{"id":2218,"date":"2026-11-12T23:12:17","date_gmt":"2026-11-12T17:42:17","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2218"},"modified":"2026-11-12T23:12:20","modified_gmt":"2026-11-12T17:42:20","slug":"icse-class-10-physics-2019-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2019-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Physics 2019 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-2019-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-2019-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\">14<\/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 2019 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\">2019<\/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 2019<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q1-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>The diagram below shows a claw hammer used to remove a nail: (i) To which class of lever does it belong?<\/p>\n<p>(ii) Give one more example of the same class of lever mentioned by you in (i) for which the mechanical advantage is greater than one. <strong>[2]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-1_303_551_1432_342.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 1(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) This claw hammer and nail arrangement belongs to first class of lever. Here, the fulcrum (hammer rests at this point) is closer to the output force than to the input force (force applied by the user).<\/p>\n<p>(ii) Example of class 1 lever where mechanical advantage is greater than one is bottle opener. A small force applied on the bottle opener, exerts a larger force on the cap.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q1-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Two bodies A and B have masses in the ratio 5:1 and their kinetic energies are in the ratio 125:9. Find the ratio of their velocities. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. (b) Given, ratio of masses of bodies \\(m_{A}: m_{B}=5: 1\\) Ratio of their kinetic energies, \\(K_{A}: K_{B}=125: 9\\) The formula for kinetic energy \\(K=\\frac{1}{2} m v^{2}\\) Using the formula for bodies A snd B, we get<\/p>\n<div class=\"mathjax-overflow\">\\[\nK_{A}=\\frac{1}{2} m_{A V}{ }^{2}{ }_{A} \\text { and } K_{B}=\\frac{1}{2} m_{B V}{ }^{2}{ }_{B}\n\\]<\/div>\n<p>Dividing \\(\\mathrm{K}_{\\mathrm{A}}\\) by \\(\\mathrm{K}_{\\mathrm{B}}\\), we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{K_{A}}{K_{B}}=\\frac{m_{A} v_{A}^{2}}{m_{B} v_{B}^{2}}\n\\]<\/div>\n<p>Now using the data given above, we get<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{125}{9}=\\frac{5}{1} \\times \\frac{v_{A}^{2}}{v_{B}^{2}} \\text { or } \\frac{v_{A}^{2}}{v_{B}^{2}}=\\frac{25}{9} \\text { or } \\frac{v_{A}}{v_{B}}=\\frac{5}{3}\n\\]<\/div>\n<p>Thus the ratio of velocities \\(\\mathrm{v}_{\\mathrm{A}}: \\mathrm{v}_{\\mathrm{B}}=5: 3\\).<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q1-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>(i) Name the physical quantity which is measured in calories. (ii) How is calorie related to the S.I unit of that quantity? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Heat energy is measured in calories.<\/p>\n<p>(ii) S.I unit of the quality is Joule and the relation between them is<\/p>\n<div class=\"mathjax-overflow\">\\[\n1 \\text { Calorie = 4.12 J }\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q1-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>(i) Define couple. (ii) State the S.I. unit of moment of couple. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Couple: Couple is the set of two equal and opposite forces having different line of action. A couple produces rotation without translation. Forces are separated by a distance.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-1_281_538_1976_1148.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 1(d) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) S.I unit of the moment of couple is Newton meter (Nm).<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q1-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>(i) Define critical angle. (ii) State one important factor which affects the critical angle of a given medium. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Critical Angle: When the angle of incidence of a light ray in a denser medium is such that the angle of refraction becomes 90\u00b0, the angle of incidence is called the critical angle. The basic condition for this is to occur light to be incident from denser to rarer medium.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-2_311_564_432_301.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 1(e) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) Important factor which affects the critical angle is the ratio of the refraction indices of the two media.