<\/p>\n<div class=\"mathjax-overflow\">\\[\nQ=m S \\Delta T \\tag{i}\n\\]<\/div>\n<p>For block A,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Mass } &amp; =2 m, \\\\\n\\text { Specific heat capacity } &amp; =S_{\\mathrm{P}} \\\\\n\\text { temperature } &amp; =\\Delta T\n\\end{aligned}\n\\]<\/div>\n<p>For block B,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Mass } &amp; =m \\\\\n\\text { Specific heat capacity } &amp; =S_{\\mathrm{Q}} \\\\\n\\text { temperature } &amp; =\\Delta T\n\\end{aligned}\n\\]<\/div>\n<p>As per question,<\/p>\n<div class=\"mathjax-overflow\">\\[\nQ_{\\mathrm{P}}=Q_{\\mathrm{Q}} \\tag{ii}\n\\]<\/div>\n<p>\u2234 from eqns. (i) and (ii), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{rlrl}\n\\Rightarrow &amp; &amp; 2 m S_{\\mathrm{P}} \\Delta T &amp; =m S_{\\mathrm{Q}} \\Delta T \\\\\n\\Rightarrow &amp; &amp; 2 S_{\\mathrm{P}} &amp; =\\mathrm{S}_{\\mathrm{Q}} \\\\\nS_{\\mathrm{P}} &amp; =\\frac{1}{2} S_{\\mathrm{Q}}\n\\end{array}\n\\]<\/div>\n<p>It means that specific heat capacity of block P is half of the capacity of block Q.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q4-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>When a current carrying conductor is placed in a magnetic field, it experiences a mechanical force. What should be the angle between the magnetic field and the length of the conductor so that the force experienced is :<\/p>\n<p>(i) Zero<\/p>\n<p>(ii) Maximum ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. It is given that when a current carrying conductor is placed in a magnetic field, it experiences a mechanical force \\((F=i B L \\sin \\theta)\\).<\/p>\n<p>(i) So, the force experienced will be zero when the angle between the magnetic field and the length of the conductor is 0\u00b0.<\/p>\n<p>(ii) And the force experienced will be maximum when the angle between the magnetic field and the length of conductor is 90\u00b0.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q4-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>A nucleus \\({ }_{84} \\mathrm{X}^{202}\\) of an element emits an alpha particle followed by a beta particle. The final nucleus is \\({ }_{\\boldsymbol{a}} \\boldsymbol{Y}^{\\boldsymbol{b}}\\). Find \\(\\boldsymbol{a}\\) and \\(\\boldsymbol{b}\\). <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. According to the question, a nucleus \\({ }_{84} \\mathrm{X}^{202}\\) of an element emits an alpha particle followed by a beta particle<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; { }_{84} \\mathrm{X}^{202} \\xrightarrow{\\alpha \\text {-emision }}{ }_{82} \\mathrm{Z}^{198}+{ }_{2}^{4} \\mathrm{He} \\\\\n&amp; { }_{82} \\mathrm{Z}^{198} \\xrightarrow{\\beta \\text {-emision }}{ }_{83} \\mathrm{Y}^{198}+{ }_{-1}^{0} e \\\\\n&amp; \\therefore\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q4-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>The diagram below shows a loop of wire carrying current I :<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 4(e) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-4_367_300_1212_1267.jpg\"\/><\/div>\n<p>(i) What is the magnetic polarity of the loop that faces us ?<\/p>\n<p>(ii) With respect to the diagram how can we increase the strength of the magnetic field produced by this loop ? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) The magnetic polarity of the loop, that faces us will be clockwise, i.e., south pole.<\/p>\n<p>(ii) We can increase the strength of the magnetic field produced by the loop in the given diagram, by.<\/p>\n<p>1. increasing number of turns of coil.<\/p>\n<p>2. increasing the current.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-section-heading\">\n<div class=\"xc-section-heading-icon\">\ud83d\udcdd<\/div>\n<div class=\"xc-section-heading-body\"><div class=\"xc-section-heading-label\">ICSE Physics 2020<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"SECTION_II\"><\/span>SECTION II<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">40 MARKS \u2022 Attempt any four questions from this Section.<\/div><\/div>\n<div class=\"xc-section-heading-stat\"><div class=\"xc-section-heading-stat-num\">40<\/div><div class=\"xc-section-heading-stat-label\">Marks<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">5<\/div><div><div class=\"xc-set-title\">Question 5<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q5-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The figure below shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at the position A is zero. At the position B the pendulum bob is raised by 5 m.