{"id":2129,"date":"2026-11-12T21:09:22","date_gmt":"2026-11-12T15:39:22","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2129"},"modified":"2026-11-12T21:10:44","modified_gmt":"2026-11-12T15:40:44","slug":"icse-class-10-physics-2024-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2024-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Physics 2024 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-2024-question-paper-with-solutions\/#SECTION_A\" title=\"SECTION A\">SECTION A<\/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-2024-question-paper-with-solutions\/#SECTION_B\" title=\"SECTION B\">SECTION B<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2024-question-paper-with-solutions\/#ICSE_Physics_2024_Paper_FAQs\" title=\"ICSE Physics 2024 Paper FAQs\">ICSE Physics 2024 Paper FAQs<\/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\">29<\/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 2024 Question Paper with Solutions<\/strong>, including MCQs, numericals, diagrams and answers from Sections A and B. 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\">2024<\/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\">Two 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>Answers to this Paper must be written on the paper provided separately.<\/p>\n<p>You will not be allowed to write during first 15 minutes.<\/p>\n<p>This time is to be spent in reading the question paper.<\/p>\n<p>The time given at the head of this Paper is the time allowed for writing the answers.<\/p>\n<p>Section A is compulsory. Attempt any four questions from Section B.<\/p>\n<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 2024<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"SECTION_A\"><\/span>SECTION A<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\">15 Marks \u2022 15 MCQs<\/div><\/div><\/div>\n<div class=\"xc-qtext\"><p>Choose the correct answers to the questions from the given options.<\/p>\n<p>(Do not copy the questions, write the correct answers only.)<\/p><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>When a bell fixed on a cycle rings, then the energy conversion that takes place is<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>gravitational potential energy to sound energy<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>kinetic energy to sound energy<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>sound energy to electrical energy<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>sound energy to mechanical energy<\/span><\/div>\n<\/div>\n<div 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>Option (b) is correct.<\/p>\n<p>Explanation: Bell fixed on a cycle rings when pressure by finger is applied. So, mechanical energy is converted into sound energy.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of force produced is 1 Nm , then the minimum force required is<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>5 N<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>10 N<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>20 N<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>0.2 N<\/span><\/div>\n<\/div>\n<div 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>Option (a) is correct<\/p>\n<p>Explanation: Moment of a force = Force x perpendicular distance from the line of action of the force<\/p>\n<p>Or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n1=F \\times 0.2\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nF=5 \\mathrm{~N}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>A force &#8216;F&#8217; moves a load from A to C as shown in the figure below. For the calculation of the work done, which of these lengths would you use as the displacement?<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Force moves a load from A to C diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-1_301_518_1890_324.jpg\"\/><\/div><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>3 m<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>4 m<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>5 m<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>7 m<\/span><\/div>\n<\/div>\n<div 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>Option (c) is correct.<\/p>\n<p>Explanation: Displacement should be in the direction of force.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iv<\/div>\n<div class=\"xc-qtext\"><p>A radioactive nucleus containing 128 nucleons emits a \\(\\beta\\)-particle. After \\(\\beta\\)-emission the number of nucleons present in the nucleus will be<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>128<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>129<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>124<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>127<\/span><\/div>\n<\/div>\n<div 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>Option (a) is correct.<\/p>\n<p>Explanation: Number of nucleons i.e., mass no. does not change due to \\(\\beta\\) emission Only atomic number will increase by 1.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">v<\/div>\n<div class=\"xc-qtext\"><p>Assertion (A): Ultraviolet radiations are scattered more as compared to the microwave radiations.<\/p>\n<p>Reason (R): Wavelength of ultraviolet radiation is more than the wavelength of microwave radiation.<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Both A and R are true.<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>A is true but R is false.<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>A is false but R is true.<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>Both A and R are false.