{"id":4597,"date":"2026-03-14T21:57:56","date_gmt":"2026-03-14T16:27:56","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=4597"},"modified":"2026-03-14T21:58:04","modified_gmt":"2026-03-14T16:28:04","slug":"expect-7-physics-questions-jee","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/expect-7-physics-questions-jee\/","title":{"rendered":"Expect these 7 Physics Questions in JEE 2026"},"content":{"rendered":"\n

All students preparing themselves for JEE Main 2025 should have already started their preparation from class 11. The simple mantra for success here is to have the complete knowledge of JEE Main 2025 Syllabus. It\u2019s not easy to cover the entire syllabus within the time for most of the students as most of them start without planning and with the time constraints end up covering multiple topics in a single day.
Physics is one of the greatest areas for scoring in JEE if someone prepares well. The most important aspect is to have the basic understanding of each concept, theory, and formula along with problem-solving skills. So, here are few questions which are usually common in JEE Main exam and you can expect these in JEE 2024 as well.<\/p>\n\n\n\n

Rotational Motion<\/strong> <\/p>\n\n\n\n

Angular momentum commonly appears in rotational motion topic. There is a high probability that you can find some questions from Angular Momentum & its application in JEE main exam. Previous year\u2019s question paper consisted of the following question.<\/p>\n\n\n\n


Question:<\/strong>
In Bohr\u2019s model of a hydrogen atom, an electron revolves around a proton in a circular orbit of radius $5.29 \\times 10^{-11}$ m with a speed of $2.18 \\times 10^6$ m\/s. (a) what is the acceleration of the electron in this model of the hydrogen atom? (b) What is the magnitude and direction of the net force that acts on the electron?
Concept: <\/strong>
(a)<\/strong> The acceleration of the electron is given as:
a = v2<\/sup>\/r Here, v is the speed with which the electron revolves and r is the radius of the circular orbit of the electron.
(b)<\/strong> The force acting on the electron is the centripetal force with magnitude, F= ma Here, m is the mass of the electron, and is equal to $9.1 \\times 10^{-31}$ k The force is the centripetal force, and is directed towards the center of the circular orbit.
Solution:<\/strong>
(a) Substitute $5.29 \\times 10^{-11}$ m for r and $2.18 \\times 10^6 $m\/s for v in equation a = v2<\/sup>\/r ,
$a = \\frac{(2.18 \\times 10^6 \\quad m\/s)^2}{(5.29 \\times 10^{-11} \\quad m)}$
$= 8.98 \\times 10^{22}$ m\/s2<\/sup>
Therefore, the acceleration of the electron is $8.98 \\times 10^{22} $m\/s2<\/sup> .
(b) Substitute $8.98 \\times 10^{22}$ m\/s2<\/sup> for a and $9.1 \\times 10^{-31}$ kg for m in equation F= ma,<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

Therefore, the force acting on the electron is of magnitude $8.1 \\times 10^{-8}$ N, and directed towards the center of the circular orbit.<\/p>\n\n\n\n

Semiconductor Devices <\/strong><\/p>\n\n\n\n

It\u2019s another very important topic where questions can come from \u201cLogic Gates\u201d in JEE 2024. The question can be as below:<\/p>\n\n\n\n

Question<\/strong>
<\/strong>The combination of NAND Gates shown here under (figure) are equivalent to<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n
    \n
  1. A) An OR gate and an AND gate respectively<\/li>\n\n\n\n
  2. B) An AND gate and a NOT gate respectively<\/li>\n\n\n\n
  3. C) An AND gate and an OR gate respectively<\/li>\n\n\n\n
  4. D) An OR gate and a NOT gate respectively<\/li>\n<\/ol>\n\n\n\n

