{"id":196107,"date":"2022-06-18T08:00:24","date_gmt":"2022-06-18T02:30:24","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=196107"},"modified":"2023-04-09T16:57:59","modified_gmt":"2023-04-09T11:27:59","slug":"jee-main-physics-previous-year-questions-with-solutions","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/jee-main-physics-previous-year-questions-with-solutions\/","title":{"rendered":"JEE Main Physics Previous Year Questions With Solutions"},"content":{"rendered":"<p><a href=\"https:\/\/www.aakash.ac.in\/jee-mains-results?utm_source=seobanner&amp;utm_medium=jeemain&amp;utm_campaign=jeemain2023result\" target=\"_blank\" rel=\"noopener\"><img decoding=\"async\" src=\"https:\/\/d20x1nptavktw0.cloudfront.net\/wordpress_media\/2023\/02\/1300x420-1140x368.jpg\" alt=\"jee main exam\" width=\"100%\" data-entity-type=\"file\" data-entity-uuid=\"d4e023ef-9ff8-4b8f-b582-b9892a7d2953\" \/><\/a><br \/>\n<span style=\"font-weight: 400;\">JEE Mains is a PAN India level exam, and if you dream of qualifying for it, you have to be the best with your preparations. It comprises three sections: Physics, <\/span><span style=\"font-weight: 400;\">Chemistry<\/span><span style=\"font-weight: 400;\">, and <\/span><span style=\"font-weight: 400;\">Mathematics<\/span><span style=\"font-weight: 400;\">. And, it is necessary to attempt all three sections if you want to be enlisted as one of the toppers.<\/span><\/p>\n<p>Also See:\u00a0<a href=\"https:\/\/www.aakash.ac.in\/blog\/jee-main-2022-session-1-day-2-exam-live-updates-paper-analysis-difficulty-level-student-reactions\/\" target=\"_blank\" rel=\"noopener\" data-wpel-link=\"internal\">JEE Main 2022 Exam Live Updates<\/a><\/p>\n<p><span style=\"font-weight: 400;\">To ensure that you are preparing on the right track, check your preparations for each section separately. This article aims to help you out with your practices. Here are some important Physics questions with solutions from the previous years&#8217; <\/span><a href=\"https:\/\/www.aakash.ac.in\/jee-main-exam\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">JEE Mains exam<\/span><\/a><span style=\"font-weight: 400;\"> papers. Have a look.<\/span><\/p>\n<h4><strong><a href=\"https:\/\/www.aakash.ac.in\/jee-crash-courses\" target=\"_blank\" rel=\"noopener\">Click Here to Register for JEE Main Crash Courses 2022 by Aakash Byju&#8217;s\u00a0<\/a><\/strong><\/h4>\n<h4><b>Important Physics Questions with Solutions from <\/b><a href=\"https:\/\/www.aakash.ac.in\/jee-archive\/\" target=\"_blank\" rel=\"noopener\"><b>JEE Mains Previous Years\u2019 Papers<\/b><\/a><\/h4>\n<p>1. <span style=\"font-weight: 400;\">A Plano-convex lens with a refractive index of one and a focal length f<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\"> and a Plano concave lens with a refractive index of two and focal length f<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> are in proximity. If the radius of curvature of spherical faces of both lenses is R each and f<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\"> = 2f<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">, what is the relation between <\/span><span style=\"font-weight: 400;\">1<\/span> <span style=\"font-weight: 400;\">and <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">?<\/span><\/p>\n<p>1f2 &#8211; 1f1 = (2- 1)(1 &#8211; 1-R)<\/p>\n<p><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">f<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> = (<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">&#8211; 1)(<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">-R<\/span><span style=\"font-weight: 400;\"> &#8211; <\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">(<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">&#8211; 1)<\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">(<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">&#8211; 1)<\/span><span style=\"font-weight: 400;\">2R<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> 2<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">&#8211; <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">= 1<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the required relation is 2<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">&#8211; <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">= 1.