<\/span><\/div>\n<\/div>\n<div 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>Option (b) is correct.<\/p>\n<p>Explanation: As wavelength decreases, scattering increases.<\/p>\n<p>Wavelength of ultraviolet is less than the wavelength of microwave.<\/p>\n<p>Hence, scattering of ultraviolet is more than that of microwave.<\/p>\n<p>Hence, the assertion is true but reason is false.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">vi<\/div>\n<div class=\"xc-qtext\"><p>When the stem of vibrating tuning fork is pressed on a table, the tabletop starts vibrating. These vibrations are definitely an example of<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>resonance<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>natural vibrations<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>forced vibrations<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>damped vibrations<\/span><\/div>\n<\/div>\n<div 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>Option (c) is correct.<\/p>\n<p>Explanation: When stem of a vibrating tuning fork is pressed on a table, the table top starts vibrating due to the periodic force applied by the tuning fork. This is an example of forced vibration.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">vii<\/div>\n<div class=\"xc-qtext\"><p>Which of the following is a class III lever?<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Pair of scissors<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>Wheelbarrow<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>Crowbar<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">d<\/span><span>Human forearm<\/span><\/div>\n<\/div>\n<div 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>Option (d) is correct.<\/p>\n<p>Explanation: For class III lever force is applied in between the fulcrum and the load.<\/p>\n<p>In human forearm, elbow is the pivot, bicep muscle applies the force and load is placed on the palm. Hence, it is a class III lever.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">viii<\/div>\n<div class=\"xc-qtext\"><p>The specific resistance of a conductor depends on its<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>length<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>material<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>area of cross section<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>radius<\/span><\/div>\n<\/div>\n<div 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>Option (b) is correct.<\/p>\n<p>Explanation: Specific resistance is independent of the dimension of the conductor. It depends on the material of the conductor.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ix<\/div>\n<div class=\"xc-qtext\"><p>Identify the option that displays the correct wiring with correct colour code<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><div><div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Electrical wiring option a\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-1_378_430_1774_1148.jpg\"\/><\/div><\/div><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><div><div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Electrical wiring option b\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-1_372_432_2155_1146.jpg\"\/><\/div><\/div><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><div><div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Electrical wiring option c\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-2_375_412_239_314.jpg\"\/><\/div><\/div><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><div><div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Electrical wiring option d\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-2_373_411_617_314.jpg\"\/><\/div><\/div><\/div>\n<\/div>\n<div 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>Option (a) is correct.<\/p>\n<p>Explanation: From mains, blue wire (NEUTRAL) should directly reach the appliance and the brown wire (LIVE) should reach the appliance through a switch.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">x<\/div>\n<div class=\"xc-qtext\"><p>The potential difference between terminals of a cell in a closed electric circuit is<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>terminal voltage<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>electro motive force<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>voltage drop<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>none of these<\/span><\/div>\n<\/div>\n<div 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>Option (a) is correct.<\/p>\n<p>Explanation: Potential difference between terminals of a cell, in open circuited condition, is termed as electromotive force. Potential difference between terminals of a cell, in closed circuited condition, is termed as terminal voltage.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">xi<\/div>\n<div class=\"xc-qtext\"><p>During melting of ice at 0\u00b0C the<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>energy is released and temperature remains constant.<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>energy is absorbed and temperature remains constant.<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>energy is released and temperature decreases.<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>energy is absorbed and temperature increases.<\/span><\/div>\n<\/div>\n<div 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>Option (b) is correct.<\/p>\n<p>Explanation: During melting, ice requires latent heat (a form of energy) and temperature remains constant at its melting point.