{"id":2167,"date":"2026-11-12T21:32:08","date_gmt":"2026-11-12T16:02:08","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2167"},"modified":"2026-11-12T21:32:09","modified_gmt":"2026-11-12T16:02:09","slug":"icse-class-10-physics-2023-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-physics-2023-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Physics 2023 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-2023-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-2023-question-paper-with-solutions\/#SECTION_B\" title=\"SECTION B\">SECTION B<\/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\">26<\/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 2023 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\">2023<\/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>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 2023<\/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 answer only).<\/p>\n<p>[15]<\/p><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Clockwise moment produced by a force bout a fulcrum is considered to be:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Positive<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>Negative<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>Zero<\/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 (b) is correct<\/p>\n<p>Explanation: By convention, anticlockwise moment is considered positive and clockwise moment is considered negative.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>When the speed of a moving object is doubled, then its kinetic energy.<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>remains the same<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>decreases<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>is doubled<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">d<\/span><span>becomes four times<\/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: Kinetic energy \\(=\\frac{1}{2} m v^{2}\\)<\/p>\n<p>So, kinetic energy will be four limes, if the speed ( \\(v\\) ) is doubled.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>The energy conversion in a washing machine is from ____.<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>magnetic to electrical<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>electrical to mechanical<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>electrical to magnetic<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>mechanical to electrical<\/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: In washing machine, there is a motor which converts electrical energy into mechanical energy.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>Which of the following radiations suffer maximum deflection in a magnetic field?<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Alpha radiations<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>Beta radiations<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>Gamma radiations<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>X-radiations<\/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: \\(\\beta\\) particles are fast moving negatively charged particle (electrons) with very high e\/m ratio.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">v<\/div><div class=\"xc-qtext\"><p>Speed of blue light in water is:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>more than green light<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>more than orange light<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>more than violet light<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>more than red light<\/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: Refractive index of a medium for blue light is less than that of a violet light.<\/p>\n<p>Since,<\/p>\n<div class=\"mathjax-overflow\">\\[\nv=\\frac{c}{\\mu}\n\\]<\/div>\n<p>So, velocity of blue light is more than that of violet light.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vi<\/div><div class=\"xc-qtext\"><p>A concave lens produces only ____ image.<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>real, enlarged<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>virtual, enlarged<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>virtual, diminished<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>real, diminished<\/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: Since, concave lens diverges incident ray, hence the refracted rays do not meet.<\/p>\n<p>So, virtual image is formed.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-2_293_401_2248_384.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 1(vi) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vii<\/div><div class=\"xc-qtext\"><p>When a body vibrates under a periodic force, then vibrations of the body are always:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>natural vibrations<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>damped 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>resonant 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><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">viii<\/div><div class=\"xc-qtext\"><p>Two notes are produced form two different musical instruments, such that they have same loudness and same pitch. The produced notes differ in their.