<\/span><\/p>\n<h4><strong>Watch our video on JEE Main 2022: Last Min Tips to Crack JEE Exam | JEE Exam Day: Do&#8217;s and Don&#8217;ts | JEE Exam Hall Tips<\/strong><\/h4>\n<div class=\"jeg_video_container jeg_video_content\"><iframe loading=\"lazy\" title=\"JEE Main 2022: Last Min Tips to Crack JEE Exam | JEE Exam Day: Do&#039;s and Don&#039;ts | JEE Exam Hall Tips\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vj5iBOn1u9o?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/div>\n<p><span style=\"font-weight: 400;\">2. A cable of mass 5g and the length of 1 m is fixed at both ends. The tension in the cable is 8.0 N. The cable is set into vibration using an extrinsic vibrator of frequency 100 Hz. How close to the separation between sequential nodes on the cable?<\/span><\/p>\n<p>Velocity of wave on cable, V = T = 85 x 1000<\/p>\n<p><span style=\"font-weight: 400;\"> V = 40 m\/s<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Now, wavelength of wave <\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">40<\/span><span style=\"font-weight: 400;\">100<\/span><span style=\"font-weight: 400;\"> m<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, separation between successive nodes, <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">20<\/span><span style=\"font-weight: 400;\">100<\/span><span style=\"font-weight: 400;\"> m = 20 cm<\/span><\/p>\n<p><span style=\"font-weight: 400;\">3. Two guns, X and Y, can fire bullets at 2km\/s and 4 km\/s. They are ejected in all feasible directions from a point parallel to the ground. What will be the ratio of utmost areas engulfed by the bullets ejected by two guns on the ground?<\/span><\/p>\n<p>R = u\u00b2 sin 2g and A = R2<\/p>\n<p><span style=\"font-weight: 400;\"> A <\/span><span style=\"font-weight: 400;\">\u221d<\/span><span style=\"font-weight: 400;\"> R<\/span><span style=\"font-weight: 400;\">2<\/span> <span style=\"font-weight: 400;\">\u221d<\/span><span style=\"font-weight: 400;\"> u<\/span><span style=\"font-weight: 400;\">4<\/span><\/p>\n<p><span style=\"font-weight: 400;\">A\u2081<\/span><span style=\"font-weight: 400;\">A\u2082<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">u<\/span><span style=\"font-weight: 400;\">\u2081\u2074<\/span><span style=\"font-weight: 400;\">u\u2082\u2074<\/span><span style=\"font-weight: 400;\"> = [<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">4<\/span><span style=\"font-weight: 400;\">]<\/span><span style=\"font-weight: 400;\">4<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">16<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the required ratio is 1:16.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">4. The TV transmission tower has a height of 140m. The height of the receiving aerial is 40m. What is the maximal distance upto that can broadcast this tower&#8217;s signals in the LOS (Line of Sight) system? [Given: Radius of Earth = 6.4 x 10<\/span><span style=\"font-weight: 400;\">6<\/span><span style=\"font-weight: 400;\"> m]<\/span><\/p>\n<p>Maximum distance upto that can broadcast signals is dmax = 2RhT + 2Rhg<\/p>\n<p><span style=\"font-weight: 400;\"> d<\/span><span style=\"font-weight: 400;\">max<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">2 x 6.4 x 106<\/span><span style=\"font-weight: 400;\"> [<\/span><span style=\"font-weight: 400;\">104<\/span><span style=\"font-weight: 400;\"> + <\/span><span style=\"font-weight: 400;\">40<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> d<\/span><span style=\"font-weight: 400;\">max<\/span><span style=\"font-weight: 400;\"> = 65 km<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the required maximum distance is 65 km.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">5. If the magnetic field of a horizontal electromagnetic wave is given by<\/span><\/p>\n<p><span style=\"font-weight: 400;\">B = 100 x 10<\/span><span style=\"font-weight: 400;\">-6<\/span><span style=\"font-weight: 400;\"> sin [2<\/span><span style=\"font-weight: 400;\"> x 2 x 10<\/span><span style=\"font-weight: 400;\">15<\/span><span style=\"font-weight: 400;\">(t &#8211; <\/span><span style=\"font-weight: 400;\">x<\/span><span style=\"font-weight: 400;\">c<\/span><span style=\"font-weight: 400;\">)],<\/span><\/p>\n<p><span style=\"font-weight: 400;\">what will be the maximum electric field associated with it? [Given: Speed of light = 3 x 10<\/span><span style=\"font-weight: 400;\">8<\/span><span style=\"font-weight: 400;\"> m\/s]<\/span><\/p>\n<p>E0 = B0 x C<\/p>\n<p><span style=\"font-weight: 400;\"> E<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\"> = 100 x 10<\/span><span style=\"font-weight: 400;\">-6<\/span><span style=\"font-weight: 400;\"> x 3 x 10<\/span><span style=\"font-weight: 400;\">8<\/span><span style=\"font-weight: 400;\"> = 3 x 10<\/span><span style=\"font-weight: 400;\">4<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the maximum electric field associated is 3 x 10<\/span><span style=\"font-weight: 400;\">4<\/span><span style=\"font-weight: 400;\"> N\/C.