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">xii<\/div>\n<div class=\"xc-qtext\"><p>Linear magnification(m) produced by a concave lens is<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>\\(m&lt;1\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>\\(m&gt;1\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>\\(m=1\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>\\(m=2\\)<\/span><\/div>\n<\/div>\n<div 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>Option (a) is correct.<\/p>\n<p>Explanation: Concave lens always produces virtual, erect image, smaller than the object. Hence, the magnification is less than 1.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">xiii<\/div>\n<div class=\"xc-qtext\"><p>A radioactive element is placed in an evacuated chamber. Then the rate of radioactive decay will<\/p><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Decrease<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>Increase<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>Remain unchanged<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>Depend on the surrounding temperature<\/span><\/div>\n<\/div>\n<div 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>Option (c) is correct.<\/p>\n<p>Explanation: Rate of radioactive elements is determined by the properties of the nucleus of an atom. so, it does not change by environment (external factors).<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">xiv<\/div>\n<div class=\"xc-qtext\"><p>The graph given below shows heat energy supplied against change in temperature when no energy is lost to the surrounding. The slope of this graph will give<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Heat energy supplied against change in temperature graph\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-2_420_603_2078_282.jpg\"\/><\/div><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Specific heat capacity<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>Latent heat of fusion<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>Latent heat of vaporization<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">d<\/span><span>Heat capacity<\/span><\/div>\n<\/div>\n<div 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>Option (d) is correct.<\/p>\n<p>Explanation: Slope the graph gives the amount of heat required for unit rise of temperature. This is termed as the heat capacity.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">xv<\/div>\n<div class=\"xc-qtext\"><p>A block of glass is pushed into the path of the light as shown below. Then the converging point X will<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Glass block inserted in path of converging light\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-2_376_443_394_1195.jpg\"\/><\/div><\/div>\n<\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>Move away from the slab<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>Move towards the slab<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>Not shift<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>Move towards the left side of the lens<\/span><\/div>\n<\/div>\n<div 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>Option (a) is correct.<\/p>\n<p>Explanation: The light passing through glass slab will bend towards the normal and after coming out in the air, the light will converge out at a point away from the glass slab.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Solution ray diagram for glass slab question\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-6_299_619_1087_1106.jpg\"\/><\/div><\/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\">15 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>(a) In the following atoms, which one is a radioisotope? Give one use of this isotope. \\(\\mathrm{O}^{16}, \\mathrm{C}^{14}, \\mathrm{~N}^{14}, \\mathrm{He}^{4}\\)<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Name the class of the lever shown in the picture below: [3]\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Lever diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-2_236_494_1180_1169.jpg\"\/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) \\(\\mathrm{C}^{14}\\) is a radioisotope because of the instability of the nucleus. Used for carbon dating.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Class II lever.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>Fill in the blanks: [2]<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> When a stone tied to a string is rotated in a horizontal plane, the tension in the string provides \\_\\_\\_\\_ force necessary for circular motion.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Work done by this force at any instant is\n<br\/>\\_\\_\\_\\_.\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Centripetal<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> zero\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>A non uniform beam of weight 120 N pivoted at one end is shown in the diagram below.<\/p>\n<p>Calculate the value of F to keep the beam in equilibrium. [2]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Non uniform beam equilibrium diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-2_360_644_1841_1105.jpg\"\/><\/div><\/div>\n<\/div>\n<div 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>The weight of the beam will produce a clockwise moment and the applied force F will produce a clockwise moment.<\/p>\n<p>Equating the moments of the force,<\/p>\n<div class=\"mathjax-overflow\">\\[\n120 \\times 0.2=F \\times 0.8\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nF=30 \\mathrm{~N}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iv<\/div>\n<div class=\"xc-qtext\"><p>Meera chose to use a block and tackle system of &#8216;9&#8217; pulleys instead of a single movable pulley to lift a heavy load. [2]<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> What is the advantage of using a block and tackle system over a single movable pulley?