{"id":13673,"date":"2018-04-09T17:40:49","date_gmt":"2018-04-09T12:10:49","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13673"},"modified":"2025-10-09T17:06:38","modified_gmt":"2025-10-09T11:36:38","slug":"cbse-sample-papers-class-11-physics","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/","title":{"rendered":"CBSE Sample Papers Class 11 Physics 2025"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 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\" style=\"cursor:inherit\">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:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#CBSE_Sample_Papers_Class_11_Physics_2024-25\" >CBSE Sample Papers Class 11 Physics 2024-25<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_%E2%80%93_Physics_Sample_Paper_2024-25\" >Class 11 &#8211; Physics Sample Paper (2024-25)<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Physics_Sample_Paper_Section_A\" >Physics Sample Paper Section A<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Physics_Sample_Paper_Section_B\" >Physics Sample Paper Section B<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Physics_Sample_Paper_Section_C\" >Physics Sample Paper Section C<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Physics_Sample_Paper_Section_D\" >Physics Sample Paper Section D<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_%E2%80%93_Physics_Sample_Paper_Solution_2024-25\" >Class 11 &#8211; Physics Sample Paper Solution (2024-25)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_Physics_Sample_Paper_Solution_Section_A\" >Class 11 Physics Sample Paper Solution Section A<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_Physics_Sample_Paper_Solution_Section_B\" >Class 11 Physics Sample Paper Solution Section B<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_Physic_Sample_Paper_Solution_Section_C\" >Class 11 Physic Sample Paper Solution Section C<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_Physics_Sample_Paper_Section_D_Solution\" >Class 11 Physics Sample Paper Section D Solution<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#Class_11_Physics_Sample_Paper_Section_E_Solution\" >Class 11 Physics Sample Paper Section E Solution<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/#CBSE_Sample_Papers_for_Class_11\" >CBSE\u00a0Sample Papers\u00a0for Class 11<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_Class_11_Physics_2024-25\"><\/span><strong>CBSE Sample Papers Class 11 Physics 2024-25<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>CBSE Sample Papers Class 11 Physics myCBSEguide provides CBSE Class 11 Sample Papers of Physics for the year 2024-2025 with solutions in PDF format for free download. The CBSE Sample Papers for all \u2013 NCERT books and based on CBSE latest syllabus must be downloaded and practiced by students. Class 11 Physics New Sample Papers follow the blueprint of that year only. A student must check the latest syllabus and marking scheme. Sample papers for Class 11 Physics and other subjects are available for download as PDF in-app too. myCBSEguide provides sample papers with solutions for the year 2025.<\/p>\n<p style=\"text-align: center;\"><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1340\/type\/2\">Download Physics Sample Papers\u00a0as PDF<\/a><\/strong><\/p>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Class_11_%E2%80%93_Physics_Sample_Paper_2024-25\"><\/span><strong>Class 11 &#8211; Physics Sample Paper (2024-25)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<hr \/>\n<p><b>Maximum Marks: 70<br \/>\nTime Allowed: : 3 hours<\/b><\/p>\n<hr \/>\n<p><b>General Instructions:<\/b><\/p>\n<ol>\n<li>There are 33 questions in all. All questions are compulsory.<\/li>\n<li>This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.<\/li>\n<li>Section A contains sixteen questions, twelve MCQ and four Assertion Reasoning based of 1 mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains two case study-based questions of four marks each and Section E contains three long answer questions of five marks each.<\/li>\n<li>There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.<\/li>\n<li>Use of calculators is not allowed.<\/li>\n<\/ol>\n<hr \/>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"text-align: center; clear: both; display: block;\">\n<h3><span class=\"ez-toc-section\" id=\"Physics_Sample_Paper_Section_A\"><\/span><b>Physics Sample Paper Section A<\/b><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<\/li>\n<li>Young\u2019s modulus of steel is 1.9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>11<\/sup> N\/m<sup>2<\/sup>. When expressed in CGS units of dynes\/cm<sup>2<\/sup>, it will be equal to (1N = 10<sup>5<\/sup> dyne, 1m<sup>2<\/sup> = 10<sup>4<\/sup> cm<sup>2<\/sup>)\n<div style=\"margin-left: 20px;\">a) 1.9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>10<\/sup><\/div>\n<div style=\"margin-left: 20px;\">b) 1.9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>11<\/sup><\/div>\n<div style=\"margin-left: 20px;\">c) 1.9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>12<\/sup><\/div>\n<div style=\"margin-left: 20px;\">d) 1.9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>13<\/sup><\/div>\n<\/li>\n<li>The human ear is most sensitive to sound in the frequency range of:\n<div style=\"margin-left: 20px;\">a) 1000 to 2000 Hz<\/div>\n<div style=\"margin-left: 20px;\">b) 200 to 400 Hz<\/div>\n<div style=\"margin-left: 20px;\">c) 10000 to 20000 Hz<\/div>\n<div style=\"margin-left: 20px;\">d) 20 to 20000 Hz<\/div>\n<\/li>\n<li>Centre of gravity can be defined:\n<div style=\"margin-left: 20px;\">a) as that point where the total gravitational torque on the body is greater than zero<\/div>\n<div style=\"margin-left: 20px;\">b) as the center of mass<\/div>\n<div style=\"margin-left: 20px;\">c) as that point where the total gravitational torque on the body is zero<\/div>\n<div style=\"margin-left: 20px;\">d) as that point where the total gravitational force on the body is zero<\/div>\n<\/li>\n<li>What will be the height of a liquid column in a capillary tube on the surface of the moon?\n<div style=\"margin-left: 20px;\">a) 1\/6th of what was on earth&#8217;s surface<\/div>\n<div style=\"margin-left: 20px;\">b) Six times that on the earth&#8217;s surface<\/div>\n<div style=\"margin-left: 20px;\">c) It will remain unchanged<\/div>\n<div style=\"margin-left: 20px;\">d) 36 times that on earth&#8217;s surface<\/div>\n<\/li>\n<li>The formula for gravitational potential energy associated with two particles of masses m<sub>1\u00a0<\/sub>and m<sub>2<\/sub> separated by distance\u00a0r is given by\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}{\\rm{v}} = &#8211; \\frac{G{m_1}}{r}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}{\\rm{v}} = &#8211; \\frac{G{m_1}{m_2}}{r}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}\\mathrm{v}=-\\frac{G m_2}{r}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\mathrm{v}=-\\frac{G m_1 m_2}{2 r}{\/tex}<\/span><\/div>\n<p>To practice more questions &amp; prepare well for exams, download <strong>myCBSEguide App<\/strong>. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use <strong>Examin8 App<\/strong> to create similar papers with their own name and logo.<\/li>\n<li>A source of sound gives 5 beats per second when sounded with another source of frequency 100 second<sup>-1<\/sup>. The second harmonic of the source, together with a source of frequency 205 sec<sup>-1<\/sup> gives 5 beats per second. What is the frequency of the source?\n<div style=\"margin-left: 20px;\">a) 100 second<sup>-1<\/sup><\/div>\n<div style=\"margin-left: 20px;\">b) 105\u00a0second <sup>-1<\/sup><\/div>\n<div style=\"margin-left: 20px;\">c) 205 second<sup>-1<\/sup><\/div>\n<div style=\"margin-left: 20px;\">d) 95 second<sup>-1<\/sup><\/div>\n<\/li>\n<li>Motion along a straight line is called __________.\n<div style=\"margin-left: 20px;\">a) parabolic motion<\/div>\n<div style=\"margin-left: 20px;\">b) circular motion<\/div>\n<div style=\"margin-left: 20px;\">c) oscillatory motion<\/div>\n<div style=\"margin-left: 20px;\">d) rectilinear motion<\/div>\n<\/li>\n<li>A transverse wave is represented by y = sin (<span class=\"math-tex\">{tex}\\omega{\/tex}<\/span>t &#8211; kx)\u00a0For what value of the wavelength is the wave velocity equal to the maximum particle velocity?