<\/span><\/p>\n<h4><strong><a href=\"https:\/\/www.aakash.ac.in\/jee-main-mock-test\" target=\"_blank\" rel=\"noopener\">Click Here to Register for JEE Main Mock Tests 2022 by Aakash Byju&#8217;s\u00a0<\/a><\/strong><\/h4>\n<p><span style=\"font-weight: 400;\">6. A hydrogen atom initially at the ground state is excited by absorbing a photon of wavelength 980<\/span><span style=\"font-weight: 400;\">\u212b<\/span><span style=\"font-weight: 400;\">. What will be the atom&#8217;s radius in the excited state in the form of Bohr&#8217;s radius a<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\">?<\/span><\/p>\n<p>Energy of photon = 12500980 = 12.75 eV<\/p>\n<p><span style=\"font-weight: 400;\"> Electrons will be excited to the n = 4 level.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">As R <\/span><span style=\"font-weight: 400;\">\u221d<\/span><span style=\"font-weight: 400;\"> n<\/span><span style=\"font-weight: 400;\">2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the radius of the atom will be 16a<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\">.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">7. The unbending diatomic ideal gas via an adiabatic process at room temperature. The connection between volume and temperature of this process is TV<\/span><span style=\"font-weight: 400;\">x<\/span><span style=\"font-weight: 400;\"> = constant. What will be the value of x?<\/span><\/p>\n<p><span style=\"font-weight: 400;\">For adiabatic process,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">T<\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\"> = constant<\/span><\/p>\n<p><span style=\"font-weight: 400;\">For diatomic process, <\/span><span style=\"font-weight: 400;\"> &#8211; 1 = <\/span><span style=\"font-weight: 400;\">7<\/span><span style=\"font-weight: 400;\">5<\/span><span style=\"font-weight: 400;\"> &#8211; 1<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> x =\u00a0 <\/span><span style=\"font-weight: 400;\">7<\/span><span style=\"font-weight: 400;\">5<\/span><span style=\"font-weight: 400;\"> &#8211; 1 = <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">5<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the value of x is <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">5<\/span><span style=\"font-weight: 400;\">.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">8. Two equal resistances are connected in series to a battery and consume 60 W of electric power. If these resistances are now fastened in parallel combinations to the same battery, what will be the amount of electric power they consume?<\/span><\/p>\n<p>In series combination, equivalent resistance = 2R<\/p>\n<p><span style=\"font-weight: 400;\"> 60 W power consumed = <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">2R<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In parallel combination, equivalent resistance = <\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\">2<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> New power, P<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">(<\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">)<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> P<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\"> = 4P = 240 W<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the amount of electric power consumed by them = 240 W.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">9. Suppose a circular coil of wire containing current, I, forms a magnetic dipole. The magnetic flux across a boundless plane that contains the circular coil and excludes the circular coil area is given by <\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">. The magnetic flux throughout the area is given by <\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\">. What will be the relation between <\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\"> and <\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\">?<\/span><\/p>\n<p>As magnetic field lines always form a closed loop, every magnetic field line creating magnetic flux in the inner region must pass all over the outer region. The magnetic flux across the area is given by 0. Since the flux in two regions are in opposite directions,<\/p>\n<p><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\"> = &#8211; <\/span><span style=\"font-weight: 400;\">0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the required relation is <\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\"> + <\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\"> = 0.