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Why should she connect more number of pulleys in the upper fixed block?\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Mechanical advantage using a block and tackle system of 9 pulleys is 9 whereas the mechanical advantage of a single moveable pulley is 2.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Increase of number of pulley increases mechanical advantage.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">v<\/div>\n<div class=\"xc-qtext\"><p>Sumit does 600 J of work in 10 min and Amit does 300 J of work in 20 min. Calculate the ratio of the powers delivered by them. [2]<\/p>\n\n<\/div>\n<\/div>\n<div 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>\\(\\mathrm{P}_{\\text {Sumit }} \/ \\mathrm{P}_{\\text {Amit }}=\\frac{\\frac{\\mathrm{W}_{\\text {Sumit }}}{t_{\\text {Sumit }}}}{\\frac{W_{\\text {Amit }}}{t_{\\text {Amit }}}}=\\frac{\\frac{600}{10}}{\\frac{300}{20}}=4: 1\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">vi<\/div>\n<div class=\"xc-qtext\"><p>5 bulbs are connected in series in a room. One bulb is fused. It is removed and remaining 4 bulbs are again connected in series to the same circuit. What will be the effect on the following physical quantities? (Increases, Decreases, Remain Same)<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Resistance\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Intensity of light [2]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Resistance decreases<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Current (Intensity of light) increases as resistance decreases\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">vii<\/div>\n<div class=\"xc-qtext\"><p>Rohan conducted experiments on echo in different media. He observed that a minimum distance of &#8216; \\(x\\) &#8216; metres is required for the echo to be heard in oxygen and &#8216; \\(y\\) &#8216; metres in benzene. Compare &#8216; \\(x\\) &#8216; and &#8216; \\(y\\) &#8216;. Justify your answer.<\/p>\n<p>Speed of sound in oxygen: \\(340 \\mathrm{~ms}^{-1}\\)<\/p>\n<p>Speed of sound in benzene: \\(200 \\mathrm{~ms}^{-1}\\) [2]<\/p><\/div>\n<\/div>\n<div 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>Applying the formula<\/p>\n<div class=\"mathjax-overflow\">\\[\nd=v \\times \\frac{0.1}{2}\n\\]<\/div>\n<p>For Oxygen,<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=340 \\times \\frac{0.1}{2}=17 \\mathrm{~m}\n\\]<\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\">For Benzene,\n<div class=\"mathjax-overflow\">\\[\ny=200 \\times \\frac{0.1}{2}=10 \\mathrm{~m}\n\\]<\/div>\n<br\/>So,\n<div class=\"mathjax-overflow\">\\[\nx: y=17: 10\n\\]<\/div>\n<\/div>\n<\/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\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>(a) In a reading glass what is the position of the object with respect to the convex lens used?<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Why can we not use concave lens for the same purpose? [2]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) In between optical centre and focus<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Because concave lens produces reduced size image.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>A fuse is rated 5 A. Can it be used with a geyser rated 1540 W, 220 V. Write Yes or No. Give supporting calculations to justify your answer. [2]<\/p><\/div>\n<\/div>\n<div 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>No.<\/p>\n<p>The geyser draws \\(\\mathrm{I}=\\frac{\\mathrm{P}}{\\mathrm{V}}=\\frac{1540}{220}=7 \\mathrm{~A}\\).<\/p>\n<p>Hence, 5A rated fuse will blow off for the geyser.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>State two factors affecting the speed of rotation of the coil in a D.C. motor. [2]<\/p><\/div>\n<\/div>\n<div 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>Speed of rotation may be increased by<\/p>\n<p>Increasing the number of turns of the armature.<\/p>\n<p>Increasing the applied battery e.m.f.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iv<\/div>\n<div class=\"xc-qtext\"><p>How much heat is required to convert 500 g of ice at 0\u00b0C to water at 0\u00b0C? The latent heat of fusion of ice is \\(330 \\mathrm{Jg}^{-1}\\). [2]<\/p><\/div>\n<\/div>\n<div 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>Required heat \\(\\mathrm{Q}=\\mathrm{mL}=500 \\times 330=165000 \\mathrm{~J}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">v<\/div>\n<div class=\"xc-qtext\"><p>Copy and complete the nuclear reaction by filling in the blanks. [2]<\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }_{92} \\mathrm{U}^{235}+{ }_{0} \\mathrm{n}^{1} \\longrightarrow{ }_{56} \\mathrm{Ba}^{\\cdots}+\\ldots \\mathrm{Kr}^{92}+3{ }_{0} \\mathrm{n}^{1}\n\\]<\/div>\n<\/div>\n<\/div>\n<div 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>\\({ }_{92} \\mathrm{U}^{235}+{ }_{0} \\mathrm{n}^{1} \\rightarrow{ }_{46} \\mathrm{Ba}^{(41)}+{ }_{36} \\mathrm{Kr}^{92}+3{ }_{0} \\mathrm{n}^{1}\\)<\/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 2024<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"SECTION_B\"><\/span>SECTION B<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\">4<\/div><div><div class=\"xc-set-title\">Question 4<\/div><div class=\"xc-set-sub\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>The image of a candle flame placed at a distance of 36 cm from a spherical lens, is formed on a screen