<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q2-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>An electromagnetic radiation is used for photography in fog. (i) Identify the radiation.<\/p>\n<p>(ii) Why is this radiation mentioned by you, ideal for this purpose? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Radiation used for photography in fog is infrared radiation.<\/p>\n<p>(ii) Infra-red radiation is a long wavelength and low frequency radiation. Hence, the energy associated with them is also low, so they can penetrate deep into the fog.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q2-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) What is the relation between the refractive index of water with respect to air \\(\\left({ }_{\\mathbf{a}} \\mu_{\\mathbf{w}}\\right)\\) and the refractive index of air with respect to water ( \\({ }_{\\mathrm{w}} \\mu_{\\mathrm{a}}\\) ). (ii) If the refractive index of water with respect to air \\(\\left({ }_{\\mathbf{a}} \\mu_{\\mathbf{w}}\\right)\\) is \\(\\frac{5}{3}\\). Calculate the refractive index of air with respect to water \\(\\left({ }_{\\mathbf{w}} \\boldsymbol{\\mu}_{\\mathbf{a}}\\right)\\). <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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=\"mathjax-overflow\">\\[\n{ }_{\\mathrm{a}} \\mu_{\\mathrm{w}}=\\frac{1}{{ }_{\\mathrm{w}} \\mu_{\\mathrm{a}}}\n\\]<\/div>\n<p>(ii) Given, refractive index of water, with respect to air \\({ }_{\\mathrm{a}} \\mu_{\\mathrm{w}}=\\frac{5}{3}\\)<\/p>\n<p>\u2234 refractive index of air with respect to water<\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }_{w} \\mu_{a}=\\frac{3}{5}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q2-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The specific heat capacity of a substance A is \\(\\mathbf{3 , 8 0 0} \\mathbf{J ~ k g}^{\\boldsymbol{-} \\mathbf{1}} \\mathbf{K}^{\\boldsymbol{-} \\mathbf{1}}\\) and that of a substance B is \\(400 \\mathrm{~J} \\mathrm{~kg}^{-\\mathbf{1}} \\mathrm{K}^{1}\\). Which of the two substances is a good conductor of heat? Give a reason for your answer. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. From definition, specific heat capacity of a substance is defined as the amount of heat that raises the temperature of 1 kg substance through 1K. Given specific heat capacity of substance A \\(=3800 \\mathrm{~J} \\mathrm{~kg}^{-1} \\mathrm{~K}^{-1}\\) and specific heat capacity of substance \\(\\mathrm{B}=400 \\mathrm{~J} \\mathrm{~kg}^{-1} \\mathrm{~K}^{-1}\\). Thus, the amount of heat required to raise the temperature of substance A is much more than that required for substance B. The substance B is a good conductor of heat.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q2-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>A man playing a flute is able to produce notes of different frequencies. If he closes the holes near his mouth, will the pitch of the note produced, increase or decrease? Give a reason. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 flute is played, the pitch of the note is determined by the length of tube between the mouthpiece and the first open hole. Opening holes to shorten the tube makes the pitch rise, and closing holes to lengthen the tube makes the pitch fall.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q2-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>The diagram below shows a light source P embedded in a rectangular glass block ABCD of critical angle 42\u00b0. Complete the path of the ray PQ till it emerges out of the block. [Write necessary angles.] <strong>[2]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-2_284_485_1100_1233.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 2(e) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Incidence angle is the angle that incident ray make with the normal to the surface.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-2_519_653_1488_1090.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 2(e) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q3-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>(i) If the lens is placed in water instead of air, how does its focal length change? (ii) Which lens, thick or thin has greater focal length? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Let \\(f_{a}\\) be the focal length of lens in air and has the refractive index \\({ }_{w} \\mu_{g}\\). Apply lens maker formula,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{f_{0}}=\\left({ }_{0} \\mu_{g}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\right) \\tag{i}\n\\]<\/div>\n<p>Let \\(f_{w}\\) be the focal length of lens in water when emerged in water where in the refractive index is \\({ }_{w} \\mu_{g}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{f_{\\omega}}=\\left({ }_{\\omega} \\mu_{g}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\right) \\tag{ii}\n\\]<\/div>\n<p>From equation (i) and (ii)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{f_{w}}{f_{a}}=\\frac{{ }_{a} \\mu_{g}-1}{{ }_{w} \\mu_{g}-1}\n\\]<\/div>\n<p>Since, \\(\\left({ }_{a} \\mu_{g}-1\\right)>\\left({ }_{w} \\mu_{g}-1\\right)\\) So, \\(f_{w}>f_{a}\\) Hence , the focal length will be more in water than in air.<\/p>\n<p>(ii) Thin lens has greater focal length.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q3-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>Two waves of the same pitch have amplitudes in the ratio 1:3. What will be the ratio of their: (i) intensities and<\/p>\n<p>(ii) frequencies? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 amplitude ratio of two waves<\/p>\n<div class=\"mathjax-overflow\">\\[\nA_{1}: A_{2}=1: 3\n\\]<\/div>\n<p>As intensity I:A \\({ }^{2}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\therefore \\quad \\frac{I_{1}}{I_{2}}=\\frac{A_{1}^{2}}{A_{2}^{2}}=\\frac{1^{2}}{3^{2}}\n\\]<\/div>\n<p>\\(\\therefore \\quad I_{1}: I_{2}=1: 9\\)<\/p>\n<p>(ii) Ratio of frequencies \\(=1: 1\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q3-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>How does an increase in the temperature affect the specific resistance of a: (i) Metal and<\/p>\n<p>(ii) Semiconductor? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. With the increase in temperature, the specific resistance<\/p>\n<p>(i) of metals increases.<\/p>\n<p>(ii) of semiconductors decreases.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q3-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>(i) Define resonant vibrations. (ii) Which characteristic of sound, makes it possible to recognize a person by his voice without seeing him? <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Each and every object in universe has its natural frequency. If any external object is made to vibrate at the same frequency of the given object then the amplitude of vibration becomes large and the system is said to be in resonance. These are known as resonant vibrations and the frequency is called resonant frequency.<\/p>\n<p>(ii) Frequency makes it possible for us to recognise a person by his voice without seeing him.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q3-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>Is it possible for a hydrogen \\(\\left(\\begin{array}{l}1 \\\\ 1\\end{array} \\mathrm{H}\\right)\\) nucleus to emit an alpha particle? Give a reason for your answer. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Since, \\(\\alpha\\)-particles has \\({ }_{2} \\mathrm{He}^{4}\\) nuclei. Thus the \\(\\alpha\\)-particles contain two protons and two neutrons. Hence, emission of \\(\\alpha\\)-particle reduces the atomic number by two and atomic weight by four. Thus, \\(\\left({ }_{1}^{1} \\mathrm{H}\\right)\\) hydrogen nucleus cannot emit an \\(\\alpha\\)-particle<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q4-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Calculate the effective resistance across AB: <strong>[2]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-3_290_734_474_1049.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 4(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Equivalent network<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-3_185_803_837_1015.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 4(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Since, \\(R_{2}\\) and \\(R_{3}\\) are in series, sot their equivalent resistance \\(R_{23}=R_{2}+R_{3}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n=5 \\Omega+4 \\Omega=9 \\Omega\n\\]<\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-3_214_716_1201_1058.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 4(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Again \\(R_{2,3} \\| R_{4}\\), so their equivalent resistance \\(R^{\\prime}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{R^{\\prime}}=\\frac{1}{9}+\\frac{1}{3}\n\\]<\/div>\n<p>or<\/p>\n<div class=\"mathjax-overflow\">\\[\nR^{\\prime}=\\frac{9}{4} \\Omega\n\\]<\/div>\n<p>Finally \\(R_{1}\\) and \\(R^{\\prime}\\) are in series.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-3_101_577_1750_1100.