<\/span><\/p>\n<p><strong>Check our visual story on <a href=\"https:\/\/www.aakash.ac.in\/blog\/web-stories\/how-to-prepare-for-jee-main-physics-strategies-important-tips\/\" target=\"_blank\" rel=\"noopener\">How to Prepare for JEE Main Physics: Strategies &amp; Important Tips<\/a><\/strong><\/p>\n<blockquote class=\"wp-embedded-content\" data-secret=\"P1ATgQSBZU\"><p><a href=\"https:\/\/www.aakash.ac.in\/blog\/web-stories\/how-to-prepare-for-jee-main-physics-strategies-important-tips\/\">How to Prepare for JEE Main Physics: Strategies &#038; Important Tips<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; clip: rect(1px, 1px, 1px, 1px);\" title=\"&#8220;How to Prepare for JEE Main Physics: Strategies &#038; Important Tips&#8221; &#8212; Aakash Blog\" src=\"https:\/\/www.aakash.ac.in\/blog\/web-stories\/how-to-prepare-for-jee-main-physics-strategies-important-tips\/embed\/#?secret=MILzrivjkJ#?secret=P1ATgQSBZU\" data-secret=\"P1ATgQSBZU\" width=\"360\" height=\"600\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-weight: 400;\">10. A 60 HP electric motor lifts a lift having a maximal total load range of 2000 kg. If the frictional force on the lift is 4000 N, find the elevator&#8217;s speed at full load.<\/span><\/p>\n<p>The speed of elevator at full load = (4000 x v) + (mg x v)<\/p>\n<p><span style=\"font-weight: 400;\">(4000 x v) + (mg x v) = P<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> v x (4000 + mg) = P<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> v = <\/span><span style=\"font-weight: 400;\">P<\/span><span style=\"font-weight: 400;\">(4000 + mg)<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">60 x 746<\/span><span style=\"font-weight: 400;\">4000+20000<\/span><\/p>\n<p><span style=\"font-weight: 400;\"> v = 1.86 m\/s<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the required speed is 1.9 m\/s.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">11. Two moles of an absolute gas with <\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">p<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">5<\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\"> are mixed with 3 moles of another absolute gas with <\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">p<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">4<\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\">. What will be the value of <\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">p<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\"> for the mixture?<\/span><\/p>\n<p>The value of CpCv for the mixture = mixture<\/p>\n<p><span style=\"font-weight: 400;\">mixture<\/span><span style=\"font-weight: 400;\">= <\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">p<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">+<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">p<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">+<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\">2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">mixture<\/span><span style=\"font-weight: 400;\">= <\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\"> + <\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\">2-1<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\"> + <\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">-1<\/span><\/p>\n<p><span style=\"font-weight: 400;\">On rearranging,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">+<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">mix<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\"> + <\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">-1<\/span><\/p>\n<p><span style=\"font-weight: 400;\">5<\/span><span style=\"font-weight: 400;\">mix<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\"> + <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">3<\/span><\/p>\n<p><span style=\"font-weight: 400;\">mixture<\/span><span style=\"font-weight: 400;\">= <\/span><span style=\"font-weight: 400;\">17<\/span><span style=\"font-weight: 400;\">12<\/span><span style=\"font-weight: 400;\"> = 1.42<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the value of <\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">p<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">v<\/span><span style=\"font-weight: 400;\"> for the mixture is 1.42.