\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}\\frac{\\pi A}{2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) <span class=\"math-tex\">{tex}\\pi A{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">c) 2 <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>A<\/div>\n<div style=\"margin-left: 20px;\">d) A<\/div>\n<\/li>\n<li>Pressures inside two soap bubbles are 1.01 atm and 1.03 atm, ratio between their volumes is\n<div style=\"margin-left: 20px;\">a) 3 : 1<\/div>\n<div style=\"margin-left: 20px;\">b) none of these<\/div>\n<div style=\"margin-left: 20px;\">c) 27 : 1<\/div>\n<div style=\"margin-left: 20px;\">d) 127 : 101<\/div>\n<\/li>\n<li>There is no atmosphere on the moon, because\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>it is closer to the earth and also it has the inactive inert gases in it.<\/li>\n<li>it is too far from the sun and has very low pressure in its outer surface.<\/li>\n<li>escape velocity of gas molecules is greater than their root mean square velocity.<\/li>\n<li>escape velocity of gas molecules is less than their root mean square velocity.<\/li>\n<\/ol>\n<div style=\"margin-left: 20px;\">a) Option ii<\/div>\n<div style=\"margin-left: 20px;\">b) Option i<\/div>\n<div style=\"margin-left: 20px;\">c) Option iii<\/div>\n<div style=\"margin-left: 20px;\">d) Option iv<\/div>\n<\/li>\n<li>Two rings of radii R and nR made from the same wire have the ratio of moments of inertia about an axis passing through their centre equal to 1 : 8. The value of n is\n<div style=\"margin-left: 20px;\">a) 2<\/div>\n<div style=\"margin-left: 20px;\">b) 4<\/div>\n<div style=\"margin-left: 20px;\">c) <span class=\"math-tex\">{tex}2 \\sqrt{2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span><\/div>\n<\/li>\n<li>When water is heated from 0\u00b0C to 10\u00b0C its volume\n<div style=\"margin-left: 20px;\">a) increases continuously<\/div>\n<div style=\"margin-left: 20px;\">b) first decreases and then increases<\/div>\n<div style=\"margin-left: 20px;\">c) decreases continuously<\/div>\n<div style=\"margin-left: 20px;\">d) first increases and then decreases<\/div>\n<\/li>\n<li><strong>Assertion:<\/strong> n small balls each of mass m colliding plastically each second on surface which velocity u. The force experienced by the surface is 2mnu.<br \/>\n<strong>Reason:<\/strong> On elastic collision, the ball rebounds with the same velocity.<\/p>\n<div style=\"margin-left: 20px;\">a) Assertion and reason both are correct statements and reason is correct explanation for assertion.<\/div>\n<div style=\"margin-left: 20px;\">b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.<\/div>\n<div style=\"margin-left: 20px;\">c) Assertion is correct statement but reason is wrong statement.<\/div>\n<div style=\"margin-left: 20px;\">d) Assertion is wrong statement but reason is correct statement.<\/div>\n<\/li>\n<li><strong>Assertion:<\/strong> The value of AQ is always zero in adiabatic process.<br \/>\n<strong>Reason:<\/strong> Adiabatic process is always a cyclic process.<\/p>\n<div style=\"margin-left: 20px;\">a) Assertion and reason both are correct statements and reason is correct explanation for assertion.<\/div>\n<div style=\"margin-left: 20px;\">b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.<\/div>\n<div style=\"margin-left: 20px;\">c) Assertion is correct statement but reason is wrong statement.<\/div>\n<div style=\"margin-left: 20px;\">d) Assertion is wrong statement but reason is correct statement.<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> If earth were a hollow sphere, gravitational field intensity at any point inside the earth would be zero.<br \/>\n<strong>Reason (R):<\/strong> Net force on a body inside the sphere is zero.<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> In projectile motion, the angle between the instantaneous velocity and acceleration at the highest point is 180\u00b0.<br \/>\n<strong>Reason (R):<\/strong> At the highest point, velocity of projectile will be in horizontal direction only.<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\">\n<h3><span class=\"ez-toc-section\" id=\"Physics_Sample_Paper_Section_B\"><\/span><b>Physics Sample Paper Section B<\/b><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<\/li>\n<li>At what temperature (in \u00b0C) will the speed of sound in air be 3 times its value at 0\u00b0C?<\/li>\n<li>In the relation p = <span class=\"math-tex\">{tex}( a \/ b ) e ^ { &#8211; ( a z \/ \\theta )}{\/tex}<\/span>, p is the pressure, z is the distance, and <span class=\"math-tex\">{tex}\\theta{\/tex}<\/span> is the temperature. What is the dimensional formula of b?<\/li>\n<li>A gas bubble, from an explosion underwater, oscillates with a period T proportional to p<sup>a<\/sup> d<sup>b<\/sup> E<sup>c<\/sup>, where p is the static pressure, d is the density of water and E is the total energy of the explosion. Find the values of a, b and c.<\/li>\n<li>A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. For sliding the block A on B, a horizontal force of 12 N is required to be applied on it. How much maximum horizontal force can be applied on B so that both A and B move together? Also find out the acceleration produced by this force.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1637065523-m7zmsb.jpg\" alt=\"\" width=\"149\" height=\"65\" data-imgur-src=\"rfvpH0a.png\" \/><\/li>\n<li>The distance of the planet Jupiter from the sun is 5.2 times that of the earth. Find the period of the Jupiter\u2019s revolution around the sun?\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>What is the difference between inertial mass and gravitational mass of a body?<\/li>\n<li style=\"text-align: center; clear: both; display: block;\">\n<h3><span class=\"ez-toc-section\" id=\"Physics_Sample_Paper_Section_C\"><\/span><b>Physics Sample Paper Section C<\/b><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<\/li>\n<li>Eight rain drops of radius 1 mm each falling down with terminal velocity of 5 cms<sup>-1<\/sup> coalesce to form a bigger drop. Find the terminal velocity of the bigger drop.<\/li>\n<li>Develop a relation between the co-efficient of linear expansion, co-efficient superficial expansion and coefficient of cubical expansion of a solid.<\/li>\n<li>A parachutist bails out from an aeroplane and after dropping through a distance of 40 m, he opens the parachute and decelerates at 2 ms<sup>-2<\/sup>. If he reaches the ground with a speed of 2 ms<sup>-1<\/sup>, how long is he in the air? At what height did he bailout from the plane?<\/li>\n<li>A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium? (Take g = 10 ms<sup>-2<\/sup>). Neglect the mass of the rope.<br \/>\n<img decoding=\"async\" style=\"width: 108px; height: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1613448893-gu6b37.jpg\" alt=\"\" data-imgur-src=\"LevYvR1.png\" \/><img decoding=\"async\" style=\"width: 66px; height: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1613448896-6p2r3t.jpg\" alt=\"\" data-imgur-src=\"y8AOuQu.png\" \/><img decoding=\"async\" style=\"width: 90px; height: 115px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1613448898-4w7guq.jpg\" alt=\"\" data-imgur-src=\"wUvW03Q.png\" \/><\/li>\n<li>Explain, why?\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>500 J of work is done on a gas to reduce its volume by compression adiabatically. What is the change in internal energy of the gas?<\/li>\n<li>The coolant in a chemical or a nuclear plant, i.e. the liquid used to prevent the different parts of a plant from getting too hot should have high specific heat.<\/li>\n<li>The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.<\/li>\n<\/ol>\n<\/li>\n<li>Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms<sup>-2<\/sup>. What is the net force on the man? If the coefficient of static friction between the man&#8217;s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)<br \/>\n<strong><img decoding=\"async\" id=\"Picture 58\" style=\"height: 76px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/impq\/11\/phy\/ch05\/3m\/image092.jpg\" \/><\/strong><\/li>\n<li>Briefly discuss how Pascal&#8217;s law is affected by gravity. Hence obtain Pascal&#8217;s law of transmission of pressure.