placed at a distance of 72 cm from the lens. Calculate the focal length of the lens and its power. [3]<\/p><\/div>\n<\/div>\n<div 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>Using lens formula,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{v}-\\frac{1}{u}=\\frac{1}{f}\n\\]<\/div>\n<p>Or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{72}+\\frac{1}{36}=\\frac{1}{f}\n\\]<\/div>\n<p>Or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{3}{72}=\\frac{1}{f}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nf=24 \\mathrm{~cm}\n\\]<\/div>\n<p>Power of the lens \\(=\\frac{1}{f}=\\frac{100}{24}=4.17 \\mathrm{D}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>Below is an incomplete table showing the arrangement of electromagnetic spectrum in the increasing order of their wavelength. Complete the table. [3]<\/p>\n<div class=\"mathjax-overflow\"><table><tbody>\n<tr><td>Gamma ray<\/td><td>X-ray<\/td><td>U V rays<\/td><td>Visible rays<\/td><td>Infrared<\/td><td>A<\/td><td>Radio waves<\/td><\/tr>\n<\/tbody><\/table><\/div>\n\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Identify the radiation A.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Name the radiation used to detect fracture in bones.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Name one property common to both A and Radio waves.\n<\/div>\n\n<\/div>\n<\/div>\n<div 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>(a) Microwave<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> X-rays\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Both have same speed in vacuum\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>(a) Why do we use red colour as a danger signal on the top of a skyscraper?<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> The diagram below shows the path of a blue ray through the prism:\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>1.<\/strong> Calculate the critical angle of the material of the prism for blue colour.\n<\/div>\n\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong> What is the measure of the angle of this prism (A)?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>3.<\/strong> Which colour should replace the blue ray, for the ray to undergo Total Internal Reflection?\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Blue ray through prism diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-3_259_549_411_1183.jpg\"\/><\/div>\n<br\/>[4]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Red colour is used as danger signal because it is scattered least due to its longest wavelength in visible range. Hence, it is visible from a very long distance.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> 1. Critical angle is 43\u00b0\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Solution prism diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-7_286_606_1766_348.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong> From the above figure, Angle of prism \\(=43^{\\circ}\\)\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>3.<\/strong> If the blue ray is replaced by indigo or violet, there will be total internal reflection.\n<br\/>Critical angles for indigo and violets are less than that of blue. Hence, the angle of incidence 43\u00b0 will be greater than the critical angle and total internal reflection will occur.\n<\/div>\n\n<\/div>\n<\/div>\n<\/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\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>(a) Refractive index of glass with respect to water is \\(\\frac{9}{8}\\).<\/p>\n<p>Find the refractive index of water with respect to glass.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Name the principle used to find the value in part (a).\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> If we change the temperature of water, then will the ratio \\(\\frac{9}{8}\\) remain the same? Write Yes or No.[3]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Refractive index of glass with respect to water<\/p>\n<div class=\"mathjax-overflow\">\\[\n=\\frac{9}{8}\n\\]<\/div>\n<p>Hence, the Refractive index of water with respect<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { to glass }=\\frac{8}{9}\n\\]<\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> The principle of reversibility\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> No.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>Light travels a distance of &#8216; \\(10 x^{\\prime}\\) units in time &#8216; \\(t_{1}\\) &#8216; in vacuum and it travels a distance of &#8216; \\(x\\) &#8216; units in time &#8216; \\(t_{2}\\) &#8216; in a denser medium. Using this information answer the question that follows:<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> &#8216;Light covers a distance of &#8216; \\(20 x^{\\prime}\\) units in time &#8216; \\(t_{1}\\) &#8216; in diamond. &#8216;State true or false.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Calculate the refractive index of the medium in terms of &#8216; \\(t_{1}\\) &#8216; and &#8216; \\(t_{2}\\) &#8216;. [3]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) The statement is false.<\/p>\n<p>Diamond is a denser medium compared to vacuum. Hence, in it light cannot cover longer distance than that in vacuum.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Speed of light in vacuum \\(=\\frac{10 x}{t_{1}}\\)\n<br\/>Speed of light in the medium \\(=\\frac{x}{t_{2}}\\)\n<br\/>So, refractive index of the medium \\(=\\frac{\\frac{10 x}{t_{1}}}{\\frac{x}{t_{2}}}\\)\n<div class=\"mathjax-overflow\">\\[\n=\\frac{10 t_{2}}{t_{1}}\n\\]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>A monochromatic ray of light is incident on an equilateral prism placed at minimum deviation position with an angle of incidence 45\u00b0 as shown in the diagram.