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 4(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>So, their equivalent resistance \\(R\\) in given as<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; R=R^{\\prime}+R_{1}=\\frac{9}{4}+8=\\frac{41}{4} \\Omega \\\\\n&#038; R=10.25 \\Omega \\\\\n&#038; \\quad \\frac{41}{4} \\Omega=10.25 \\Omega\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q4-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) State whether the specific heat capacity of a substance remains the same when its state changes from solid to liquid. (ii) Give one example to support your answer. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Specific heat capacity of a substance changes when its state changes from solid to liquid.<\/p>\n<p>(ii) Specific heat capacity tells you how much energy one must put in per unit mass in order to<\/p>\n<p>raise temperature by 1\u00b0C. When you come out of shower you often feel cold. This is because the water on you is evaporating, for this heat flows from your body to the body droplets in order to change their phase from water to gas. You are losing heat and thus feel cold.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q4-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>A magnet kept at the centre of two coils A and B is moved to and fro as shown in the diagram. The two galvanometers show deflection.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-4_293_670_635_289.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 4(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-4_278_497_972_360.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 4(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>State with a reason whether: \\(x>y\\) or \\(x<y\\). [x and y are magnitudes of deflection.] <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. \\(x<y\\), as the number of turns of coil in B is greater than that in A, hence the larger amount of current is produced in B and hence greater deflection.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q4-d\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">d<\/div><div class=\"xc-qtext\"><p>(i) Why is a nuclear fusion reaction called a thermo nuclear reaction? (ii) Complete the reaction: <strong>[2]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }^{3} \\mathrm{He}_{2}+{ }^{2} \\mathrm{H}_{1} \\longrightarrow{ }^{4} \\mathrm{He}_{2} \\text {&#8230;&#8230;&#8230;&#8230;..+ Energy }\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 nuclear fusion reaction, the fusion of two light atomic nuclei into a single heavier nuclei by the collision of two interacting particles at extremely high temperatures, with the consequent release of a relatively large amount of energy which defines the thermo nuclear reaction.<\/p>\n<p>(ii) \\({ }^{3} \\mathrm{He}_{2}+{ }^{2} \\mathrm{H}_{1} \\rightarrow{ }^{4} \\mathrm{He}_{2}+{ }^{1} \\mathrm{H}_{1}+\\) energy<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q4-e\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">e<\/div><div class=\"xc-qtext\"><p>State two ways to increase the speed of rotation of a D.C. motor. <strong>[2]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Speed of D.C. motor can be increased by<\/p>\n<p>1. varying the flux and by varying the current through field winding.<\/p>\n<p>2. varying supply voltage.<\/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 2019<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q5-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>A body of mass 10 Kg is kept at a height of 5 m. It is allowed to fall and reach the ground. (i) What is the total mechanical energy possessed by the body at the height of 2 m assuming it is a frictionless medium?<\/p>\n<p>(ii) What is the kinetic energy possessed by the body just before hitting the ground? Take g = \\(\\mathbf{1 0 ~ m} \/ \\mathbf{s}^{\\mathbf{2}}\\). <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Total mechanical energy (M.E.) of the body is conserved that is constant throughout to fall the given height.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n\\text { At point } \\mathrm{A},(K E)_{\\mathrm{A}}=0 \\text { and }(P E)_{\\mathrm{A}} &#038; =m g \\times 5 \\\\\n&#038; =10 \\times 10 \\times 5 \\mathrm{~J} \\\\\n&#038; =500 \\mathrm{~J}\n\\end{aligned}\n\\]<\/div>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n(M E)_{\\mathrm{A}} &#038; =(P E)_{\\mathrm{A}}+(K E)_{\\mathrm{A}} \\\\\n&#038; =500 \\mathrm{~J}+0=500 \\mathrm{~J}\n\\end{aligned}\n\\]<\/div>\n<p>Since, the mechanical energy is constant at every point. Hence, the mechanical energy at height 2 m will be \\((M E)_{\\mathrm{B}}=500 \\mathrm{~J}\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-4_403_482_1554_1176.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 5(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) Just before hitting the ground, its kinetic energy will be maximum and potential energy will be zero.<\/p>\n<p>\\(\\therefore K . E .