\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>To what height should a cylindrical vessel be filled with a homogeneous liquid to make the force, with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel?<\/li>\n<li style=\"text-align: center; clear: both; display: block;\">\n<h3><span class=\"ez-toc-section\" id=\"Physics_Sample_Paper_Section_D\"><\/span><b>Physics Sample Paper Section D<\/b><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<\/li>\n<li><b>Read the text carefully and answer the questions:<\/b><br \/>\nElastic potential energy is Potential energy stored as a result of the deformation of an elastic object, such as the stretching of a spring. It is equal to the\u00a0work\u00a0done to stretch the spring, which depends upon the spring constant k as well as the distance stretched<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1607601210-kjq9xr.jpg\" alt=\"\" data-imgur-src=\"9S4txUh.png\" \/><\/p>\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li>If stretch in spring of force constant k is doubled, then the ratio of final to initial forces is:\n<div style=\"margin-left: 20px;\">a) 4:1<\/div>\n<div style=\"margin-left: 20px;\">b) 1:4<\/div>\n<div style=\"margin-left: 20px;\">c) 2:1<\/div>\n<div style=\"margin-left: 20px;\">d) 1:2<\/div>\n<\/li>\n<li>A light body and a heavy body have the same kinetic energy. which one has greater linear momentum?\n<div style=\"margin-left: 20px;\">a) light body<\/div>\n<div style=\"margin-left: 20px;\">b) both heavy and light body<\/div>\n<div style=\"margin-left: 20px;\">c) none of these<\/div>\n<div style=\"margin-left: 20px;\">d) heavy body<\/div>\n<\/li>\n<li>A spring is cut into two equal halves. How is the spring constant of each half affected?\n<div style=\"margin-left: 20px;\">a) becomes double<\/div>\n<div style=\"margin-left: 20px;\">b) none of these<\/div>\n<div style=\"margin-left: 20px;\">c) becomes 1\/4th<\/div>\n<div style=\"margin-left: 20px;\">d) becomes half<\/div>\n<\/li>\n<li style=\"list-style: none; display: block;\">\n<p style=\"text-align: center;\"><b>OR<\/b><\/p>\n<p>When spring is compressed, its potential energy:<\/p>\n<div style=\"margin-left: 20px;\">a) none of these<\/div>\n<div style=\"margin-left: 20px;\">b) decrease<\/div>\n<div style=\"margin-left: 20px;\">c) first\u00a0increase then decrease<\/div>\n<div style=\"margin-left: 20px;\">d) increase<\/div>\n<\/li>\n<li>What type of energy is stored in the spring of a watch?\n<div style=\"margin-left: 20px;\">a) potential energy<\/div>\n<div style=\"margin-left: 20px;\">b) none of these<\/div>\n<div style=\"margin-left: 20px;\">c) mechanical energy<\/div>\n<div style=\"margin-left: 20px;\">d) kinetic energy<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li><b>Read the text carefully and answer the questions:<\/b><br \/>\nIn a gas the particles are always in a state of random motion, all the particles move at different speed constantly colliding and changing their speed and direction, as speed increases it will result\u00a0in an\u00a0increase in its kinetic energy.<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 208px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1607601157-ctn57y.jpg\" alt=\"\" data-imgur-src=\"R0xUBvM.png\" \/><\/p>\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li>If the temperature of the gas increases from 300 K to 600 K then the average kinetic energy becomes:\n<div style=\"margin-left: 20px;\">a) same<\/div>\n<div style=\"margin-left: 20px;\">b) becomes double<\/div>\n<div style=\"margin-left: 20px;\">c) becomes half<\/div>\n<div style=\"margin-left: 20px;\">d) none of these<\/div>\n<\/li>\n<li>What is the average velocity of the molecules of an ideal gas?\n<div style=\"margin-left: 20px;\">a) Infinite<\/div>\n<div style=\"margin-left: 20px;\">b) Same<\/div>\n<div style=\"margin-left: 20px;\">c) None of these<\/div>\n<div style=\"margin-left: 20px;\">d) Zero<\/div>\n<\/li>\n<li>Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will ________.\n<div style=\"margin-left: 20px;\">a) decrease<\/div>\n<div style=\"margin-left: 20px;\">b) none of these<\/div>\n<div style=\"margin-left: 20px;\">c) increase<\/div>\n<div style=\"margin-left: 20px;\">d) remains same<\/div>\n<\/li>\n<li>Find the ratio of average kinetic energy per molecule of Oxygen and\u00a0Hydrogen:\n<div style=\"margin-left: 20px;\">a) 1:1<\/div>\n<div style=\"margin-left: 20px;\">b) 4:1<\/div>\n<div style=\"margin-left: 20px;\">c) 1:2<\/div>\n<div style=\"margin-left: 20px;\">d) 1:4<\/div>\n<\/li>\n<li style=\"list-style: none; display: block;\">\n<p style=\"text-align: center;\"><b>OR<\/b><\/p>\n<p>The velocities of the three\u00a0molecules are 3v, 4v, and 5v. calculate their root mean square velocity?<\/p>\n<div style=\"margin-left: 20px;\">a) 4.0 v<\/div>\n<div style=\"margin-left: 20px;\">b) 4.02 v<\/div>\n<div style=\"margin-left: 20px;\">c) 4.08 v<\/div>\n<div style=\"margin-left: 20px;\">d) 4.04 v<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p>To practice more questions &amp; prepare well for exams, download <strong>myCBSEguide App<\/strong>. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use <strong>Examin8 App<\/strong> to create similar papers with their own name and logo.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section E<\/b><\/li>\n<li>A person normally weighing 50 kg stands on a mass less platform which oscillates up and down harmonically at a frequency of 2.0 s<sup>\u20131\u00a0<\/sup>and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Will there be any change in weight of the body, during the oscillation? Figure In extensible string.<\/li>\n<li>If answer to part (a) is yes, what will be the maximum and\u00a0minimum reading in the machine and at which position?<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = Acos(<span class=\"math-tex\">{tex}\\omega{\/tex}<\/span>t + <span class=\"math-tex\">{tex}\\phi{\/tex}<\/span>) If the initial (t = 0) position of the particle is 1 cm and its initial velocity is <span class=\"math-tex\">{tex}\\omega{\/tex}<\/span>\u00a0cm \/s, then what are its amplitude and initial phase angle? The angular frequency of the particle is <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span> s<sup>-1<\/sup>. If instead of the cosine function, we choose the sine function to describe the SHM, x = Bsin(<span class=\"math-tex\">{tex}\\omega{\/tex}<\/span>t + <span class=\"math-tex\">{tex}\\phi{\/tex}<\/span>), then what are the amplitude and initial phase of the particle with the above initial conditions?<\/li>\n<li>A marble rolls along a table at a constant speed of 1.00 m\/s and then falls off the edge of the table to the floor 1.00 m below,\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>How long does the marble take to reach the floor?<\/li>\n<li>At what horizontal distance from the edge of the table does the marble land?<\/li>\n<li>What is its velocity as it strikes the floor?<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>What is the angle between <span class=\"math-tex\">{tex}\\overrightarrow A{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\overrightarrow B{\/tex}<\/span> if <span class=\"math-tex\">{tex}\\overrightarrow A{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\overrightarrow B{\/tex}<\/span> denote the adjacent sides of a parallelogram drawn form a point and the area of the parallelogram is<span class=\"math-tex\">{tex}\\frac { 1 } { 2 } A B{\/tex}<\/span>?<\/li>\n<li>State and prove triangular law of vector addition.<\/li>\n<\/ol>\n<\/li>\n<li>A particle of mass m is released from point P at x = x<sub>0<\/sub>\u00a0on the X-axis from origin O and falls vertically along the Y-axis, as shown in Fig.<br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 148px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1685082708-zb8nc9.jpg\" alt=\"\" data-imgur-src=\"lfXCevJ.png\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Find the torque t acting on the particle at a time t when it is at point Q with respect to O.