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Equilateral prism at minimum deviation diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-3_288_442_1772_1193.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Copy the diagram and complete the path of the ray PQ.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> State two factors on which the angle of deviation depends. [4]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Completed ray path through prism\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-7_289_627_989_1143.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Angle of deviation depends on\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\">Angle of incidence\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\">Refractive index\n<\/div>\n\n<\/div>\n<\/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\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>(a) Define Centre of Gravity. [3]<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> A hollow ice cream cone has height 6 cm.\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>1.<\/strong> Where is the position of its centre of gravity from the broad base?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong> Will its position change when it is filled completely with honey? Write Yes or No.\n<\/div>\n\n<\/div>\n\n\n<\/div>\n<\/div>\n<div 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>(a) Centre of gravity: Centre of gravity is the point where the mass of the whole body appears to be concentrated.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> 1. The centre of gravity from the broad base is at a height \\(\\frac{h}{3}=\\frac{6}{3}=2 \\mathrm{~cm}\\).\n<br\/>2. Yes.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Two marbles rolling along two paths diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-4_291_616_274_274.jpg\"\/><\/div>\n<p>Two identical marbles A and B are rolled down along Path 1 and Path 2 respectively. Path 1 is frictionless and Path 2 is rough.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Which marble will surely reach the next peak?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Along which path\/s the mechanical energy will be conserved?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Along which path\/s is the law of conservation of energy obeyed?\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Marble A<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Along both the paths.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Along both the paths.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>Given are two pulleys. [4]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Two pulleys diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-4_476_233_986_466.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Copy and complete the labelled diagram connecting the two pulleys with a tackle to obtain Velocity Ratio \\(=2\\).\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> If Load = 48 kgf and efficiency is 80\\% then calculate:\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>1.<\/strong> Mechanical Advantage.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong> Effort needed to lift the load.\n<\/div>\n\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a)<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Completed pulley connection diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-7_455_220_1865_1150.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> 1.\n<div class=\"mathjax-overflow\">\\[\n\\text { Efficiency }=\\frac{\\text { M.A. }}{\\text { V.R. }}\n\\]<\/div>\n<br\/>Or,\n<div class=\"mathjax-overflow\">\\[\n0.8=\\frac{\\mathrm{M} . \\mathrm{A} .}{2}\n\\]<\/div>\n<br\/>\u2234\n<div class=\"mathjax-overflow\">\\[\n\\text { M.A. = } 1.6\n\\]<\/div>\n<\/div>\n\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong>\n<div class=\"mathjax-overflow\">\\[\n\\text { Effort }=\\frac{\\text { Load }}{\\text { M.A. }}\n\\]<\/div>\n<br\/>Or,\n<div class=\"mathjax-overflow\">\\[\n\\text { Effort }=\\frac{48}{1.6}\n\\]<\/div>\n<br\/>\u2234\n<div class=\"mathjax-overflow\">\\[\n\\text { Effort = } 30 \\text { kgf }\n\\]<\/div>\n<\/div>\n<\/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\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>(a) Name the waves used in SONAR. [3]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Lata standing between two cliffs echo diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-4_210_625_1874_314.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Cliff A\n<br\/>In the above diagram Lata stands between two cliffs and claps her hands.\n<br\/>Determine the time taken by her to hear the first echo.\n<br\/>Speed of sound in air \\(320 \\mathrm{~ms}^{-1}\\).\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Ultrasonic wave<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Lata cannot clearly distinguish the sound reflected from cliff A. Since the distance is less than the minimum distance required to hear a distinct echo.\n<br\/>So, she will hear the sound reflected from cliff B as \\(1^{\\text {st }}\\) echo.\n<br\/>So, she will hear the \\(1^{\\text {st }}\\) echo after time \\(t=\\frac{2 d}{v}\\) \\(=\\frac{2 \\times 160}{320}=1 \\mathrm{~s}\\).