<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>Waveform<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>Frequency<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>Wavelength<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>Speed<\/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 produced notes can be determined by the qualities that are governed by waveform.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ix<\/div><div class=\"xc-qtext\"><p>When a current I flows through a wire of resistance \\(\\boldsymbol{R}\\) for time \\(\\boldsymbol{t}\\) then the electrical energy produced is given by:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">a<\/span><span>\\(\\mathrm{I}^{2} \\mathrm{R} t\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>\\(\\mathrm{IR}^{2} t\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>IR \\(t\\)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>\\(\\mathrm{IR} t^{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: It follows from Joule&#8217;s law of heating. \\(I^{2} R t\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">x<\/div><div class=\"xc-qtext\"><p>Choose the correct relation for e.m.f. ( \\(\\varepsilon\\) ) and terminal voltage V:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>\\(\\varepsilon=\\mathrm{V}\\) (always)<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>\\(\\mathrm{V}>\\varepsilon\\) [always]<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>\\(\\mathrm{V}<\\varepsilon\\) [When the cell is in use]<\/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 (c) is correct<\/p>\n<p>Explanation: \\(\\varepsilon=V+i R\\)<\/p>\n<p>When \\(i \\neq 0, \\varepsilon>V\\)<\/p>\n<p>When \\(i=0, \\varepsilon=V\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xi<\/div><div class=\"xc-qtext\"><p>If the strength of the current flowing through a wire is increased, the strength of the magnetic field produced by it:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>decreases<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>increases<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>remains the same<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>first increases then decreases<\/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: Strength of current \\(\\propto\\) Magnetic field strength produced.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xii<\/div><div class=\"xc-qtext\"><p>Specific latent heat of a substance:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>is directly proportional to the mass<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>is directly proportional to the change in the temperature<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>depends on the material<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>is inversely proportional to the mass<\/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><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xiii<\/div><div class=\"xc-qtext\"><p>Specific heat capacity of a substance X is 1900 \\(\\mathbf{J k g}^{-\\mathbf{1}}{ }^{\\boldsymbol{\\circ}} \\mathbf{C}^{\\mathbf{- 1}}\\) means:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>Substance X absorbs 1900 J for 1\u00b0C rise in temperature<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">b<\/span><span>1 kg of substance X absorbs 1900 J heat for 1\u00b0C rise in temperature<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>1 kg of substance X absorbs 1900 J heat to increase the temperature<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>1 kg of substance X absorbs 1900 J heat to cool down by 1\u00b0C<\/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: It is clear from the magnitude and the unit of specific heat capacity that 1900 Joule heat is required to rise the temperature of 1 kg substance by 1\u00b0C.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xiv<\/div><div class=\"xc-qtext\"><p>When a ray of light travels normal to the given surface, then the angle of refraction is:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>180\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>90\u00b0<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">c<\/span><span>0\u00b0<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">d<\/span><span>45\u00b0<\/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: The ray travels undeviated. Hence angle of incidence = angle of refraction = 0\u00b0.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-2_230_360_1234_1236.