<\/li>\n<li>Find the angular momentum L of the particle about O at this time t.<\/li>\n<li>Show that\u00a0<span class=\"math-tex\">{tex}\\tau=\\frac{d L}{d t}{\/tex}<\/span>\u00a0in this example.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>A uniform disc of radius\u00a0<em>R<\/em>, is resting on a table on its rim. The coefficient of friction between disc and table is <span class=\"math-tex\">{tex}\\mu{\/tex}<\/span>\u00a0(Figure).<br \/>\n<img decoding=\"async\" style=\"width: 176px; height: 67px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1613364811-nsj4td.jpg\" alt=\"\" data-imgur-src=\"HNQ3RlM.png\" \/><br \/>\nNow the disc is pulled with a force\u00a0<span class=\"math-tex\">{tex}\\vec { F }{\/tex}<\/span>\u00a0as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?<\/li>\n<\/ol>\n<h2 style=\"text-align: center; page-break-before: always;\"><span class=\"ez-toc-section\" id=\"Class_11_%E2%80%93_Physics_Sample_Paper_Solution_2024-25\"><\/span><strong>Class 11 &#8211; Physics Sample Paper Solution (2024-25)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>You are getting not only the question paper but also a complete solution with a step-by-step explanation. The solution of the question paper is based on the latest guidelines issued by CBSE for the session 2024-25.<\/p>\n<p>To get more model papers with solutions, you can download the myCBSEguide app or visit <a href=\"https:\/\/mycbseguide.com\/dashboard\/\">https:\/\/mycbseguide.com\/dashboard\/<\/a>. CBSE has reduced the course content for class 11 Physics. This question paper and solution is based on the updated and reduced syllabus of CBSE for 2024-25.<\/p>\n<hr \/>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Class_11_Physics_Sample_Paper_Solution_Section_A\"><\/span><strong>Class 11 Physics Sample Paper Solution <\/strong><b>Section A <\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol style=\"padding-left: 20px;\">\n<li>(c) 1.9 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>12<\/sup><br \/>\n<b>Explanation: <\/b> According to the problem,<br \/>\nYoung&#8217;s modulus, Y = 1.9\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>11<\/sup> N\/m<sup>2<\/sup><br \/>\n1N in SI system of units = 10<sup>5<\/sup> dyne in C.G.S system.<br \/>\nHence, Y = 1.9\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>11<\/sup>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>5<\/sup> dyne\/m<sup>2<\/sup><br \/>\nIn C.G.S length is measured in unit cm, so we should also convert m into cm.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Y =\u00a0<span class=\"math-tex\">{tex}1.9 \\times 10^{11}\\left(\\frac{10^{5} \\mathrm{dyne}}{10^{4} \\mathrm{cm}^{2}}\\right){\/tex}<\/span>\u00a0[<span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a01m = 100 cm]\n= 1.9\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>12<\/sup> dyne\/cm<sup>2<\/sup><\/li>\n<li>(d) 20 to 20000 Hz<br \/>\n<b>Explanation: <\/b> Humans are most sensitive to (i.e. able to discern at lowest intensity) frequencies between 2,000 and 5,000 Hz. The human ear can respond to minute pressure variations in the air if they are in the audible frequency range, roughly 20 Hz &#8211; 20 kHz. This incredible sensitivity is enhanced by an effective amplification of the sound signal by the outer and middle ear structures.<\/li>\n<li>(c) as that point where the total gravitational torque on the body is zero<br \/>\n<b>Explanation: <\/b> The point at which the entire weight of a body may be thought of as centered so that if supported at this point the body would balance perfectly, so it can also be defined as that point where the total gravitational torque on the body is zero.<\/li>\n<li>(b) Six times that on the earth&#8217;s surface<br \/>\n<b>Explanation: <\/b> The height to which the liquid rises in a capillary tube is inversion all proportional to the acceleration\u00a0due to gravity.<br \/>\n<span class=\"math-tex\">{tex}h=\\frac{2 S \\cos \\theta}{r \\rho g}{\/tex}<\/span><br \/>\nthus if all other parameters are fixed<br \/>\n<span class=\"math-tex\">{tex}h \\propto\\frac{1}{g}{\/tex}<\/span><br \/>\nOn the surface of the moon, acceleration due to gravity is\u00a0<span class=\"math-tex\">{tex}1\\over6{\/tex}<\/span>th as that on the surface of the earth. Therefore the water will rise to a\u00a0height\u00a06 times that on earth&#8217;s surface.<\/li>\n<li>(b) <span class=\"math-tex\">{tex}{\\rm{v}} = &#8211; \\frac{G{m_1}{m_2}}{r}{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> <img decoding=\"async\" style=\"width: 150px; height: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/cbse\/review\/Image_18.png\" alt=\"\" \/><br \/>\nChange in gravitational potential energy of a system is defined as the -ve of the work done by the gravitational force as the configuration of the system is changed.<br \/>\n<span class=\"math-tex\">{tex}{U_f}\\; &#8211; {\\rm{ }}{U_i}\\; = {\\rm{ }}{W_{gr}}=V(Potential Energy){\/tex}<\/span><br \/>\nChange in gravitational potential energy of two point masses m<sub>1<\/sub>\u00a0and m<sub>2<\/sub>\u00a0as their separation is changed from r<sub>1\u00a0<\/sub>to r<sub>2<\/sub>\u00a0is given by<br \/>\n<span class=\"math-tex\">{tex}U\\left(r_2\\right)-U\\left(r_1\\right)=G m_1 m_2\\left[\\frac{1}{r_1}-\\frac{1}{r_2}\\right]{\/tex}<\/span><br \/>\nIf, at infinite separation, gravitational potential energy is assumed to be zero, then the gravitational potential energy of the above two point mass system at separation r,<br \/>\n<span class=\"math-tex\">{tex}U(r) = \u00a0&#8211; G\\frac{{m_1}{m_2}} {r}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore V = \u00a0&#8211; G\\frac{{m_1}{m_2}} { r}{\/tex}<\/span><br \/>\nTo practice more questions &amp; prepare well for exams, download <strong>myCBSEguide App<\/strong>. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use <strong>Examin8 App<\/strong> to create similar papers with their own name and logo.<\/li>\n<li>(b) 105\u00a0second <sup>-1<\/sup><br \/>\n<b>Explanation: <\/b> Frequency of the source = 100 <span class=\"math-tex\">{tex}\\pm{\/tex}<\/span>\u00a05 = 105 or 95 Hz<br \/>\nThe second harmonic of the source = 210 Hz or 190 Hz<br \/>\nAs the second harmonic gives 5 beats\/sec with a sound of frequency 205 Hz, the second harmonic should be 210 Hz.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Frequency of the source = 105 Hz<\/li>\n<li>(d) rectilinear motion<br \/>\n<b>Explanation: <\/b> Rectilinear motion is another name for straight-line motion. This type of motion describes the movement of a particle or a body. A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.<\/li>\n<li>(c) 2 <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>A<br \/>\n<b>Explanation: <\/b> Maximum particle velocity =\u00a0Wave velocity<br \/>\n<span class=\"math-tex\">{tex}\\omega A=\\frac{\\omega}{k}{\/tex}<\/span><br \/>\nor\u00a0<span class=\"math-tex\">{tex}k=\\frac{2 \\pi}{\\lambda}=\\frac{1}{A}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore \\lambda{\/tex}<\/span> =\u00a02 <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>A<\/li>\n<li>(c) 27 : 1<br \/>\n<b>Explanation: <\/b> Excess pressures in the two bubbles will be (1.01 -1) or 0.01 atm and (1.03 &#8211; 1) or 0.03 atm.<br \/>\n<span class=\"math-tex\">{tex}\\frac{p_{1}}{p_{2}}=\\frac{r_{2}}{r_{1}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\frac{r_{1}}{r_{2}}=\\frac{p_{2}}{p_{1}}=\\frac{0.03}{0.01}{\/tex}<\/span>\u00a0= 3<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac{V_{1}}{V_{2}}=\\left(\\frac{r_{1}}{r_{2}}\\right)^{3}{\/tex}<\/span>\u00a0= 3<sup>3<\/sup> = 27 : 1<\/li>\n<li>(d) Option iv<br \/>\n<b>Explanation: <\/b> The required escape velocity of gas molecules is less than their root mean square velocity.<\/li>\n<li>(a) 2<br \/>\n<b>Explanation: <\/b> As radius of second ring is n\u00a0times, length and hence mass of wire used is also n\u00a0times.