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">12. In the shown circuit of the following, the battery is of EMF 4 and internal resistance 8 <\/span><span style=\"font-weight: 400;\">. Find the potential difference between points x and y as shown in the circuit. [Given: R = 12 <\/span><span style=\"font-weight: 400;\">.]<\/span><\/p>\n<p>I = 420 = 210<\/p>\n<p><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">x<\/span><span style=\"font-weight: 400;\">&#8211; <\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">y<\/span><span style=\"font-weight: 400;\"> = IR = I x 12<\/span><\/p>\n<p><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">x<\/span><span style=\"font-weight: 400;\">&#8211; <\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">y<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">10<\/span><span style=\"font-weight: 400;\"> x 12<\/span><\/p>\n<p><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">x<\/span><span style=\"font-weight: 400;\">&#8211; <\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">y<\/span><span style=\"font-weight: 400;\"> = 2.4 V<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the potential difference between points x and y is 2.4 V.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">13. A body cools from 100\u2103 to 90\u2103 in 20 minutes. Find the time to cool from 110\u2103 to 100\u2103.<\/span><\/p>\n<p>By Newton\u2019s law of cooling,<\/p>\n<p><span style=\"font-weight: 400;\">dQ<\/span><span style=\"font-weight: 400;\">dt<\/span><span style=\"font-weight: 400;\"> \u221d (T &#8211; <\/span><span style=\"font-weight: 400;\">T<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\">)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Here, (T &#8211; <\/span><span style=\"font-weight: 400;\">T<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\">) is the temperature difference between the body and the surroundings. If this difference is more, then the rate of cooling is higher. Or cooling is fast.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, to cool from 110\u2103 to 100\u2103 takes less than 20 minutes compared to the time taken to cool from 100\u2103 to 90\u2103.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">14. For a transistor, the emitter current is 40 mA, and the collector current is 35 mA. What will be the <\/span><span style=\"font-weight: 400;\"> of this transistor?<\/span><\/p>\n<p>= ICIb = IcIe- Ic<\/p>\n<p><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">35<\/span><span style=\"font-weight: 400;\">40-35<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">35<\/span><span style=\"font-weight: 400;\">5<\/span><span style=\"font-weight: 400;\"> = 7<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the <\/span><span style=\"font-weight: 400;\"> of this transistor is 7.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">15. Choose the correct option (s) regarding the Hydrogen spectrum<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">A is the series limit of Lyman.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">B is the third line of Balmer.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">C is the second line of Paschen.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">All the above are correct.<\/span><\/li>\n<\/ol>\n<ol>\n<li><b><\/b><span style=\"font-weight: 400;\"> The correct option is (d).<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">A \u2192 Series limit of Lyman.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">B \u2192 3rd line of Balmer.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">C \u2192 2nd line of Paschen.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">16. Two satellites revolve around a planet in orbit. The ratio of their periods is 1:8. Find the ratio of their angular velocities.<\/span><\/p>\n<p>= 2T<\/p>\n<p><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">T<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">T<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">8<\/span><span style=\"font-weight: 400;\">1<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the ratio of their angular velocities is 8:1.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">17. A 220 volts AC supply is given to the main circuit of the transformer, and the output of 12 volts DC is abstracted using a rectifier. If the subsidiary number of turns was 24, find the number of turns in the main coil.