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 1(xiv) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">xv<\/div><div class=\"xc-qtext\"><p>Small air bubbles rising up a fish tank appear silvery when viewed from some particular angle is due to the:<\/p><\/div><\/div>\n<div class=\"xc-options\">\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">a<\/span><span>reflection<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">b<\/span><span>refraction<\/span><\/div>\n<div class=\"xc-opt\"><span class=\"xc-opt-key\">c<\/span><span>dispersion<\/span><\/div>\n<div class=\"xc-opt correct\"><span class=\"xc-opt-key\">d<\/span><span>total internal reflection<\/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: Air is there inside bubble. When light travelling through water (denser than air) is incident on the bubble (inside the bubble the medium is rarer) at a suitable angle (more than critical angle), the total internal reflection takes place and the bubble appears silvery.<\/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 Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>(a) When does the nucleus of an atom tend to become radioactive?<\/p>\n<p>(b) Name a single pulley in which displacement of load and effort is not the same.<\/p>\n<p>(c) State one advantage of this pulley. [3]<\/p><\/div><\/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 the number of neutrons become much greater than the number of protons \\(\\left(\\frac{n}{p}>1\\right)\\) then the nucleus tend to become radioactive.<\/p>\n<p>(b) In single movable pulley the displacement of load is not equal to the displacement of effort.<\/p>\n<p>In this pulley, displacement of effort \\(=2 \\mathrm{x}\\) displacement of load.<\/p>\n<p>(c) Advantage of single movable pulley:<\/p>\n<p>Its mechanical advantage is 2 i.e. the load can be lifted by applying an effort equal to half the load.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>(a) What is the position of centre of gravity of a triangular lamina?<\/p>\n<p>(b) When this triangular lamina is suspended freely from any one vertex, what is the moment of force produced by its own weight in its rest position? [2]<\/p><\/div><\/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 position of intersections of the medians i.e. the centroid is the triangle is the centre of gravity of triangular media.<\/p>\n<p>(b) Algebraic sum of the moments will be zero since the line of action coincides with the axis of rotation.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>The diagram shows wheel O pivoted at point A. Three equal force \\(\\mathbf{F}_{\\mathbf{1}}, \\mathbf{F}_{\\mathbf{2}}\\) and \\(\\mathbf{F}_{\\mathbf{3}}\\) act at point B on the wheel.<\/p>\n<p>(a) Which force will produce maximum moment about A?<\/p>\n<p>(b) Give a reason for your answer in (a). [2]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-3_395_269_238_449.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 2(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/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) Force \\(\\mathrm{F}_{1}\\) will produce maximum moment about A.<\/p>\n<p>(b) Perpendicular distance of \\(\\mathrm{F}_{1}\\) is maximum.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-3_358_335_2250_415.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 2(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Perpendicular distance of \\(\\mathrm{F}_{1}\\) from A is AQ.<\/p>\n<p>Perpendicular distance of \\(\\mathrm{F}_{2}\\) from A is AR.<\/p>\n<p>Perpendicular distance of \\(\\mathrm{F}_{3}\\) from A is AP.<\/p>\n<div class=\"mathjax-overflow\">\\[\nA Q>A R>A P\n\\]<\/div>\n<p>Hence, the Force \\(\\mathrm{F}_{1}\\) will produce maximum moment about point A.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>(a) What should be the angle between the direction of force and the directions of displacement, for work to be negative?<\/p>\n<p>(b) Name the physical quantity obtained using the formula \\(\\frac{U}{h}\\), where \\(U\\) is the potential energy and \\(h\\) is the height. [2]<\/p><\/div><\/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 angle should be 180\u00b0.<\/p>\n<p>Work done \\(=\\mathrm{FS} \\cos \\theta\\).<\/p>\n<p>When \\(\\theta=180^{\\circ}\\left(\\cos 180^{\\circ}=-1\\right)\\), the work done is &#8211; FS i.e., negative work done.<\/p>\n<p>Here, F is the applied force and S is the displacement.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">v<\/div><div class=\"xc-qtext\"><p>Calculate the power spent by a crane while lifting a load of mass 2000 kg, at velocity of 1.5 \\(\\mathbf{m s}^{-\\mathbf{1}} .\\left(\\mathbf{g}=\\mathbf{1 0} \\mathbf{m s}^{-\\mathbf{2}}\\right)\\) [2]<\/p><\/div><\/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>Power= Force \u00d7 velocity<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{array}{ll}\n\\text { Power }=m g \\times v \\\\\n\\text { Or, } &#038; \\text { Power }=2000 \\times 10 \\times 1.5 \\\\\n\\therefore &#038; \\text { Power }=30000 \\text { Watt }\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vi<\/div><div class=\"xc-qtext\"><p>A metal foot ruler is held at the edge of a table. It is pressed at its free end and then released. It vibrates.<\/p>\n<p>(a) Name the vibrations produced.<\/p>\n<p>(b) State one way to increase the frequency of these vibrations. [2]<\/p><\/div><\/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) Damped vibration, since the amplitude will decrease with time.