=500 \\mathrm{~J}\\). (as whole of its P.E. has been converted into K.E.)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q5-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A uniform meter scale is in equilibrium as shown in the diagram:<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-4_182_1165_2221_416.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 5(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(i) Calculate the weight of the meter scale.<\/p>\n<p>(ii) Which of the following options is correct to keep the ruler in equilibrium when 40 gf wt is shifted to \\(\\mathbf{0 ~ c m}\\) mark? F is shifted towards 0 cm. or F is shifted towards 100 cm. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_168_660_318_251.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 5(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>The sum of the moments of the masses is always zero about the centre of mass. Let weight of the rode be \\(m g f\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n40 \\times 25+m \\times 30=m \\times 70\n\\]<\/div>\n<p>or \\(m \\times 40=40 \\times 25 \\Rightarrow m=25 g f\\) \u2234 mass of rod = 25gf<\/p>\n<p>(ii) If 40gf wt is shifted towards 0 mark<\/p>\n<p>F is shifted towards 0 mark to make the system in equilibrium.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_183_715_886_224.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 5(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q5-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The diagram below shows a pulley arrangement: (i) Copy the diagram and mark the direction of tension on each strand of the string.<\/p>\n<p>(ii) What is the velocity ratio of the arrangement?<\/p>\n<p>(iii) If the tension acting on the string is T, then what is the relationship between . T and effort E?<\/p>\n<p>(iv) If the free end of the string moves through a distance x, find the distance by which the load is raised. <strong>[4]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_366_312_1535_428.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 5(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) \\(T=\\) tension force by which rope affects \\(E\\) \\(T^{\\prime}=T^{\\prime \\prime}=\\) tension forces by which rope affects the load W<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_527_672_2056_247.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 5(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) Velocity ratio of this pulley system is 2.<\/p>\n<p>(iii) If the load is not moving, in that case \\(T=E\\).<\/p>\n<p>(iv) If the free end of rope moves by a distance \\(x\\), Load moves by a distance \\(=\\frac{\\text { Effort arm }}{\\text { MA }}=\\frac{x}{2}\\)<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q6-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>How does the angle of deviation formed by a prism change with the increase in the angle of incidence? Draw a graph showing the variation in the angle of deviation with the angle of incidence at a prism surface. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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=\\) angle of incidence \\(\\delta=\\) angle of deviation As the angle of incidence is increased, the angle of deviation also increases.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_306_504_908_1165.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 6(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_341_564_1220_1136.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 6(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>\\(\\delta_{m}=\\) angle of minimum deviation<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q6-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A virtual, diminished image is formed when an Object is placed between the optical centre and the principal focus of a lens. (i) Name the type of lens which forms the above image.<\/p>\n<p>(ii) Draw a ray diagram to show the formation of the image with the above stated characteristics. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) This type of image formation takes place due to concave lens. (ii)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-5_284_569_1989_1160.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 6(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q6-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm. (i) What is the nature of the image so formed?<\/p>\n<p>(ii) Calculate the distance of the image from the lens.<\/p>\n<p>(iii) Calculate the magnification of the image. <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Given, object distance \\(\\boldsymbol{u}=\\mathbf{- 2 4 ~ c m}\\) focal length, \\(f=8 \\mathrm{~cm}\\)<\/p>\n<p>(i) Since, \\(u>f\\), hence nature of image so formed, will be (i) real, (ii) inverted (iii) diminished<\/p>\n<p>(ii) To find image distance \\(v=\\) ?