\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>(a) Complete the following radioactive reaction:<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\mathrm{X} \\rightarrow \\mathrm{Y}+{ }_{2}^{4} \\mathrm{He} \\rightarrow{ }_{91}^{234} \\mathrm{Z}+{ }_{-1}^{0} \\mathrm{e}\n\\]<\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Uranium is available in two forms U-235 and U-238. Which of the two isotopes of Uranium is more fissionable? [3]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) \\({ }_{90}^{234} \\mathrm{X} \\rightarrow{ }_{88}^{230} \\mathrm{Y}+{ }_{2}^{4} \\mathrm{He} \\rightarrow{ }_{91}^{234} \\mathrm{Z}+{ }_{-1}^{0} \\mathrm{e}\\)<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> U-235\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>In the given diagram, a vibrating tuning fork is kept near the mouth of a burette filled with water. The length of the air column is adjusted by opening the tap of the burette. At a length of 5 cm of the air column, a loud sound is heard. [4]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Tuning fork and burette resonance experiment\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-4_602_393_427_1221.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Name the phenomenon illustrated by the above experiment.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Why is a loud sound heard at this particular length?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> If the present tuning fork is replaced with a tuning fork of higher frequency, should the length of the air column increase or decrease to produce a loud sound? Give a reason.\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Resonance<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Natural frequency of the air column matched with the frequency of the tuning fork.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Length of air column should be decrease. Since frequency is inversely proportional to the air column.\n<\/div>\n<\/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\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>The voltage &#8211; current readings of a certain material are shown in the table given below: [3]<\/p>\n<div class=\"mathjax-overflow\"><table><tbody>\n<tr><td>Voltage (V)<\/td><td>10 V<\/td><td>20 V<\/td><td>30 V<\/td><\/tr>\n<tr><td>Current (I)<\/td><td>2 A<\/td><td>3 A<\/td><td>4 A<\/td><\/tr>\n<\/tbody><\/table><\/div>\n<p>Study the table.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> State whether the conductor used is ohmic or non-ohmic.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Justify your answer.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> State Ohm&#8217;s law.\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Non-ohmic<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> The resistance is not constant. It increases as voltage and current increases.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Ohm&#8217;s law:\n<br\/>The flow of current through the conductor is directly proportional to the potential difference established across the conductor provide its physical conditions (like, temperature, pressure, density, volume etc.) are constant.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>Below is the diagram of a transformer: [3]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Transformer diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-4_451_525_1922_1155.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Identify the type of transformer.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> In this type of transformer which of the wire is thicker, the primary or the secondary? Give a reason.\n<\/div>\n\n\n<\/div>\n<\/div>\n<div 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>(a) Step down transformer<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Wire of secondary coil is thicker because secondary current is higher than primary current.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>Study the diagram: [4]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Resistor circuit diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-5_412_568_277_298.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(a)<\/strong> Calculate the total resistance of the circuit.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Calculate the current drawn from the cell.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> State whether the current through \\(10 \\Omega\\) resistor is greater than, less than or equal to the current through the \\(12 \\Omega\\) resistor.\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) \\(10 \\Omega\\) and \\(6 \\Omega\\) are in series. Hence, equivalent resistance \\(R_{1}=16 \\Omega\\).<\/p>\n<p>\\(12 \\Omega\\) and \\(4 \\Omega\\) are in series. Hence, equivalent resistance \\(R_{2}=16 \\Omega\\).<\/p>\n<p>\\(\\mathrm{R}_{1}\\) and \\(\\mathrm{R}_{2}\\) are in parallel. Hence, the equivalent resistance \\(=R_{\\mathrm{eq}}=8 \\Omega\\).<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Current \\(\\mathrm{I}=\\frac{\\mathrm{V}}{\\mathrm{R}_{e q}}=\\frac{4 V}{8 \\Omega}=0.5 \\mathrm{~A}\\)\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Current in both the resistors are equal because both the arms have same resistance value.\n<\/div>\n<\/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\">10 Marks<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">i<\/div>\n<div class=\"xc-qtext\"><p>85 g of water at 30\u00b0C is cooled to 5\u00b0C by adding certain mass of ice. Find the mass of ice required. [3] [Specific heat capacity of water \\(=4.2 \\mathrm{Jg}^{-1 \\circ} \\mathrm{C}^{-1}\\), Specific latent heat of fusion \\(=336 \\mathrm{Jg}^{-1}\\) ] [3]<\/p><\/div>\n<\/div>\n<div 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>Let the amount of ice \\(=m\\) gram<\/p>\n<p>Heat lost by water \\(=85 \\times 4.2 \\times(30-5)\\)<\/p>\n<p>Heat lost by ice \\(=m \\times 336+m \\times 4.2 \\times(5-0)\\)<\/p>\n<p>Applying principle of calorimetry,<\/p>\n<div class=\"mathjax-overflow\">\\[\n85 \\times 4.2 \\times(30-5)=m \\times 336+m \\times 4.2 \\times(5-0)\n\\]<\/div>\n<p>Or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n8925=357 m\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nm=25 \\mathrm{~g}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">ii<\/div>\n<div class=\"xc-qtext\"><p>(a) Why does it become pleasantly warm when the lakes start freezing?