<\/span><\/p>\n<ol start=\"17\">\n<li><b><\/b> <span style=\"font-weight: 400;\">N<\/span><span style=\"font-weight: 400;\">P<\/span><span style=\"font-weight: 400;\">N<\/span><span style=\"font-weight: 400;\">S<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">P<\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">S<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">N<\/span><span style=\"font-weight: 400;\">P<\/span><span style=\"font-weight: 400;\">24<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">220<\/span><span style=\"font-weight: 400;\">12<\/span><\/p>\n<p><span style=\"font-weight: 400;\">N<\/span><span style=\"font-weight: 400;\">P<\/span><span style=\"font-weight: 400;\">= 440<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence, the number of turns in the primary coil is 440.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">18. An electromagnetic wave propagates in a medium, where <\/span><span style=\"font-weight: 400;\">r<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">r<\/span><span style=\"font-weight: 400;\"> = 2. If the speed of light in such a medium is x x 10<\/span><span style=\"font-weight: 400;\">7<\/span><span style=\"font-weight: 400;\"> m\/s, find the value of x.<\/span><\/p>\n<p>n = rr = 2<\/p>\n<p><span style=\"font-weight: 400;\">v = <\/span><span style=\"font-weight: 400;\">c<\/span><span style=\"font-weight: 400;\">n<\/span><span style=\"font-weight: 400;\"> = <\/span><span style=\"font-weight: 400;\">3 x 10\u2078<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> = 15 x 10<\/span><span style=\"font-weight: 400;\">7<\/span><span style=\"font-weight: 400;\"> m\/s\u00a0 [c = speed of light = <\/span><span style=\"font-weight: 400;\">3 x 10\u2078<\/span><span style=\"font-weight: 400;\"> m\/s]<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Therefore, the value of x = 15.<\/span><\/p>\n<h3>Conclusion<\/h3>\n<p><span style=\"font-weight: 400;\">Although Physics is a typical subject and the toughest among the three JEE Mains subjects, practice can help you qualify it with an excellent score. It is a golden word that can solve every problem. While preparing, please focus on the concepts behind the questions and strike the answer or formulae according to them. Understanding the concepts is equally important as knowing the formula to crack JEE Mains. Therefore, before jumping on mock papers, go through your syllabus and notes and revise them properly and carefully.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Be calm, positive, and concentrate on your aim. And in case you get stuck with any JEE Mains question or problem, contact your peers, teachers, or experts. Otherwise, <\/span><a href=\"https:\/\/www.aakash.ac.in\/our-centres\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">find Aakash near you!<\/span><\/a><\/p>\n<p><strong>Also Read:<\/strong><\/p>\n<p><a href=\"https:\/\/www.aakash.ac.in\/blog\/jee-main-physics-chapter-wise-important-questions\/\" target=\"_blank\" rel=\"noopener\">JEE Main Physics Chapter-wise Important Questions<\/a><\/p>\n<p><a href=\"https:\/\/www.aakash.ac.in\/blog\/jee-main-physics-exam-important-questions-to-evaluate-your-preparation\/\" target=\"_blank\" rel=\"noopener\">JEE Main Physics Exam 2022: Important Questions to Evaluate Your Preparation<\/a><\/p>\n<p><a href=\"https:\/\/www.aakash.ac.in\/blog\/what-type-of-questions-are-expected-in-jee-main-physics\/\" target=\"_blank\" rel=\"noopener\">What Type of Questions Are Expected in JEE Main 2022 Physics<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>JEE Mains is a PAN India level exam, and if you dream of qualifying for it, you have to be the best with your preparations. It comprises three sections: Physics, Chemistry, and Mathematics. And, it is necessary to attempt all three sections if you want to be enlisted as one of the toppers. Also See:\u00a0JEE [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":196135,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3719],"tags":[],"class_list":["post-196107","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-jee"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>JEE Main Physics Previous Year Questions With Solutions<\/title>\n<meta name=\"description\" content=\"This article highlights some important Physics questions with solutions from the previous years&#039; JEE Mains exam papers.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aakash.ac.in\/blog\/jee-main-physics-previous-year-questions-with-solutions\/\" \/>\n<meta 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