<\/p>\n<p>Using lens formula,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\frac{1}{f}=\\frac{1}{v}-\\frac{1}{u} \\\\\n&#038; \\frac{1}{8}=\\frac{1}{v}-\\frac{1}{(24)}=\\frac{1}{v}+\\frac{1}{24}\n\\end{aligned}\n\\]<\/div>\n<p>or \\(\\frac{1}{v}=\\frac{1}{8}-\\frac{1}{24}=\\frac{3-1}{24}=\\frac{2}{24}=\\frac{1}{2}\\) or \\(v=12 \\mathrm{~cm}\\)<\/p>\n<p>(iii) magnification, \\(m=\\frac{v}{u}=\\frac{12 \\mathrm{~cm}}{-24 \\mathrm{~cm}}=-0.5\\) negative sign indicates, inverted image.<\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q7-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>It is observed that during march-past we hear a base drum distinctly from a distance compared to the side drums. (i) Name the characteristic of sound associated with the above observation.<\/p>\n<p>(ii) Give a reason for the above observation. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) It is the pitch (or frequency) of sound which makes us listen to base drum distinctly from a distance.<\/p>\n<p>(ii) The base drums have greater amplitude of vibration, so the intensity of sound is high, so it becomes audible at larger distance and its frequency distinguishes it from side drums.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q7-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>A pendulum has a frequency of 4 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 6 vibrations of the pendulum. If the velocity of sound in air is 340 m\/s, find the distance between the cliff and the observer. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Frequency of pendulum \\(=4\\) vibrations per second \u2234 Time elapsed in 4 vibration \\(=1 \\mathrm{~s}\\) \u2234 Time elapsed after 6 vibrations \\(=\\frac{1}{4} \\times 6=\\frac{3}{2} \\mathrm{~s}\\) This means observer hears the echo after 1.5 s. Given speed of sound \\(=340 \\mathrm{~m} \/ \\mathrm{s}\\). \u2234 Distance between cliff and the observer<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{\\text { speeed \u00d7 time }}{2}=\\frac{340 \\times 1.5}{2}=\\frac{510}{2}=255 \\mathrm{~m} .\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q7-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>Two pendulums C and D are suspended from a wire as shown in the figure given below. Pendulum C is made to oscillate by displacing it from its mean Position. It is seen that D also starts oscillating.<\/p>\n<p>(i) Name the type of oscillation, C will execute.<\/p>\n<p>(ii) Name the type of oscillation, D will execute.<\/p>\n<p>(iii) If the length of D is made equal to C then what difference will you notice in the oscillations of D?<\/p>\n<p>(iv) What is the name of the phenomenon when the length of D is made equal to C? <strong>[4]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-6_225_419_464_1197.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 7(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Pendulum C executes simple harmonic oscillations.<\/p>\n<p>(ii) Pendulum D will also execute force vibrations in phase with first one.<\/p>\n<p>(iii) When the length of pendulum D is increased the oscillations of D will increase.<\/p>\n<p>(iv) Resonance.<\/p><\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q8-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>(i) Write one advantage of connecting electrical appliances in parallel combination. (ii) What characteristics should a fuse wire have?<\/p>\n<p>(iii) Which wire in a power circuit is connected to the metallic body of the appliance? <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) If one of the devices in parallel combination fuses then the other devices keep working without being affected.<\/p>\n<p>(ii) Characteristics of a fuse wire are:<\/p>\n<p>(1) it should have low melting point,<\/p>\n<p>(2) high conductivity<\/p>\n<p>(3) low resistance<\/p>\n<p>(iii) Earth wire is connected to the metallic body of the appliances in a power circuit.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q8-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The diagram below shows a dual control switch circuit connected to a bulb. (i) Copy the diagram and complete it so that the bulb is switched ON.<\/p>\n<p>(ii) Out of A &amp; B which one is the live wire and which one is the neutral wire? <strong>[3]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-6_317_505_1735_1164.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 8(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-6_286_532_2115_1150.