<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac{I_{1}}{I_{2}}=\\frac{M R^{2}}{n M(n R)^{2}}=\\frac{1}{n^{3}}=\\frac{1}{8}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\therefore n=2{\/tex}<\/span><\/li>\n<li>(b) first decreases and then increases<br \/>\n<b>Explanation: <\/b> Density of water is maximum at 4<sup>o<\/sup>C. So during heating from 0<sup>o<\/sup>C to 10<sup>o<\/sup>C, its density increases at first and then it decreases resulting in the decrease in volume at first and then increase in its volume.<\/li>\n<li>(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.<br \/>\n<b>Explanation: <\/b> In elastic collision, kinetic energy remains conserved therefore the ball rebounds with the same velocity. According to Newton\u2019s second law.<br \/>\nF <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> t = change in linear momentum.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0F\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a01 = m\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>n(u + u) or F = 2mnu<\/li>\n<li>(c) Assertion is correct statement but reason is wrong statement.<br \/>\n<b>Explanation: <\/b> Assertion is correct statement but reason is wrong statement.<\/li>\n<li>(a) Both A and R are true and R is the correct explanation of A.<br \/>\n<b>Explanation: <\/b> <img decoding=\"async\" style=\"width: 190px; height: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/1632052297-26jkyw.jpg\" alt=\"\" data-imgur-src=\"LwM95Qd.png\" \/><br \/>\nA uniform shell of matter exerts no gravitational force on a particle located inside it. This also means if a particle is located inside a uniform solid sphere of matter at a distance r\u00a0from its centre, the gravitational force exerted on the particle is due only to the mass M&#8217;\u00a0that lies within a sphere of radius r. This mass M&#8217;\u00a0is given by M&#8217; = <span class=\"math-tex\">{tex}\\rho \\frac{4 \\pi r^{3}}{3}{\/tex}<\/span>, where <span class=\"math-tex\">{tex}\\rho{\/tex}<\/span>\u00a0is the density of the sphere.<\/li>\n<li>(d) A is false but R is true.<br \/>\n<b>Explanation: <\/b> At the highest point, the instantaneous velocity is acting horizontally and acceleration of projectile (acceleration due to gravity) is acting vertically downward. Therefore, angle between velocity and acceleration at the highest point is 90\u00b0.<\/li>\n<li style=\"text-align: center; display: block;\">\n<h2><span class=\"ez-toc-section\" id=\"Class_11_Physics_Sample_Paper_Solution_Section_B\"><\/span><b>Class 11 Physics Sample Paper Solution Section B <\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<\/li>\n<li>We know that, speed of sound, v\u00a0<span class=\"math-tex\">{tex} \\propto \\sqrt{T} {\/tex}<\/span>, T being absolute temperature.\u200b\u200b\u200b<br \/>\nwhere, T is in kelvin.<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\frac{V_{t}}{V_{0}}=\\sqrt{\\frac{273+t}{273+0}} {\/tex}<\/span>\u00a0= 3 (say, the required temperature is t\u2103)<br \/>\n<span class=\"math-tex\">{tex} \\Rightarrow \\frac{273+t}{273} {\/tex}<\/span>\u00a0= 9<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0t = (9\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0273) &#8211; 273 = 2184\u00b0C<\/li>\n<li>Since,\u00a0<span class=\"math-tex\">{tex}e ^ { &#8211; ( a z \/ \\theta ) }{\/tex}<\/span>\u00a0is dimensionless (exponential function), we have\u00a0<span class=\"math-tex\">{tex}a z \/ \\theta = 1{\/tex}<\/span><br \/>\nor,\u00a0<span class=\"math-tex\">{tex}a = \\frac { \\theta } { Z } = \\frac { K } { L } = \\left[ \\mathrm { L } ^ { &#8211; 1 } \\mathrm { K } \\right]{\/tex}<\/span><br \/>\nWe find that a\/b = dimensions of pressure (because <span class=\"math-tex\">{tex}e ^ { &#8211; ( a z \/ \\theta ) }{\/tex}<\/span>\u00a0is dimensionless)<br \/>\na\/b = [ML<sup>-1<\/sup>T<sup>-2<\/sup>]\nTherefore,<br \/>\n<span class=\"math-tex\">{tex}b = \\frac { a } { \\left[ \\mathrm { ML } ^ { &#8211; 1 } \\mathrm { T } ^ { &#8211; 2 } \\right] } = \\frac { \\left[ \\mathrm { L } ^ { &#8211; 1 } \\mathrm { K } \\right] } { \\left[ \\mathrm { ML } ^ { &#8211; 1 } \\mathrm { T } ^ { &#8211; 2 } \\right] }{\/tex}<\/span><br \/>\n= [M<sup>-1<\/sup>T<sup>2<\/sup>K]<\/li>\n<li>Let T = K p<sup>a<\/sup> d<sup>b<\/sup> E<sup>c<\/sup>,<br \/>\nwhere K = a dimensionless constant.<br \/>\nPutting the dimensions of various quantities,<br \/>\nT = [ML<sup>-1<\/sup> T<sup>-2<\/sup>]<sup>a<\/sup> [ML<sup>-3<\/sup>]<sup>b<\/sup> [ML<sup>2<\/sup> T<sup>-2<\/sup>]<sup>c<\/sup><br \/>\nor M<sup>0<\/sup>L<sup>0<\/sup>T = M<sup>a +b + c<\/sup> L<sup>~a ~3b + 2c<\/sup> T<sup>-2a &#8211; 2c<\/sup><br \/>\nEquating the powers of M, L and T on both sides, we get<br \/>\na + b + c = 0, &#8211; a &#8211; 3b + 2c = 0, &#8211; 2a &#8211; 2c = 1<br \/>\nOn solving,<br \/>\na =\u00a0<span class=\"math-tex\">{tex}-\\frac{5}{6}{\/tex}<\/span>, b =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>, c =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{3}{\/tex}<\/span><\/li>\n<li>Here m<sub>1<\/sub> = 4 kg, m<sub>2<\/sub>, = 5 kg<br \/>\nForce applied on block A = 12 N<br \/>\nThis force must atleast be equal to the kinetic friction applied on A by B.<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span> 12 = f<sub>k<\/sub> = <span class=\"math-tex\">{tex}\\mu_k{\/tex}<\/span> R = <span class=\"math-tex\">{tex}\\mu_k{\/tex}<\/span> m<sub>1<\/sub> g<br \/>\nor 12 = <span class=\"math-tex\">{tex}\\mu_k{\/tex}<\/span> <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 4 g<br \/>\nor <span class=\"math-tex\">{tex}\\mu_{k}=\\frac{12}{4 g}=\\frac{3}{g}{\/tex}<\/span><br \/>\nThe block B is on a smooth surface. Hence to move A and B together, the (maximum) force F that can be applied on B is equal to the frictional forces applied on A by B and applied on B by A.<br \/>\nF = <span class=\"math-tex\">{tex}\\mu_k{\/tex}<\/span> m<sub>1<\/sub>g + <span class=\"math-tex\">{tex}\\mu_k{\/tex}<\/span> m<sub>2<\/sub> g = <span class=\"math-tex\">{tex}\\mu_k{\/tex}<\/span> (m<sub>1<\/sub> + m<sub>2<\/sub>)g<br \/>\n<span class=\"math-tex\">{tex}=\\frac{3}{g}{\/tex}<\/span>(4 + 5)g = 27 N<br \/>\nAs this force moves both the blocks together on a smooth table, so the acceleration produced is<br \/>\n<span class=\"math-tex\">{tex}a=\\frac{F}{m_{1}+m_{2}}=\\frac{27}{4+5}{\/tex}<\/span> = 3 ms<sup>-2<\/sup><\/li>\n<li style=\"clear: both;\">Te = 1 year R<sub>J\u00a0<\/sub>= 5.2 R<sub>e<\/sub><br \/>\nAccording to the Kepler&#8217;s third law<br \/>\nT<sup>2<\/sup>\u00a0is directly proportional R<sup>3<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\left( \\frac { T _ { J } } { \\overline { T } e } \\right) ^ { 2 } = \\left( \\frac { R _ { J } } { \\mathrm { Re } } \\right) ^ {3 }{\/tex}<\/span><br \/>\nT<sub>J <\/sub>= (5.2) <span class=\"math-tex\">{tex}\\frac 32{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 1year<br \/>\nT<sub>J<\/sub> = 11.86 year<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>The inertial mass of a body is a measure of its inertia and is given by the ratio of the external force applied on it to the acceleration produced in it. The gravitational mass of a body, on the other hand, is a measure of the gravitational pull acting on it due to the earth. The gravitational mass is measured by a common balance.