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 5(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-4_351_560_2203_1139.jpg\"\/><\/div>\n<p>(i) What is the potential energy of the pendulum at the position B ?<\/p>\n<p>(ii) What is the total mechanical energy at point C ?<\/p>\n<p>(iii) What is the speed of the bob at the position A when released from B ?<\/p>\n<p>(Take \\(\\mathbf{g}=\\mathbf{1 0 ~ \\mathbf { m s } ^ { &#8211; \\mathbf { 2 } }}\\) and there is no loss of energy.) <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Consider the given diagram,<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 5(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-5_351_560_640_305.jpg\"\/><\/div>\n<p>It is given that Mas of simple pendulum, \\(m=200 \\mathrm{~g}=0.2 \\mathrm{~kg}\\). Potential energy at position A is zero, i.e., \\(P E_{\\mathrm{A}}=0\\).<\/p>\n<p>(i) Potential energy at position \\(\\mathrm{PE}, P E_{\\mathrm{B}}=m g h\\) \\(=0.2 \\times 10 \\times 5=10 \\mathrm{~J}\\).<\/p>\n<p>(ii) Whenever the bob of the pendulum rotates, then mechanical energy remains same at all the points according to the law of conservation of mechanical energy.<\/p>\n<p>\u2234 Mechanical energy at point \\(C=10 \\mathrm{~J}\\).<\/p>\n<p>(iii) As the bob is rotating, so the potential energy of bob at position B will get converted into kinetic energy at point A.<\/p>\n<p>\u2235 KE at position \\(\\mathrm{A}, K E_{\\mathrm{A}}=\\frac{1}{2} m v^{2}\\) \\(\\Rightarrow \\quad 10=\\frac{1}{2} \\times 0.2 \\times v^{2}\\) \\(\\Rightarrow \\quad 100=v^{2}\\) \\(\\Rightarrow \\quad v=10 \\mathrm{~ms}^{-1}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q5-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) With reference to the direction of action, how does a centripetal force differ from a centrifugal force during uniform circular motion ?<\/p>\n<p>(ii) Is centrifugal force the force of reaction of centripetal force ?<\/p>\n<p>(iii) Compare the magnitudes of centripetal and centrifugal force. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) With reference to the direction of motion, both centrifugal force and centripetal force acts in opposite direction. Centripetal force acts towards the centre while centrifugal force acts radially outwards.<\/p>\n<p>(ii) No, the centrifugal force is not the force of reaction at centripetal force. It is because the action-reaction pair of forces acts on two bodies,, but here it is acting on a single body.<\/p>\n<p>(iii) The magnitudes of centripetal and centrifugal force are equal.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q5-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A block and tackle system of pulleys has velocity ratio 4.<\/p>\n<p>(i) Draw a neat labelled diagram of the system indicating clearly the points of application and direction of load and effort.<\/p>\n<p>(ii) What will be its V.R. if the weight of the movable block is doubled ? <strong>[4]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 5(c) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-5_487_326_552_1304.jpg\"\/><\/div>\n<p>So, this is the required diagram.<\/p>\n<p>(ii) \u2234 Velocity ratio is always equals to \\(n\\).<\/p>\n<p>So, if the weight of movable block is doubled, then its mechanical advantage will get effected but the velocity ratio will remain constant.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">6<\/div><div><div class=\"xc-set-title\">Question 6<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q6-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>A diver in water looks obliquely at an object AB in air.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 6(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-5_297_689_1332_1073.jpg\"\/><\/div>\n<p>(i) Does the object appear taller, shorter or of the same size to the diver ?<\/p>\n<p>(ii) Show the path of two rays AC and AD starting from the tip of the object as it travels towards the diver in water and hence obtain the image of the object. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Consider the given diagram,<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 6(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-5_259_516_1912_1159.jpg\"\/><\/div>\n<p>(i) and (ii)<\/p>\n<p>As we know that when a ray of light enters from air to water, i.e., from rarer to denser medium, it bends toward the normal. So, the above diagram will become as follows : air water 3<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q6-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Complete the path of the ray AB through the glass prism in PQR till it emerges out of the prism. Given the critical angle of the glass as 42\u00b0. <strong>[3]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 6(b) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-6_463_302_936_432.jpg\"\/><\/div><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 6(b) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-6_531_455_1402_359.jpg\"\/><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q6-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A lens of focal length 20 cm forms an inverted image at a distance 60 cm from the lens.<\/p>\n<p>(i) Identify the lens.<\/p>\n<p>(ii) How far is the lens present in front of the object&#8217; ?<\/p>\n<p>(iii) Calculate the magnification of the image. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) As per the information given in the question, we can conclude that the given lens is a convex lens because the formed image is inverted.<\/p>\n<p>(ii) Given,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; f=20 \\mathrm{~cm} \\\\\n&amp; v=60 \\mathrm{~cm}\n\\end{aligned}\n\\]<\/div>\n<p>According to lens formula,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{f}=\\frac{1}{v}-\\frac{1}{u}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\Rightarrow \\quad \\frac{1}{u}=\\frac{1}{v}-\\frac{1}{f}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{u}=\\frac{1}{60}-\\frac{1}{20}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{u}=\\frac{-2}{60}\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\nu=-30 \\mathrm{~cm}\n\\]<\/div>\n<p>(iii) Magnification, \\(m=\\frac{v}{u}=\\frac{60}{-30}=-2\\)<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">7<\/div><div><div class=\"xc-set-title\">Question 7<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q7-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Give reasons for the following : 3 During the day :<\/p>\n<p>(i) Clouds appear white.<\/p>\n<p>(ii) Sky appears blue. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) As clouds scatter all the colours of sunlight (i.e., VIBGYOR scatter) in the sky during the day, therefore, due to this reason, the clouds appear white.<\/p>\n<p>(ii) Sky appears blue because when the light enters from the sun into the sky, the air layers scatter only blue part of light more than any other colour.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q7-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) Name the system which enables us to locate underwater objects by transmitting ultrasonic waves and detecting the reflecting impulse.<\/p>\n<p>(ii) What are acoustically measurable quantities related to pitch and loudness ? <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) SONAR, i.e., Sound Navigation and Ranging is a system which enables us to locate underwater objects by transmitting ultrasonic waves and detecting the reflecting impulse.<\/p>\n<p>(ii) Frequency and Amplitude are respectively measurable quantities related to pitch and loudness.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q7-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>(i) When a tuning fork [vibrating] is held close to ear, one hears a faint hum. The same [vibrating tuning fork] is held such that its stem is in contact with the table surface, then one hears a loud sound. Explain.<\/p>\n<p>(ii) A man standing in front of a vertical cliff fires a gun. He hears the echo after 3.5 seconds. On moving closer to the cliff by 84 m, he hears the echo after 3 seconds. Calculate the distance of the cliff from the initial position of the man. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) When a tuning fork is held close to ear, it has some sound energy. But when it is held such that its stem is in contact with the table surface, then due to the larger surface area of table top, it produces more vibrations with greater sound energy.<\/p>\n<p>(ii) Let the man is standing infront of cliff at a distance of \\(d \\mathrm{~cm}\\) from the cliff.