<\/p>\n<p>(b) Length may be reduced to increase the frequency.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">vii<\/div><div class=\"xc-qtext\"><p>&#8216;A geyser is rated 240 W &#8211; 220 V&#8217;. Explain the meaning of this statement. [2]<\/p><\/div><\/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>When the geyser is connected to 220V supply and it will consume 240 Joule energy per second.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"xc-set\">\n<div class=\"xc-set-header\"><div class=\"xc-set-icon\">3<\/div><div><div class=\"xc-set-title\">Question 3<\/div><div class=\"xc-set-sub\">ICSE Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>(a) Is it possible for a concave lens to form an image of size two times that of the object? Write Yes or No.<\/p>\n<p>(b) What will happen to the focal length of the lens if a part of the lens is covered with an opaque paper? [2]<\/p><\/div><\/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) No. Concave lens always process virtual image of smaller size.<\/p>\n<p>(b) There will be no change in focal length. Since, the refractive index and the radius of curvature do not change. Only intensity of light will change.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>(a) Which electrical component protects the electric circuit in case of excess current and which can also be used as a switch?<\/p>\n<p>(b) Name the wire to which this electrical component is connected in an electric circuit. [2]<\/p><\/div><\/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) Electrical component will be fuse. Miniature Circuit breaker (MCB) can be used as a switch.<\/p>\n<p>(b) Live wire.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>A copper conductor is placed over two stretched copper wires whose ends are connected to a D.C. supply as shown in the diagram.<\/p>\n<p>(a) What should be the magnetic poles at the points A and B lying on either side of the conductor to experience the force in the upward direction?<\/p>\n<p>(b) Name the law used to find these polarities.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-3_315_495_1843_1216.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 3(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>[2]<\/p><\/div><\/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) North pole at A and south pole at B.<\/p>\n<p>(b) Fleming&#8217;s left hand rule.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iv<\/div><div class=\"xc-qtext\"><p>Thermal capacities of substances A and B are same. If mass of A is more than mass of B then:<\/p>\n<p>(a) Which substance will have more specific heat capacity?<\/p>\n<p>(b) Which substance will show greater rise in temperature if the same amount of heat is supplied to both? [2]<\/p><\/div><\/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) Since, specific heat capacity \\(\\propto 1 \/\\) mass.<\/p>\n<p>The specific heat capacity of B will be more than that of A.<\/p>\n<p>(b) Thermal capacities is being same, both A and B will show same rise of temperature.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">v<\/div><div class=\"xc-qtext\"><p>How is the radioactivity of a radio isotope affected if it undergoes a chemical change? Give a reason for your answer. [2]<\/p><\/div><\/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>There will be no change.<\/p>\n<p>Reason: Radioactivity is a nuclear phenomenon that will not be affected by any chemical change.<\/p>\n<p>In chemical change, electrons take part.<\/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 2023<\/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\">ICSE Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>The diagram below shows the ray OP travelling through an equilateral prism of a certain material.<\/p>\n<p>(a) Calculate the value of \\(\\boldsymbol{i}_{\\mathbf{2}}\\), if the angle of deviation is 43\u00b0.<\/p>\n<p>(b) What is the ray QS called? [3]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-4_249_465_1045_351.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 4(i) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/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) Deviation \\(=\\delta=i_{1}+i_{2}-\\mathrm{A}\\)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\text { Or }, &#038; 43^{\\circ} &#038; =65^{\\circ}+i_{2}-60^{\\circ} \\\\\n&#038; \\therefore &#038; i_{2} &#038; =38^{\\circ}\n\\end{aligned}\n\\]<\/div>\n<p>(b) QS is called emergent ray.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>Copy the diagram given below and complete the path of the light ray PQ, as it emerges out of the prism by marking necessary angles. The critical angle of glass is 42\u00b0. [3]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-4_280_431_1480_368.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 4(ii) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/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><div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-4_293_358_795_1106.