<\/li>\n<li style=\"text-align: center; display: block;\">\n<h2><span class=\"ez-toc-section\" id=\"Class_11_Physic_Sample_Paper_Solution_Section_C\"><\/span><b>Class 11 Physic Sample Paper Solution Section C <\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<\/li>\n<li>Radius of each small drop, r = 1 mm = 0.1 cm<br \/>\nTerminal velocity of each small drop, v = 5 cms<sup>-1<\/sup><br \/>\nVolume of bigger drop = Volume of 8 small drops<br \/>\n<span class=\"math-tex\">{tex}\\frac{4}{3} \\pi R^{3}=8 \\times \\frac{4}{3} \\pi r^{3}{\/tex}<\/span><br \/>\nor R = 2r = 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.1 cm = 0.2 cm<br \/>\nTerminal velocity of each small drop is given by<br \/>\nv =\u00a0<span class=\"math-tex\">{tex}=\\frac{2}{9} \\frac{r^{2}}{\\eta}\\left(\\rho-\\rho^{\\prime}\\right) \\mathrm{g}{\/tex}<\/span>\u00a0&#8230;(i)<br \/>\nTerminal velocity of bigger drop is given by<br \/>\nV =\u00a0<span class=\"math-tex\">{tex}\\frac{2}{9} \\frac{R^{2}}{\\eta}\\left(\\rho-\\rho^{\\prime}\\right) g{\/tex}<\/span>\u00a0&#8230;(ii)<br \/>\nDividing equation (ii) by (i), we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{V}{v}=\\frac{R^{2}}{r^{2}}{\/tex}<\/span><br \/>\nor V =\u00a0<span class=\"math-tex\">{tex}v \\times \\frac{R^{2}}{r^{2}}=5 \\times \\frac{(0.2)^{2}}{(0.1)^{2}}{\/tex}<\/span><br \/>\n= 5\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a04 = 20 cms<sup>-1<\/sup><\/li>\n<li>Since, co-efficient of linear expansion,\u00a0<span class=\"math-tex\">{tex}\\alpha{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac { \\Delta L } { L \\Delta T }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Delta L{\/tex}<\/span> = change in length<br \/>\nL = length<br \/>\n<span class=\"math-tex\">{tex}\\Delta T{\/tex}<\/span> = change in temperature, for an infinitesimally small change in temperature<br \/>\n<span class=\"math-tex\">{tex}\\alpha = \\frac{dL}{LdT}{\/tex}<\/span><br \/>\nSimilarly, co-efficient of superficial expansion,\u00a0<span class=\"math-tex\">{tex}\\beta{\/tex}<\/span> = <span class=\"math-tex\">{tex}\\frac {d S } { S dT }{\/tex}<\/span><br \/>\ndS = infinitesimal change in area<br \/>\nS = original area<br \/>\ndT = infintesimaly change in temperature<br \/>\nS = L<sup>2<\/sup>,<sup>\u00a0<\/sup><span class=\"math-tex\">{tex}\\beta= \\frac{1}{L^2}\\frac{dL^2}{dT} = 2L\\frac{dL}{dT}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\beta = 2\\alpha{\/tex}<\/span><br \/>\nSimilarly, co-efficient of cubical expansion,\u00a0<span class=\"math-tex\">{tex}\\gamma = \\frac {dV}{VdT}{\/tex}<\/span><br \/>\ndV = infinitesimal change in volume<br \/>\nV = original volume<br \/>\ndT = infinitesimal change in temperature<br \/>\n<span class=\"math-tex\">{tex}\\gamma = \\frac{1}{L^3}\\frac{dL^3}{dT} = 3\\frac{1}{L}\\frac{dL}{dT} =3\\alpha{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\alpha = \\frac { \\beta } { 2 } = \\frac { \\gamma } { 3 }{\/tex}<\/span><\/li>\n<li>When the parachutist falls freely :<br \/>\nu = 0, v = 9.8 ms<sup>-2<\/sup>, s = 40m, t\u00a0= ?,\u00a0u = ?<br \/>\nAs s = ut +\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2} g t^{2}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore 40=0+\\frac{1}{2} \\times 9.8 \\times t^{2}{\/tex}<\/span><br \/>\nor t =\u00a0<span class=\"math-tex\">{tex}\\sqrt{\\frac{80}{9.8}}=\\frac{20}{7} \\mathrm{~s}{\/tex}<\/span>\u00a0= 2.86 s<br \/>\nAlso, v = u + gt = 0 + 9.8\u00a0<span class=\"math-tex\">{tex}\\times \\frac{20}{7}{\/tex}<\/span>\u00a0= 28 ms<sup>-1<\/sup><br \/>\nWhen the parachutist decelerates uniformly:<br \/>\nu = 28 ms<sup>-1<\/sup>, a = -2 ms<sup>-2<\/sup>, 5 = 2 ms<sup>-1<\/sup><br \/>\nTime taken, t =\u00a0<span class=\"math-tex\">{tex}\\frac{v \\ &#8211; \\ u}{a}=\\frac{2 \\ &#8211; \\ 28}{-2}{\/tex}<\/span>\u00a0= 13 s<br \/>\nDistance, s = ut +\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2} at^{2}{\/tex}<\/span>\u00a0= 28\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a013\u00a0<span class=\"math-tex\">{tex}-\\frac{1}{2} \\times{\/tex}<\/span>\u00a02\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0(13)<sup>2<\/sup><br \/>\n= 364 &#8211; 169 = 195 m<br \/>\nTotal time of parachutist in air<br \/>\n= 2.86 + 13 = 15.86 s<br \/>\nHeight at which parachutist bails out<br \/>\n= 40 + 195 = 235 m<\/li>\n<li>Figure (b) and (c) are known as free-body diagrams. Figure (b) is the free-body diagram of W and Figure (c) is the free-body diagram of point P.<br \/>\nConsider the equilibrium of the weight W. Clearly, T<sub>2<\/sub> = 6 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010 = 60 N. Consider the equilibrium of the point P under the action of three forces &#8211; the tensions T<sub>1<\/sub> and T<sub>2<\/sub>, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately:<br \/>\nT<sub>1<\/sub> cos<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= T<sub>2<\/sub> = 60N<br \/>\nT<sub>1<\/sub> sin<span class=\"math-tex\">{tex}\\theta{\/tex}<\/span>\u00a0= 50 N<br \/>\nWhich gives that<br \/>\n<span class=\"math-tex\">{tex}\\tan \\theta=\\frac{5}{6}{\/tex}<\/span>\u00a0or\u00a0<span class=\"math-tex\">{tex}=\\tan ^{-1}\\left(\\frac{5}{6}\\right)=40^{\\circ}{\/tex}<\/span><br \/>\nNote the answer does not depend on the length of the rope (assumed massless) nor on the point at which the horizontal force is applied.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><span class=\"math-tex\">{tex}\\because{\/tex}<\/span>\u00a0process is adiabatic<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>\u00a0= 0<br \/>\nWork done on the gas, <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>W = -500J<br \/>\nAccording to the first law of thermodynamics.<br \/>\n<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>Q\u00a0= <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>U + <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>W <span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>U\u00a0= &#8211;<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>W = 500J<\/li>\n<li>This is because heat absorbed by a substance (coolant) is directly proportional to the specific heat of the substance.<\/li>\n<li>\u00a0This is because in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of a harbour town is without extremes of hot and cold.<\/li>\n<\/ol>\n<\/li>\n<li>Conveyor belt is accelerating with 1 m\/s\u00b2. A man is stationary with respect to horizontal conveyor belt<br \/>\nHence, acceleration of man = acceleration of belt = 1m\/s\u00b2<br \/>\nHence,\u00a0Net force on the man = ma\u00a0= 65\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a01 = 65 N direction of this force is opposite to direction of conveyor belt.<br \/>\nCoefficient of friction between the man&#8217;s shoes and belt = 0.2.\u00a0Let a&#8217; is the acceleration of the belt can the man continue to be stationary relative to the belt.<br \/>\nIn equilibrium condition,<br \/>\nFriction force = ma&#8217;<br \/>\n0.2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0m <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0g = ma&#8217;<br \/>\na&#8217; = 0.2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0g = 0.2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010 = 2m\/s\u00b2<\/li>\n<li style=\"clear: both;\">PASCAL LAW :-<br \/>\nOne of the most important facts about fluid pressure is that a change in pressure at one part of the liquid will be transmitted without any change to other parts. This rule is known as Pascal&#8217;s law.<br \/>\nTake a vessel containing a liquid in the equilibrium at\u00a0rest. Consider a volume element of liquid of height h. In the absence of gravity, according to\u00a0Pascal&#8217;s law,\u00a0the pressure at the upper section A and lower section B of the volume element should be same.<br \/>\nHowever, taking gravity into consideration we find the following forces acting on the said volume element:<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Force P<sub>1<\/sub> (<span class=\"math-tex\">{tex}\\Delta A{\/tex}<\/span>) acting vertically downward at the top face A,<\/li>\n<li>Force P<sub>2<\/sub>\u00a0(<span class=\"math-tex\">{tex}\\Delta A{\/tex}<\/span>) acting vertically upward at the bottom face B, and<\/li>\n<\/ol>\n<p>Weight mg of the liquid in the volume\u00a0element acting vertically downwards. Here,\u00a0<span class=\"math-tex\">{tex}\\Delta A{\/tex}<\/span>\u00a0is the cross-section area of the given volume element.<br \/>\nFor volume element to be in equilibrium net force acting on it should be zero. Therefore, we have<br \/>\n<span class=\"math-tex\">{tex}P_1\\triangle A\u00a0+ mg &#8211; P_2\\triangle A\u00a0= 0\\ or\\ P_1\\triangle A\u00a0+ mg = P_2\\triangle A{\/tex}<\/span><br \/>\nBut m = mass of volume element =\u00a0<span class=\"math-tex\">{tex}\\Delta A{\/tex}<\/span>.h.<span class=\"math-tex\">{tex}\\rho{\/tex}<\/span>, where\u00a0<span class=\"math-tex\">{tex}\\rho{\/tex}<\/span>\u00a0= density of the liquid<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}P_1\\triangle A\u00a0+\u00a0\\triangle A h\\rho g = P_2\\triangle A{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}P_2 = P_1 + h\\rho g{\/tex}<\/span><br \/>\nAccording to the above equation, if the pressure P<sub>1<\/sub> is increased in any way, the pressure P<sub>2<\/sub>\u00a0must increase by exactly the same amount.