<\/p>\n<p>So, when he moves closer (after hearing echo) to the cliff by 84 m, then the distance will be \\((d-84) \\mathrm{cm}\\). \u2235 we know that<\/p>\n<div class=\"mathjax-overflow\">\\[\nV=\\frac{2 d}{t}\n\\]<\/div>\n<p>Since, velocity will remain same in both cases, therefore,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\frac{2 d}{t_{1}} &amp; =\\frac{2(d-84)}{t_{2}} \\\\\n\\Rightarrow \\quad \\frac{2 d}{3.5} &amp; \\left.=\\frac{2(d-84)}{3}=3.5 \\mathrm{~s} t_{2}=3 \\mathrm{~s}\\right\\} \\\\\n\\frac{2 d}{7} &amp; =\\frac{d-84}{3} \\Rightarrow 6 d=7 d-588 \\\\\n\\Rightarrow \\quad d &amp; =588 \\mathrm{~m} .\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">8<\/div><div><div class=\"xc-set-title\">Question 8<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q8-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The diagram below shows the core of a transformer and its input and output connections.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 8(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-7_215_576_1013_318.jpg\"\/><\/div>\n<p>(i) State the material used for the core.<\/p>\n<p>(ii) Copy and complete the diagram of the transformer by drawing input and output coils. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) Soft iron core are made of thin laminated sheets which are eight (8) T or U shaped.<\/p>\n<p>(ii) Since the input voltage is greater than the output voltage, therefore, the given transformer is a step-down transformer. So, the given diagram will become as follows :<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 8(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-7_250_733_1651_250.jpg\"\/><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q8-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) What are superconductors ?<\/p>\n<p>(ii) Calculate the current drawn by an appliance rated 110 W, 220 V when connected across 220 V supply.<\/p>\n<p>(iii) Name a substance whose resistance decreases with the increase in temperature. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) The substances which have almost zero resistance at low temperatures, are called superconductors.<\/p>\n<p>(ii) Current drawn by the appliance,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&amp; I=\\frac{\\text { Power }}{\\text { Voltage }} \\\\\n&amp; \\Rightarrow \\quad I=\\frac{P}{V}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\Rightarrow &amp; I=\\frac{110 \\mathrm{~W}}{220 \\mathrm{~V}}=\\frac{1}{2} \\mathrm{~A}=0.5 \\mathrm{~A} . \\\\\n\\Rightarrow &amp; I=0.5 \\mathrm{~A}\n\\end{array}\n\\]<\/div>\n<p>(iii) We know that semiconductor&#8217;s resistance decreases with the increase in temperature.<\/p>\n<p>For examples : Silicon, Graphite, etc.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q8-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 8(c) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-7_297_663_599_1155.jpg\"\/><\/div>\n<p>The diagram above shows three resistors connected across a cell of e.m.f. 1.8 V and internal resistance \\(r\\). Calculate :<\/p>\n<p>(i) Current through \\(3 \\Omega\\) resistor.<\/p>\n<p>(ii) The internal resistance \\(r\\). <strong>[4]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Consider the given diagram.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 8(c) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-7_316_630_1161_1123.jpg\"\/><\/div>\n<p>(i) Since, \\(3 \\Omega\\) and \\(1.5 \\Omega\\) resistor are in parallel combination. Therefore the potential difference across them will remain same.<\/p>\n<p>Let \\(\\mathrm{I}_{1}\\) be the current passing through \\(3 \\Omega\\) resistor and \\(\\mathrm{I}_{2}\\) be the current passing through \\(1.5 \\Omega\\) resistor.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\Rightarrow V=3 I_{1} \\text { and } V=1.5 I_{2} \\quad\\{\\because V=I R\\}\n\\]<\/div>\n<p>\u2235 potential difference across them is same,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{align*}\n\\therefore &amp; &amp; 3 I_{1} &amp; =1.5 I_{2} \\\\\n\\Rightarrow &amp; &amp; I_{1} &amp; =0.5 I_{2} \\\\\n\\Rightarrow &amp; &amp; I_{1} &amp; =\\frac{I_{2}}{2} \\Rightarrow I_{2}=2 I_{1} \\tag{i}\n\\end{align*}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{align*}\n\\because &amp; I_{1}+I_{2} &amp; =I &amp; \\{\\because \\mathrm{I}=0.3 \\mathrm{~A}\\} \\\\\n\\Rightarrow &amp; I_{1}+I_{2} &amp; =0.3 &amp; \\ldots(\\mathrm{ii}) \\tag{ii}\n\\end{align*}\n\\]<\/div>\n<p>From eqns. (i) and (ii), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\Rightarrow &amp; I_{1}+2 I_{1}=0.3 \\\\\n\\Rightarrow &amp; 3 I_{1}=0.3 \\\\\n\\Rightarrow &amp; I_{1}=0.1 \\mathrm{~A} \\text { and } I_{2}=2 \\times 0.1=0.2 \\mathrm{~A} \\\\\n\\Rightarrow &amp; I_{1}=0.1 \\mathrm{~A} \\text { and } I_{2}=0.2 \\mathrm{~A} .