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 4(ii) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>The diagram below shows two parallel rays A (Orange) &amp; B (Blue) incident from air, on airglass boundary.<\/p>\n<p>(a) Copy and complete the path of the rays A and B.<\/p>\n<p>(b) How do the speeds of these rays differ in glass?<\/p>\n<p>(c) Are the two refracted rays in glass parallel? Give a reason. [3]<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-4_250_384_2144_392.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 4(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div><\/div><\/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) Deviation of blue will be more than that of orange.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-4_354_338_1186_1292.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 4(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(b) Speed of blue is less than that of orange because the wavelength of blue is less than the orange.<\/p>\n<p>(c) Refracted rays in the glass are not parallel because of different wavelength.<\/p><\/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\">ICSE Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>A convex lens of focal length 10 cm is placed at a distance of 60 cm from a screen. How far from the lens should an object be placed so as to obtain a real image on the screen? [3]<\/p><\/div><\/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) Applying lens formula,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\frac{1}{v}-\\frac{1}{u}=\\frac{1}{f} \\\\\n&#038; \\frac{1}{60}-\\frac{1}{u}=\\frac{1}{10}\n\\end{aligned}\n\\]<\/div>\n<p>Or,<\/p>\n<p>Or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{u}=\\frac{1}{60}-\\frac{1}{10}\n\\]<\/div>\n<p>\u2234<\/p>\n<div class=\"mathjax-overflow\">\\[\nu=-12 \\mathrm{~cm}\n\\]<\/div>\n<p>So, the object should be placed at a distance 12 cm in front of the lens.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>(a) A coin kept inside water \\(\\left[\\mu=\\frac{4}{3}\\right]\\) when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water.<\/p>\n<p>(b) How is the critical angle related to the refractive index of a medium? [3]<\/p><\/div><\/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=\"mathjax-overflow\">\\[\n\\mu=\\frac{\\text { Real depth }}{\\text { Apparent depth }}\n\\]<\/div>\n<p>or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\mu-1=\\frac{\\text { Real depth }- \\text { apparent depth }}{\\text { apparent depth }}\n\\]<\/div>\n<p>or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{4}{3}-1=\\frac{3 \\mathrm{~mm}}{\\text { apparent depth }}\n\\]<\/div>\n<p>or,<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\frac{1}{3}=\\frac{3 \\mathrm{~mm}}{\\text { apparent depth }}\n\\]<\/div>\n<p>\u2234 Apparent depth \\(=9 \\mathrm{~mm}\\)<\/p>\n<p>\u2234 Real depth \\(=9 \\mathrm{~mm}+3 \\mathrm{~mm}=12 \\mathrm{~mm}\\)<\/p>\n<p>(b) \\(\\mu=\\frac{1}{\\sin \\theta_{c}}\\)<\/p>\n<p>Here, all alphabets are in their usual meaning<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>(a) Infrared radiations are used in warfare. Explain with reason, why.<\/p>\n<p>(b) A ray of light is incident at 45\u00b0 on an equilateral prism in the diagram below.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-4_313_546_2274_1143.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 5(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>1. Name the phenomenon exhibited by the ray of light when it enters and emerges out of the prism.<\/p>\n<p>2. State the cause of the above phenomenon mentioned by you. [4]<\/p><\/div><\/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) Since infrared radiation provides night vision, it is used in warfare.<\/p>\n<p>(b) (1) When ray enters in the prism, the phenomena exhibited are refraction, dispersion of light. When the ray emerges from the prism the phenomena exhibited are refraction and deviation.<\/p>\n<p>(2) The reason of the above mentioned phenomenon is due to the difference of speed of different colours of light in a medium.<\/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 Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>A block and tackle system of pulleys has velocity ratio 4.<\/p>\n<p>(a) Draw a labelled diagram of the system indicating clearly, the direction of the load and the effort.<\/p>\n<p>(b) What is the value of the mechanical advantage of the given pulley system if it is an ideal pulley system? [3]<\/p><\/div><\/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 src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-5_606_226_811_1347.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 6(i) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(b) Mechanical advantage \\(=\\frac{4 T}{T}=4\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>A metre scale of weight 50 gf can be balanced at 40 cm mark without any weight suspended on it.<\/p>\n<p>(a) If this ruler is cut at its centre then state which part [0 to 50 cm or 50 to 100 cm] of the ruler will weigh more than 25 gf.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-5_83_795_350_1020.