<br \/>\nThe pressure applied to any part of an enclosed liquid at rest is transmitted undiminished to every portion of the liquid as well as the walls of the container.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Let h be the height of the liquid column of density <span class=\"math-tex\">{tex}\\rho{\/tex}<\/span> taken in the cylindrical vessel of radius r.<br \/>\nForce exerted by the liquid on the bottom of the vessel\u00a0= Total weight of the liquid column<br \/>\n= mg =\u00a0<span class=\"math-tex\">{tex}\\pi r^{2} h \\rho g{\/tex}<\/span><br \/>\nArea of the sides of the vessel = 2\u00a0<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>rh<br \/>\nAverage pressure exerted by the liquid on the sides of the vessel<br \/>\n=\u00a0<span class=\"math-tex\">{tex}\\frac{\\text { Pressure at the top }+\\text { Pressure at the bottom }}{2}{\/tex}<\/span><br \/>\n=\u00a0<span class=\"math-tex\">{tex}=\\frac{0+h \\rho g}{2}=\\frac{1}{2} h \\rho g{\/tex}<\/span><br \/>\nForce exerted by the liquid on the sided of the vessel<br \/>\n= Pressure\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0Area =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2} h \\rho g \\times 2 \\pi r h{\/tex}<\/span><br \/>\nAs the above two forces are given to be equal, so<br \/>\n<span class=\"math-tex\">{tex}\\frac{1}{2} h \\rho g \\times 2 \\pi r h=\\pi r^{2} h \\rho g{\/tex}<\/span>\u00a0or h = r<br \/>\ni.e., Height of liquid column<br \/>\n= Radius of the cylindrical vessel.<\/li>\n<li style=\"text-align: center; display: block;\">\n<h2><span class=\"ez-toc-section\" id=\"Class_11_Physics_Sample_Paper_Section_D_Solution\"><\/span><b>Class 11 Physics Sample Paper Section D Solution<\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<\/li>\n<li>\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li style=\"clear: both;\">(c) 2:1<br \/>\n<b>Explanation: <\/b> 2:1<\/li>\n<li style=\"clear: both;\">(d) heavy body<br \/>\n<b>Explanation: <\/b> heavy body<\/li>\n<li style=\"clear: both;\">(a) becomes double<br \/>\n<b>Explanation: <\/b> becomes double<\/li>\n<li style=\"list-style: none; display: block;\">\n<p style=\"text-align: center;\"><b>OR<\/b><\/p>\n<p>(d) increase<br \/>\n<b>Explanation: <\/b> increase<\/li>\n<li style=\"clear: both;\">(a) potential energy<br \/>\n<b>Explanation: <\/b> potential energy<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li>\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li style=\"clear: both;\">(b) becomes double<br \/>\n<b>Explanation: <\/b> becomes double<\/li>\n<li style=\"clear: both;\">(d) Zero<br \/>\n<b>Explanation: <\/b> Zero<\/li>\n<li style=\"clear: both;\">(d) remains same<br \/>\n<b>Explanation: <\/b> remains same<\/li>\n<li style=\"clear: both;\">(a) 1:1<br \/>\n<b>Explanation: <\/b> 1:1<\/li>\n<li style=\"list-style: none; display: block;\">\n<p style=\"text-align: center;\"><b>OR<\/b><\/p>\n<p>(c) 4.08 v<br \/>\n<b>Explanation: <\/b> 4.08 v<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li style=\"text-align: center; display: block;\">\n<h2><span class=\"ez-toc-section\" id=\"Class_11_Physics_Sample_Paper_Section_E_Solution\"><\/span><b>Class 11 Physics Sample Paper Section E Solution<\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li style=\"text-align: left;\">Weight in weight machine will be due to the normal reaction (N) by platform. Consider the top position of platform, two forces acting on it are due to weight of person and oscillator. They both act downward.<br \/>\n<img decoding=\"async\" style=\"width: 158px; height: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/KMNkBew.png\" alt=\"\" data-imgur-src=\"KMNkBew.png\" \/><br \/>\n(mg = weight of the person with the oscillator is acting downwards, ma = force due to oscillation is acting upwards, N = normal reaction force acting upwards)<br \/>\n<span class=\"math-tex\">{tex}{\/tex}<\/span>Now for the downward motion of the system with an acceleration a,<br \/>\nma = mg &#8211; N &#8230;..(i)<br \/>\n<span class=\"math-tex\">{tex}{\/tex}<\/span>When platform lifts form its lowest position to upward<br \/>\nma = N &#8211; mg &#8230;&#8230;(ii)<br \/>\n<span class=\"math-tex\">{tex} a = \\omega ^ { 2 } A{\/tex}<\/span>\u00a0 is\u00a0value of acceleration of oscillator<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0From equation (i) we get,<br \/>\n<span class=\"math-tex\">{tex} N = m g &#8211; m \\omega ^ { 2 } A{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{\/tex}<\/span>Where A is amplitude, <span class=\"math-tex\">{tex}\\omega{\/tex}<\/span>\u00a0angular frequency and m mass of oscillator.<br \/>\n<span class=\"math-tex\">{tex}\\omega=2\\pi\\nu{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\therefore\\omega = 2 \\pi \\times 2 = 4 \\pi ~r a d \/ \\mathrm { sec }{\/tex}<\/span><br \/>\nAgain using\u00a0<span class=\"math-tex\">{tex}A = 5 ~ c m = 5 \\times 10 ^ { &#8211; 2 } m {\/tex}<\/span>\u00a0we get<br \/>\n<span class=\"math-tex\">{tex}N = 50 \\times 9.8 &#8211; 50 \\times 4 \\pi \\times 4 \\pi \\times 5 \\times 10 ^ { &#8211; 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= 50 \\left[ 9.8 &#8211; 16 \\pi ^ { 2 } \\times 5 \\times 10 ^ { &#8211; 2 } \\right]{\/tex}<\/span>N<br \/>\n<span class=\"math-tex\">{tex}= 50 \\left[ 9.8 &#8211; 80 \\times 3.14 \\times 3.14 \\times 10 ^ { &#8211; 2 } \\right]{\/tex}<\/span>N<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow N = 50 [ 9.8 &#8211; 7.89 ] = 50 \\times 1.91 = 95.50 \\mathrm { N }{\/tex}<\/span><br \/>\nSo minimum weight is 95.50 N(for downward motion of the platform)<br \/>\nFrom equation (ii), N \u2013 mg = ma<br \/>\nFor upward motion from the lowest to the highest point of oscillator,<br \/>\n<span class=\"math-tex\">{tex} N = m g + m a{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= m \\left[ 9.81 + \\omega ^ { 2 } A \\right] \\quad \\!\\!\u2007\\because a = \\omega ^ { 2 } A{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= 50 \\left[ 9.81 + 16 \\pi ^ { 2 } \\times 5 \\times 10 ^ { &#8211; 2 } \\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}= 50 [ 9.81 + 7.89 ] = 50 \u00d717.70 {\/tex}<\/span>N = 885 N<br \/>\nHence, there is a change in weight of the body during oscillation.<\/li>\n<li style=\"text-align: left;\">The maximum weight is 885 N, when platform moves from lowest to upward direction.<br \/>\nAnd the minimum weight is 95.5 N, when platform moves from the highest point to downward direction.<\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">As the marble was rolling on the table, therefore it has horizontal velocity and it will act as a projectile as soon as it leaves the edge of the table and fall freely under the effect of gravity.<br \/>\nSince, the marble is initially moving horizontally, v<sub>y0<\/sub>\u00a0= 0 and v<sub>x0<\/sub>\u00a0= 1.00 m\/s. We must consider the origin to be at the edge of the table, so that x<sub>0<\/sub> = y<sub>0<\/sub> = 0<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 189px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/5vsZ402.png\" alt=\"\" data-imgur-src=\"5vsZ402.png\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>t = ? and y = -1.00 m<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0y =\u00a0<span class=\"math-tex\">{tex}\\frac{-1}{2}{\/tex}<\/span>gt<sup>2<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow t = \\sqrt { \\frac { &#8211; 2 y } { g } } = \\sqrt { \\frac { ( &#8211; 2 ) ( &#8211; 1.00 ) } { 9.8 } }{\/tex}<\/span>= 0.452 s<\/li>\n<li>x = ?, when t = 0.452 s<br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0x = v<sub>x0<\/sub>t\u00a0= 1.00\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.452 s = 0.452 m<\/li>\n<li>v = ?,\u00a0<span class=\"math-tex\">{tex}\\theta {\/tex}<\/span>\u00a0= ? at t = 0.452 s<br \/>\nThe x-component of velocity is constant throughout the motion,<br \/>\nv<sub>x<\/sub>\u00a0= v<sub>x0<\/sub>\u00a0= 1.00 m\/s<br \/>\nThe y-component of velocity is given by<br \/>\nv<sub>y<\/sub>\u00a0= v<sub>y0<\/sub>\u00a0&#8211; gt = 0 &#8211; 9.8\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a00.452 = &#8211; 4.43 m\/s<br \/>\n<span class=\"math-tex\">{tex}\\therefore v = \\sqrt { v _ { x } ^ { 2 } + v _ { y } ^ { 2 } }{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}= \\sqrt { ( 1.00 ) ^ { 2 } + ( &#8211; 4.43 ) ^ { 2 } }{\/tex}<\/span>\u00a0= 4.54 m\/s, the magnitude of the resultant velocity of the motion.