\n\\end{array}\n\\]<\/div>\n<p>So, the current through \\(3 \\Omega\\) resistor will be, \\(I_{1}=0.1 \\mathrm{~A}\\).<\/p>\n<p>(ii) \\(\\because 3 \\Omega\\) and \\(1.5 \\Omega\\) resistors are in parallel combination,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad R \\mathrm{eq}_{1}=\\frac{3 \\times 1.5}{3+1.5}=\\frac{4.5}{4.5}=1 \\Omega\n\\]<\/div>\n<p>Now, \\(\\operatorname{Req}_{1}\\) and \\(4 \\Omega\\) are in series combination, therefore, their equivalent resistance,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\operatorname{Req}_{2}=(1+4) \\Omega=5 \\Omega\n\\]<\/div>\n<p>\u2234 Total resistance, \\(R_{\\text {tot }}=R e q_{2}+\\) Internal resistance<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\therefore &amp; R_{\\mathrm{tot}}=(5+r) \\Omega \\qquad\\text{&#8230;(iii)} \\\\\n\\because &amp; R_{\\mathrm{tot}}=\\frac{V}{I}=\\frac{1.8}{0.3}=6 \\Omega \\qquad\\text{&#8230;(iv)}\n\\end{array}\n\\]<\/div>\n<p>\u2234 from equn (iii) and (iv), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n5+r=6 \\Rightarrow r=6-5=1\n\\]<\/div>\n<p>\u21d2<\/p>\n<div class=\"mathjax-overflow\">\\[\nr=1 \\Omega\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">9<\/div><div><div class=\"xc-set-title\">Question 9<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q9-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>(i) Define heat capacity of a substance.<\/p>\n<p>(ii) Write the SI unit of heat capacity.<\/p>\n<p>(iii) What is the relationship between heat capacity and specific heat capacity of a substance ? <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) The amount of heat which is to be supplied to a given mass of a material to produce a unit change in its temperature, is known as heat capacity.<\/p>\n<p>(ii) The SI unit of heat capacity is joule per kelvin.<\/p>\n<p>(iii) \u2235 Heat capacity = mass \u00d7 specific heat capacity.<\/p>\n<p>\\(\\Rightarrow \\quad C=\\mathrm{mS}\\) So, this is the required relationship between heat capacity and specific heat capacity of a substance.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q9-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The diagram below shows the change of phases of a substance on a temperature vs time graph on heating the substance at a constant rate.<\/p>\n<p>(i) Why is the slope of CD less than slope of AB?<\/p>\n<p>(ii) What is the boiling and melting point of the substance ? <strong>[3]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 9(b) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-8_362_516_1735_327.jpg\"\/><\/div><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) In the given graph, AB represents the specific heat of solid phase and CD represents the specific heat of Liquid phase. Since, the specific heat of solid phase is always greater than that of liquid phase. So, due to this reason, the slope of CD is less than that of AB.<\/p>\n<p>(ii) \\(t_{2}\\) is the boiling point because, here solid is changing into liquid and \\(t_{1}\\) is the melting point because, here liquid is changing into gas.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q9-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A piece of ice of mass 60 g is dropped into 140 g of water at 50\u00b0C. Calculate the final temperature of water when all the ice has melted. 4 (Assume no heat is lost to the surrounding) Specific heat capacity of water \\(=4.2 \\mathrm{Jg}^{-1} \\mathrm{k}^{-1}\\) Specific latent heat of fusion of ice \\(\\boldsymbol{=} \\mathbf{3 3 6 ~ J g}^{\\boldsymbol{-} \\mathbf{1}}\\) <strong>[4]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. For Ice,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Mass } &amp; =60 \\mathrm{~g} \\\\\n\\text { Temperature } &amp; =0^{\\circ} \\mathrm{C}\n\\end{aligned}\n\\]<\/div>\n<p>For water,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Mass } &amp; =140 \\mathrm{~g} \\\\\n\\text { Temperature } &amp; =50^{\\circ} \\mathrm{C} .\n\\end{aligned}\n\\]<\/div>\n<p>Let T be the final temperature of water. Now, apply Principle of Calorimetry,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{rlrl}\n\\Rightarrow &amp; 60 \\times 336+60 \\times 4.2 \\times T=140 \\times 4.2(50-T) &amp; \\\\\n\\Rightarrow &amp; &amp; 60 \\times 336=4.2[140 \\times 50-140 T-60 T] \\\\\n&amp; \\Rightarrow &amp; \\frac{60 \\times 336}{4.2}=7000-200 T \\\\\n&amp; \\Rightarrow &amp; \\frac{60 \\times 336 \\times 10}{42}=7000-200 T \\\\\n&amp; \\Rightarrow &amp; 4800=7000-200 T \\\\\n\\Rightarrow &amp; 200 T=7000-4800 \\\\\n\\Rightarrow &amp; 200 T=2200 \\\\\n\\Rightarrow &amp; T=\\frac{2200}{200} \\\\\n\\Rightarrow &amp; T=11^{\\circ} \\mathrm{C} .