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 8(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) B is neutral and A is live<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q8-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The diagram above shows a circuit with the key k open. Calculate:<\/p>\n<p>(i) the resistance of the circuit when the key k is open.<\/p>\n<p>(ii) the current drawn from the cell when the key k is open.<\/p>\n<p>(iii) the resistance of the circuit when the key k is closed.<\/p>\n<p>(iv) the current drawn from the cell when the key k is closed. <strong>[4]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-7_249_461_580_346.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 8(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 the key k is open resistance of circuit, \\(R=(5+0.5)=5.5 \\Omega\\)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-7_246_529_942_319.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 8(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(ii) Current drawn from the cell when the key k is open is<\/p>\n<div class=\"mathjax-overflow\">\\[\nI=\\frac{E}{R+r}=\\frac{3.3}{5.5}=\\frac{3}{5}=0.6 \\mathrm{~A}\n\\]<\/div>\n<p>(iii) When the key k is closed, then \\(2 \\Omega\\) and \\(3 \\Omega\\) will be in series and further their equivalent will be in parallel to \\(5 \\Omega\\) as shown in diagram.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-7_354_584_1520_283.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 8(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-7_261_349_1915_419.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 8(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>\u2234 resistance of circuit, \\(R^{\\prime}=2.5+0 . .5=3.0 \\Omega\\)<\/p>\n<p>(iv) current drawn from the cell,when key is closed , \\(I^{\\prime}=\\frac{E}{R+r}=\\frac{3.3}{3}=1.1 A\\)<\/p><\/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 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q9-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>(i) Define Calorimetry. (ii) Name the material used for making a Calorimeter.<\/p>\n<p>(iii) Why is a Calorimeter made up of thin sheets of the above material answered in (ii)? <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Calorimetry is the process of measuring the amount of heat released or absorbed during a chemical reaction.<\/p>\n<p>(ii) Colorimeter is made of copper as it is a good conductor of heat with low specific heat capacity . So small amount of heat is absorbed or liberated during heat exchange.<\/p>\n<p>(iii) Calorimeter is made of thin sheets of copper because it decreases the heat capacity of box, thus ensuring small amount of absorption or release of heat during heat exchange.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q9-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>The melting point of naphthalene is 80\u00b0C and the room temperature is 30\u00b0C. A sample of liquid naphthalene at 100\u00b0C is cooled down to the room temperature. Draw a temperature time graph to represent this cooling. In the graph, mark the region which corresponds to the freezing process. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-7_398_691_1082_1103.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 9(b) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q9-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>104 g of water at 30\u00b0C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0\u00b0C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10\u00b0C. Find the mass of ice added. [Specific heat capacity of water = \\(4.2 \\mathrm{~J} \\mathrm{~g}^{-\\mathbf{1}}{ }^{\\circ} \\mathbf{C}^{-\\mathbf{1}}\\); Specific latent heat of fusion of ice \\(=336 \\mathrm{Jg}^{-\\mathbf{1}}\\); Specific heat capacity of copper \\(=0.4 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}\\) ] <strong>[4]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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. Here, we will use principle of calorimetry. Heat lost by initial water, calorimeter is equal to the heat absorbed by &#8216;the added ice. Thus equations becomes<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\left(m_{w} c_{w}+m_{c} c_{c}\\right) \\Delta t=m_{\\text {ice }} \\cdot L_{\\text {ice }}+m_{\\text {ice }} \\cdot C_{\\text {ice }} \\cdot \\Delta t \\\\\n&#038; \n\\text { or, } m_{\\text {ice }} &#038; =\\frac{\\left(m_{w} c_{w}+m_{c} c_{c}\\right)\\left(t_{1}-t\\right)}{L_{\\text {ice }}+C_{\\text {ice }}\\left(t-t_{2}\\right)} \\\\\n&#038; =\\frac{(104 \\times 4.2+42 \\times 0.4)(30-10)}{336+4.2 \\times(10-0)} \\\\\n&#038; =\\frac{(436.8+16.8) \\times 20}{378} \\mathrm{~g}=24 \\mathrm{~g}\n\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\">10<\/div><div><div class=\"xc-set-title\">Question 10<\/div><div class=\"xc-set-sub\">ICSE Class 10 Physics 2019<\/div><\/div><\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q10-a\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">a<\/div><div class=\"xc-qtext\"><p>Draw a neat labelled diagram of an A.C. generator. <strong>[3]<\/strong><\/p><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-8_469_621_347_309.