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 6(ii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(b) What minimum weight placed on this metre ruler can balance this ruler when it is pivoted at its centre? [3]<\/p><\/div><\/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) 0-50 cm part will weigh more because its balanced point is at 40 cm.<\/p>\n<p>(b) When the ruler is pivoted at centre, then<\/p>\n<div class=\"mathjax-overflow\">\\[\n50 \\times 10=50 \\times x\n\\]<\/div>\n<p>( \\(x\\) is the minimum weight)<\/p>\n<p>So,<\/p>\n<div class=\"mathjax-overflow\">\\[\nx=10 \\mathrm{~g}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>A car of mass 120 kg is moving at a speed 18 km\/h and it accelerates to attain a speed of 54 km\/h in 5 seconds. Calculate:<\/p>\n<p>(a) the work done by the engine.<\/p>\n<p>(b) the power of the engine.<\/p><\/div><\/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) Work done = Change in kinetic energy.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\begin{aligned}\n&#038; \\text { Initial velocity }=18 \\mathrm{~km} \/ \\mathrm{h}=18 \\times \\frac{5}{18}=5 \\mathrm{~m} \/ \\mathrm{s} \\\\\n&#038; \\text { Final velocity }=54 \\mathrm{~km} \/ \\mathrm{h}=54 \\times \\frac{5}{18}=15 \\mathrm{~m} \/ \\mathrm{s} \\\\\n&#038; \\text { So, work done }=\\frac{1}{2} 120\\left(15^{2}-5^{2}\\right)=12000 \\mathrm{~J}\n\\end{aligned}\n\\]<\/div>\n<p>(b) Power \\(=\\frac{\\text { Work done }}{\\text { time }}=\\frac{12000}{5}=2400 \\mathrm{~W}\\) [4]<\/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 Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>(a) Which characteristic of sound is affected due to the larger surface of a school bell?<\/p>\n<p>(b) Calculate the distance covered by the Ultrasonic wave having a velocity of 1.5 \\(\\mathbf{k m s}^{-\\mathbf{1}}\\) in 14 s, when it is received after reflection by the receiver of the SONAR.<\/p>\n<p>[3]<\/p><\/div><\/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) Amplitude or loudness is affected due to larger surface area.<\/p>\n<p>(b) Distance covered \\(=\\frac{\\text { velocity \u00d7 time }}{2}\\)<\/p>\n<p>Or, Distance covered \\(=\\frac{1.5 \\times 14}{2}\\)<\/p>\n<p>\u2234 Distance covered \\(=1.5 \\times 7=10.5 \\mathrm{~km}\\)<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>(a) Complete the following nuclear changes:<\/p>\n<div class=\"mathjax-overflow\">\\[\n{ }_{92}^{238} \\mathrm{P} \\rightarrow=\\mathrm{Q}+{ }_{2}^{4} \\mathrm{He} \\rightarrow=\\mathrm{R}+{ }_{-1}^{0} e\n\\]<\/div>\n<p>(b) Name the nuclear radiation which has the highest ionizing power. [3]<\/p><\/div><\/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) \\({ }_{92}^{238} \\mathrm{P} \\rightarrow{ }_{90}^{234} \\mathrm{Q}+{ }_{2}^{4} \\mathrm{He} \\rightarrow{ }_{91}^{234} \\mathrm{R}+{ }_{-1}^{0} \\mathrm{e}\\)<\/p>\n<p>Alpha particle has the highest ionising power.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>We are able to see the T.V. channels clearly when we set T.V. on auto-tuning.<\/p>\n<p>(a) Which phenomenon led to the clear visibility of the channels, due to auto tuning?<\/p>\n<p>(b) Define the above phenomenon mentioned by you.<\/p>\n<p>(c) Give any one more example of this phenomenon. [4]<\/p><\/div><\/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<p>(b) Resonance: The phenomenon when a body vibrates with increased amplitude under the influence of external periodic force exactly equal to the natural frequency of vibration of the body.<\/p>\n<p>(c) Air in sound box vibrates with higher amplitude due to vibrating string of musical instruments. This is an example of 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 Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>(a) Define specific resistance.<\/p>\n<p>(b) What happens to the specific resistance of a conductor if its length is doubled?<\/p>\n<p>(c) Name a substance whose specific resistance remains almost uncharged with the increase in its temperature. [3]<\/p><\/div><\/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) Specific resistance: Resistance offered by a conductor of unit length and unit cross section.<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\rho=\\mathrm{R} \\frac{A}{l}\n\\]<\/div>\n<p>(b) Specifies resistance does not change with change of length. It depends on the material and the temperature.<\/p>\n<p>(c) Constantan, Manganin etc.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>(a) Which nuclear radiation will travel undeviated in an electric field?<\/p>\n<p>(b) How can one stop the radiations escaping from a nuclear reactor in a nuclear power plant?<\/p>\n<p>(c) Name one internal source of background radiations. [3]<\/p><\/div><\/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) \\(\\gamma\\) radiation. It is made of photons having no charge.<\/p>\n<p>(b) To stop radiation escaping from nuclear reactor, the power plant may be shielded with thick steel or lead wall.<\/p>\n<p>(c) Biggest source of background radiation is airborne radon.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>Find the value of current I drawn from the cell.