\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">10<\/div><div><div class=\"xc-set-title\">Question 10<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2020<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q10-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>(i) Draw a neat labelled diagram of a d.c. motor.<\/p>\n<p>(ii) Write any one use of a d.c. motor. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img alt=\"ICSE Class 10 Physics 2020 Question 10(a) figure\" decoding=\"async\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/212fcc9b-c4fe-417f-ba9f-d8e610e5e433-8_615_500_1509_1214.jpg\"\/><\/div>\n<p>(ii) D.C. motor is used to convert electrical energy into mechanical energy, such as in electric fan etc.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q10-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) Differentiate between nuclear fusion and nuclear fission.<\/p>\n<p>(ii) State one safety precaution in the disposal of nuclear waste. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. (i) \\begin{tabular}[t]{|l|l|} \\hline Nuclear Fusion &amp; Nuclear Fission \\\\ \\hline 1. In this, smaller nuclei combines to form a larger one. &amp; In this, a heavier nuclei breaks into the smaller nuclei. \\\\ \\hline 2. Takes place at high temperature &amp; Takes place at very very high temperature \\\\ \\hline \\end{tabular}<\/p>\n<p>(ii) The solid radioactive waste should be sealed in red bag and placed in a cardboard radioactive waste box.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2020-q10-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>An atomic nucleus A is composed of 84 protons and 128 neutrons. The nucleus. A emits an alpha particle and is transformed into a nucleus B.<\/p>\n<p>(i) What is the composition of B ?<\/p>\n<p>(ii) The nucleus B emits a beta particle and is transformed into a nucleus C.<\/p>\n<p>What is the composition of C ?<\/p>\n<p>(iii) What is mass number of the nucleus A ?<\/p>\n<p>(iv) Does the composition of C change if it emits gamma radiations ? <strong>[4]<\/strong><\/p><\/div><\/div>\n<div aria-expanded=\"false\" class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\"><span class=\"xc-toggle-icon\">\u25bc<\/span> View Answer &amp; Solution<\/div>\n<div class=\"xc-explanation\"><div class=\"xc-expl-head\">Answer &amp; Solution<\/div><p>Ans. Nucleus A is composed of 84 protons and 128 neutrons. \u2235<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { Mass number }= &amp; \\text { Number of protons } \\\\\n&amp; \\quad+\\text { number of neutrons } \\\\\n= &amp; 84+128 \\\\\n= &amp; 212\n\\end{aligned}\n\\]<\/div>\n<p>\u2234 The composition of nucleus A will be \\({ }_{84}^{212} \\mathrm{~A}\\)<\/p>\n<p>(i) Since, nucleus A emits an \\(\\alpha\\)-particle, i.e.,<\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }_{84}^{212} \\mathrm{~A} \\xrightarrow{\\alpha \\text {-decay }}{ }_{82}^{208} \\mathrm{~B}+{ }_{2}^{4} \\mathrm{He}\n\\]<\/div>\n<p>\u2234 The composition of nucleus B will be \\({ }_{82}^{208} \\mathrm{~B}\\)<\/p>\n<p>(ii) Since, nucleus B emits a \\(\\beta\\)-particle, i.e.,<\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }_{82}^{208} \\mathrm{~B} \\xrightarrow{\\beta \\text {-decay }}{ }_{83}^{208} \\mathrm{C}+{ }_{-1}^{0} e\n\\]<\/div>\n<p>\u2234 The composition of nucleus C will be \\({ }_{83}^{208} \\mathrm{C}\\).<\/p>\n<p>(iii) The mass number of the nucleus A is 212.<\/p>\n<p>(iv) If the nucleus C emits gamma radiations, then there will be no change in its composition.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p><span class=\"rt-reading-time\" style=\"display: block;\"><span class=\"rt-label rt-prefix\">Reading Time: <\/span> <span class=\"rt-time\">35<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Physics PYQ Practice the complete ICSE Class 10 Physics 2020 Question Paper with Solutions, including numericals, diagrams [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","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":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center 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