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 10(a) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q10-b\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">b<\/div><div class=\"xc-qtext\"><p>(i) Define nuclear fission. (ii) Rewrite and complete the following nuclear reaction by filling in the atomic number of Ba and mass number of Kr: <strong>[3]<\/strong><\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }_{92}^{235} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow 144 \\mathrm{Ba}+36 \\mathrm{Kr}+3_{0}^{1} \\mathrm{n}+\\text { Energy }\n\\]<\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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) Nuclear fission : It is a type of nuclear reaction in which a heavy nucleus splits up into two lighter nuclei of comparable size. During this reaction tremendous amount of energy is released.<\/p>\n<p>(ii) \\({ }_{92}^{235} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\rightarrow{ }_{56}^{144} \\mathrm{Ba}+{ }_{36} \\mathrm{Kr}^{89}+3{ }_{0}^{1} \\mathrm{n}+\\) Energy<\/p>\n<p>In the balanced nuclear reaction atomic number should be equal on left and right side So, \\(92(\\mathrm{U})+0(\\mathrm{n})=x(\\mathrm{Ba})+36 \\mathrm{Kr}+0(\\mathrm{n})\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\Rightarrow x(\\mathrm{Ba})=56\n\\]<\/div>\n<p>Similarly, for mass no.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; 253(\\mathrm{U})+1(\\mathrm{n})=144(\\mathrm{Ba})+y(\\mathrm{Kr})+3(\\mathrm{n}) \\\\\n&#038; 236=144+y(\\mathrm{Kr})+3 \\\\\n&#038; y(\\mathrm{Kr})=236-147=89\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\" id=\"icse-physics-2019-q10-c\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">c<\/div><div class=\"xc-qtext\"><p>The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North South direction. (i) In which direction will the needle deflect when the key is closed?<\/p>\n<p>(ii) Why is the deflection produced?<\/p>\n<p>(iii) What will be the change in the deflection if the magnetic needle is taken just above the conductor AB?<\/p>\n<p>(iv) Name one device which works on this principle. <strong>[4]<\/strong><\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/a137c77a-32b7-4679-88a1-b723f8942c0e-8_284_700_720_1067.jpg\" alt=\"ICSE Class 10 Physics 2019 Question 10(c) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/div>\n<div class=\"xc-ans-toggle\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\"><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 first case when a magnetic needle is kept just below the conductor, when the key is closed the current flows from B to A. So, applying Right Hand Thumb Rule just below the conductor, the direction of magnetic field is towards East. So, needle will deflect towards East.<\/p>\n<p>(ii) The deflection is produced due to magnetic field when current is passed through the wire.<\/p>\n<p>(iii) In second case when a magnetic needle is kept just above the conductor, when the key is closed, the current flows from B to A. So, applying Right Hand Thumb Rule just above the conductor, the direction of magnetic field is towards West. So, needle will deflect towards West.<\/p>\n<p>(iv) The device which works on this principle is magnetic sensor, solenoid etc.<\/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\">14<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Physics PYQ Practice the complete ICSE Class 10 Physics 2019 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 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":""}},"iawp_total_views":1,"footnotes":""},"categories":[7],"tags":[],"class_list":["post-2218","post","type-post","status-publish","format-standard","hentry","category-physics-class-10-icse"],"modified_by":"sanjjeett","_links":{"self":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2218"}],"collection":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/comments?post=2218"}],"version-history":[{"count":2,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2218\/revisions"}],"predecessor-version":[{"id":2223,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2218\/revisions\/2223"}],"wp:attachment":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/media?parent=2218"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/categories?post=2218"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/tags?post=2218"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}