<\/p>\n<p>(a) Calculate the current I.<\/p>\n<p>(b) Calculate the terminal voltage. [4]<\/p><\/div><\/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) Two \\(15 \\Omega\\) resistors are in series. So, their equivalent resistance (R&#8217;) is \\(30 \\Omega\\).<\/p>\n<p>Now another \\(30 \\Omega\\) is in parallel with the combination.<\/p>\n<p>So, the equivalent resistance (R&#8221;) is \\(15 \\Omega\\).<\/p>\n<p>So, the current \\(=\\mathrm{I}=\\frac{E}{R^{\\prime \\prime}+r}=\\frac{3.4}{15+2}=0.2 \\mathrm{~A}\\)<\/p>\n<p>(b) Terminal voltage \\(=E-I r=3.4-0.2 \\times 2=3 \\mathrm{~V}\\)<\/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 Physics 2023<\/div><\/div><\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">i<\/div><div class=\"xc-qtext\"><p>Calculate the total amount of heat energy required to melt 200 g of ice at \\(0^{\\circ} \\mathrm{C}\\) to water at 100\u00b0C.<\/p>\n<p>(Specific latent heat of ice \\(=336 \\mathrm{~J} \\mathrm{~g}^{-\\mathbf{1}}\\), specific heat capacity of water \\(=4.2 \\mathrm{~J} \\mathrm{~g}^{-1} \\mathbf{o}^{-\\mathbf{1}}\\) ) [3]<\/p><\/div><\/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>Heat energy required for<\/p>\n<p>(1) Ice at 0\u00b0C to water at 0\u00b0C conversion<\/p>\n<div class=\"mathjax-overflow\">\\[\n=200 \\times 336=67200 \\mathrm{~J}\n\\]<\/div>\n<p>(2) Water at 0\u00b0C to water at 100\u00b0C conversion<\/p>\n<div class=\"mathjax-overflow\">\\[\n=200 \\times 4.2 \\times 100=84000 \\mathrm{~J}\n\\]<\/div>\n<p>(3)<\/p>\n<div class=\"mathjax-overflow\">\\[\n\\text { Total heat energy require } \\begin{aligned}\nd &#038; =67200+84000 \\\\\n&#038; =151200 \\mathrm{~J}\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">ii<\/div><div class=\"xc-qtext\"><p>(a) State the principle of calorimetry.<\/p>\n<p>(b) Name the material used for making a calorimeter.<\/p>\n<p>(c) Write one characteristic property of the material chosen for making a calorimeter.<\/p>\n<p>[3]<\/p><\/div><\/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) Principle of calorimetry: When two or more bodies at different temperatures are brought in contact, then heat lost by a hot body is equal to the heat gained by the cold body, provided there is no heat loss in the environment.<\/p>\n<p>(b) Copper is generally use for calorimeter.<\/p>\n<p>(c) The material used for calorimeter should have a very low specific heat capacity.<\/p><\/div>\n<\/div>\n<div class=\"xc-qblock\">\n<div class=\"xc-qhead\"><div class=\"xc-qnum\">iii<\/div><div class=\"xc-qtext\"><p>The diagram below shows a cardboard on which iron filings are kept. A wire bent in the form of a loop is seen passing through the cardboard. When current flows through it the iron filings arrange themselves as shown below.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-6_297_497_1458_1169.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 9(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>(a) State the polarities of the battery at A and B.<\/p>\n<p>(b) State the effect on the magnetic field if an iron rod is held along the axis of the coil.<\/p>\n<p>(c) State one way to:<\/p>\n<p>1.<\/p>\n<div class=\"xc-post-image xc-img-very-small\"><img src=\"https:\/\/xamcontent.com\/cisce\/wp-content\/uploads\/2026\/11\/cea1d4cd-8679-4929-a047-fbae6d9ff939-6_232_498_1964_1318.jpg\" alt=\"ICSE Class 10 Physics 2023 Question 9(iii) figure\" loading=\"lazy\" decoding=\"async\"><\/div>\n<p>Change the polarity of the coil.<\/p>\n<p>2. Decrease the strength of the magnetic field around the coil. [4]<\/p><\/div><\/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) Since at the left side of the loop, the lines of force are anti clockwise, so the current is going upward. So, terminal A should be positive and B should be negative.<\/p>\n<p>(b) If an iron rod is held along the axis, magnetic field strength will increase.<\/p>\n<p>(c) 1. To change the polarity, the direction of the current is to be reversed.<\/p>\n<p>2. To decrease the magnetic field strength, the current strength is to be decreased.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\n\n<p><\/p>\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\">26<\/span> <span class=\"rt-label rt-postfix\">minutes<\/span><\/span> ICSE Class 10 Physics PYQ Practice the complete ICSE Class 10 Physics 2023 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":3,"footnotes":""},"categories":[7],"tags":[],"class_list":["post-2167","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\/2167"}],"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=2167"}],"version-history":[{"count":3,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2167\/revisions"}],"predecessor-version":[{"id":2186,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2167\/revisions\/2186"}],"wp:attachment":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/media?parent=2167"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/categories?post=2167"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/tags?post=2167"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}