<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Water freezes to form ice. What change would you expect in the average kinetic energy of the molecules?\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> Which will contain more heat energy 1 g of ice at 0\u00b0C or 1 g water at 0\u00b0C? [3]\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) When lake freezes, it released latent heat to the atmosphere which warms the atmosphere. So, the atmosphere becomes pleasantly warm.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Since there is no change of temperature there is no change in kinetic energy.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(c)<\/strong> 1 g of ice (solid) at 1\u00b0C has extra heat in the form of its Latent Heat while 1 g of water (liquid) at 1oC has no Latent heat.\n<br\/>Hence, heat energy in 1 g of ice at 0\u00b0C has more heat energy than the 1 g water at 0\u00b0C.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\">\n<div class=\"xc-qnum\">iii<\/div>\n<div class=\"xc-qtext\"><p>(a) State one factor that affects the magnitude of induced current in an AC generator.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> Given below is a circuit to study the magnetic effect of electric current. ABCD is a cardboard kept perpendicular to the conductor XY. A magnetic compass is placed at the point P of the cardboard. \\(\\mathrm{P}_{1}\\) and \\(\\mathrm{P}_{2}\\) are the positions of the magnetic compass, before and after passing a current through XY respectively.\n<div class=\"xc-post-image xc-img-very-small\"><img decoding=\"async\" alt=\"Magnetic effect of electric current circuit diagram\" loading=\"lazy\" src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/516979a3-2044-4058-a0bf-07eebebda520-5_363_591_663_1121.jpg\"\/><\/div>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>1.<\/strong> Name the rule that is used to predict the direction of deflection of the magnetic compass.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong> State the direction of current in the conductor (X to Y or Y to X) when the circuit is complete.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>3.<\/strong> If resistance R is increased, then what will be the effect on the magnetic lines of force around the conductor? [4]\n<\/div>\n\n<\/div>\n<\/div>\n<\/div>\n<div 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>(a) Number of turns in armature may be increased.<\/p>\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>(b)<\/strong> 1. Maxwell right hand thumb rule.\n\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>2.<\/strong> Deflection being clockwise, the direction of current is from X to Y.\n<\/div>\n<div class=\"xc-clean-item\" style=\"margin:6px 0;\"><strong>3.<\/strong> As R increases, current decreases. So, strength of magnetic field decreases and hence, the density of magnetic lines of force decreases.\n<\/div>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"xc-section-heading\">\n<div class=\"xc-section-heading-icon\">\u2753<\/div>\n<div class=\"xc-section-heading-body\"><div class=\"xc-section-heading-label\">Quick Reference<\/div><h2 class=\"xc-section-heading-title\"><span class=\"ez-toc-section\" id=\"ICSE_Physics_2024_Paper_FAQs\"><\/span>ICSE Physics 2024 Paper FAQs<span class=\"ez-toc-section-end\"><\/span><\/h2><div class=\"xc-section-heading-sub\">Key details from the examination paper.<\/div><\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-qblock\"><div class=\"xc-qhead\"><div class=\"xc-qnum\">1<\/div><div class=\"xc-qtext\"><p><strong>What are the maximum marks for the ICSE Physics 2024 paper?<\/strong><\/p><p>The maximum marks are 80.<\/p><\/div><\/div><\/div>\n<div class=\"xc-qblock\"><div class=\"xc-qhead\"><div class=\"xc-qnum\">2<\/div><div class=\"xc-qtext\"><p><strong>How much time is allowed for the paper?<\/strong><\/p><p>The time allowed is two hours.<\/p><\/div><\/div><\/div>\n<div class=\"xc-qblock\"><div class=\"xc-qhead\"><div class=\"xc-qnum\">3<\/div><div class=\"xc-qtext\"><p><strong>How should Section B be attempted?<\/strong><\/p><p>Attempt any four questions from Section B.<\/p><\/div><\/div><\/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\">29<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Physics PYQ Practice the complete ICSE Class 10 Physics 2024 Question Paper with Solutions, including MCQs, numericals, [&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":2,"footnotes":""},"categories":[7],"tags":[],"class_list":["post-2129","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\/2129"}],"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=2129"}],"version-history":[{"count":10,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2129\/revisions"}],"predecessor-version":[{"id":2166,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2129\/revisions\/2166"}],"wp:attachment":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/media?parent=2129"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/categories?post=2129"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/tags?post=2129"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}