<br \/>\n<span class=\"math-tex\">{tex}\\theta = \\tan ^ { &#8211; 1 } \\left| \\frac { v _ { y } } { v _ { x } } \\right| = \\frac { 4.43 } { 1.00 } = 77.3 ^ { \\circ }{\/tex}<\/span><br \/>\nAs the marble hits the floor, its velocity is 4.54 m\/s directed 77.3\u00b0 downward with respect to\u00a0the horizontal.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>Area of a parallelogram\u00a0<span class=\"math-tex\">{tex}= | \\vec { A } \\times \\vec { B } |{\/tex}<\/span>\u00a0=\u00a0<span class=\"math-tex\">{tex}A B \\sin \\theta{\/tex}<\/span>\u00a0(<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Applying cross product)<br \/>\nGiven, area of parallelogram\u00a0<span class=\"math-tex\">{tex}= \\frac { 1 } { 2 } A B{\/tex}<\/span><br \/>\nSo, <span class=\"math-tex\">{tex}\\frac { 1 } { 2 } A B = A B \\sin \\theta{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac { 1 } { 2 } = \\sin \\theta{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\theta = \\sin ^ { &#8211; 1 } \\left( \\frac { 1 } { 2 } \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\theta = 30 ^ { \\circ }{\/tex}<\/span><\/li>\n<li>Triangular law of vector addition states that if two vectors can be represented both in magnitude and direction by the sides of a triangle taken in order then their resultant is given by the third side of the triangle taken in opposite order.<br \/>\nProof: in <span class=\"math-tex\">{tex}\\triangle{\/tex}<\/span>ADC<br \/>\n<span class=\"math-tex\">{tex}( O C ) ^ { 2 } = ( O D ) ^ { 2 } + ( D C ) ^ { 2 }{\/tex}<\/span><br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 87px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/HMndW6r.jpg\" data-imgur-src=\"HMndW6r.jpg\" \/><br \/>\n<span class=\"math-tex\">{tex}( O C ) ^ { 2 } = ( O A + A D ) ^ { 2 } + ( D C ) ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}(OC)^2\\;=\\;(OA)^2\\;+\\;(AD)^2\\;+\\;2(OA)(AD)\\;+\\;(DC)^2{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}( OC ) ^ { 2 } = \\left( P ^ { 2 } \\right) + ( Q \\cos \\theta ) 2 + 2 ( P ) ( Q \\cos \\theta ) + ( Q \\sin \\theta ) ^ { 2 }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\begin{array}{l}(OC)^2=P^2+Q^2\\left(\\sin^2\\theta+\\cos^2\\theta\\right)+2PQ\\cos\\theta\\;\\;\\;\\left(\\because\\frac{CD}{AC}=\\sin\\left(\\theta\\right)\\;,\\;\\;\\frac{AD}{AC}=\\cos\\left(\\theta\\right)\\right)\\\\\\end{array}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}( R ) ^ { 2 } = P ^ { 2 } + Q + 2 P Q \\cos \\theta \\left( \\because \\sin ^ { 2 } \\theta + \\cos ^ { 2 } Q \\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}R=\\sqrt{P^2+Q^2+2PQ\\cos\\theta}{\/tex}<\/span><\/li>\n<\/ol>\n<p>To practice more questions &amp; prepare well for exams, download <strong>myCBSEguide App<\/strong>. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use <strong>Examin8 App<\/strong> to create similar papers with their own name and logo.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>The force of gravity, F= mg\u00a0produces the torque <span class=\"math-tex\">{tex}\\tau{\/tex}<\/span>. Let <span class=\"math-tex\">{tex}\\vec{r}{\/tex}<\/span> be the position vector of Q. Then the magnitude of the torque is given by<br \/>\n<span class=\"math-tex\">{tex}\\tau=r F \\sin \\theta{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=r \\times m g \\times \\frac{x_{0}}{r}=m g x_{0}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\left[\\because \\sin \\theta=\\frac{x_{0}}{r}\\right]{\/tex}<\/span><br \/>\nThe direction of the torque is directed into the plane of paper and perpendicular to it, as shown by\u00a0<span class=\"math-tex\">{tex}\\otimes{\/tex}<\/span>.<\/li>\n<li>The magnitude of the angular momentum is\u00a0<span class=\"math-tex\">{tex}L=r p \\sin \\theta=r m v \\sin \\theta{\/tex}<\/span><br \/>\nBut the velocity v at point Q is given by\u00a0v = u + at = 0 + gt = gt<br \/>\n<span class=\"math-tex\">{tex}\\therefore \\quad L=r m g t . \\frac{x_{0}}{r}=m g x_{0} t{\/tex}<\/span><br \/>\nThe direction of angular momentum is the same as that of torque.<\/li>\n<li>Now L = mgx<sub>0<\/sub>t<br \/>\nDifferentiating both sides with respect to f, we get<br \/>\n<span class=\"math-tex\">{tex}\\frac{d L}{d t}=\\frac{d}{d t}\\left(m g x_{0} t\\right)=m g x_{0}=\\tau{\/tex}<\/span><br \/>\nHence the relation\u00a0<span class=\"math-tex\">{tex}\\tau=\\frac{d L}{d t}{\/tex}<\/span>\u00a0holds in this example.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_for_Class_11\"><\/span><strong>CBSE\u00a0Sample Papers\u00a0for Class 11<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-physics\/1340\/\"><strong>Physics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-chemistry\/1356\/\"><strong>Chemistry<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-mathematics\/1371\/\"><strong>Mathematics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-biology\/1388\/\"><strong>Biology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-accountancy\/1411\/\"><strong>Accountancy<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-economics\/1423\/\"><strong>Economics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-business-studies\/1740\/\"><strong>Business Studies<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-computer-science\/1852\/\"><strong>Computer Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-informatics-practices\/1874\/\"><strong>Informatics Practices<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-english-core\/1856\/\"><strong> English Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%95%E0%A5%8B%E0%A4%B0\/1866\/\"><strong>Hindi Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%90%E0%A4%9A%E0%A5%8D%E0%A4%9B%E0%A4%BF%E0%A4%95\/1868\/\"><strong>Hindi Elective<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-history\/1870\/\"><strong>History<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-political-science\/1880\/\"><strong>Political Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-geography\/1864\/\"><strong>Geography<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-sociology\/1882\/\"><strong>Sociology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-physical-education\/1878\/\"><strong>Physical Education<\/strong><\/a><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11\/1339\/\">Other Subjects<\/a><\/strong><\/li>\n<\/ul>\n<p>To download sample papers for class 11 Physics, Chemistry, Biology, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solutions, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Papers all are made available through <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>CBSE Sample Papers Class 11 Physics 2024-25 CBSE Sample Papers Class 11 Physics myCBSEguide provides CBSE Class 11 Sample Papers of Physics for the year 2024-2025 with solutions in PDF format for free download. The CBSE Sample Papers for all \u2013 NCERT books and based on CBSE latest syllabus must be downloaded and practiced by &#8230; <a title=\"CBSE Sample Papers Class 11 Physics 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-physics\/\" aria-label=\"More on CBSE Sample Papers Class 11 Physics 2025\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":29643,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1339,1999,1415,2031],"tags":[163,1527,12,1959,1967,1449,1340,319],"class_list":["post-13673","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cbse-sample-papers","category-class-11-sample-papers","category-physics","category-physics-sample-papers","tag-cbse-class-11","tag-cbse-question-paper","tag-cbse-sample-papers","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025","tag-class-11","tag-model-question-papers","tag-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Sample Papers Class 11 Physics 2025 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"CBSE Sample Papers Class 11 Physics in PDF format for free download. 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