    Correct Answer<\/strong>: A<\/p>\n\n\n\n

     Solution: <\/strong><\/p>\n\n\n\n

    \"\"<\/a><\/figure>\n\n\n\n

    C<\/em>=A<\/em> \u00af .B<\/em> \u00af  \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af  =A<\/em> \u00af +B<\/em> \u00af  \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af  =A<\/em>+B<\/em><\/p>\n\n\n\n

    (De Morgan\u2019s theorem) Hence output C is equivalent to OR gate.<\/p>\n\n\n\n

    \"\"<\/a><\/figure>\n\n\n\n

    C=AB \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af  .AB \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af   \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af  =AB \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af  +AB \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af   \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af \u00af  =AB+AB=AB <\/em><\/p>\n\n\n\n

    In this case output C is equivalent to AND gate.<\/p>\n\n\n\n

    Thermodynamics:<\/strong> <\/p>\n\n\n\n

    The relationship of Cp<\/sub> and Cv<\/sub> is the most common questionnaire in JEE Main exam.<\/p>\n\n\n\n

    Question. <\/strong>
     A 1-kg block of ice at 0\u00baC is placed into a perfectly insulated, sealed container that has 2 kg of water also at 0\u00baC. The water and ice completely fill the container, but the container is flexible. After some time one can accept that<\/p>\n\n\n\n

    (A) The water will freeze so that the mass of the ice will increase.<\/p>\n\n\n\n

    (B) The ice will melt so that the mass of the ice will decrease.<\/p>\n\n\n\n

    (C) Both the amount of water and the amount of ice will remain constant.<\/p>\n\n\n\n

    (D) Both the amount of water and the amount of ice will decrease<\/p>\n\n\n\n

    Solution:<\/strong> The correct option is (B) the ice will melt so that the mass of the ice will decrease. In accordance to the second law of thermodynamics, if an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases. Entropy measures the state of disorder. As the ice is more order than water, therefore the ice will melt so that the mass of the ice will decrease. From the above observation, we conclude that option (B) is correct.<\/p>\n\n\n\n

    Dynamics and<\/strong> Electrostatics – \u2018Kirchhoff\u2019s Law\u2019<\/strong><\/p>\n\n\n\n

    <\/strong>Another most common question usually you can see on every year on \u201cKirchhoff\u2019s Law\u201d.<\/p>\n\n\n\n

     Question. <\/strong>
    Calculate the value of current I in the section of networks shown in figure<\/p>\n\n\n\n

    \"\"<\/a><\/figure>\n\n\n\n

    Solution: <\/strong>
    Actually it\u2019s checking the understanding of KCL<\/p>\n\n\n\n

    At junction A: I2+8 = 15; I2=7A<\/p>\n\n\n\n

    At junction D: I3 +5 = 8; I3 = 3A<\/p>\n\n\n\n

    At junction B: I2+3 = I4; I4 =10A<\/p>\n\n\n\n

    At junction C: I1=I3 + I4; I1 =13 A<\/p>\n\n\n\n

    Magnetic Field: <\/strong><\/p>\n\n\n\n

    If you follow the last few years question of Physics in JEE Main you can find another common question is from \u201cMagnetic field in a shape\u201d.<\/p>\n\n\n\n

    Question <\/strong>
    Two straight long conductors AOB and COD are perpendicular to each other and carry currents i1<\/sub> and i2<\/sub> .The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ACBD is<\/p>\n\n\n\n

    \"\"<\/a><\/figure>\n\n\n\n

    Correct Answer:  C<\/strong><\/p>\n\n\n\n

    Solution: <\/strong>\"\"<\/a><\/p>\n\n\n\n

    <\/p>\n\n\n\n

    Electromagnetic Induction and Alternating Currents<\/strong><\/p>\n\n\n\n

    Most common question in this topic usually comes from \u201cResistor-Inductor-Capacitor\u201d.<\/p>\n\n\n\n

    Question.<\/strong><\/p>\n\n\n\n

    <\/strong>The voltage applied to a pure capacitor of 50*10-6<\/sup> F is as shown in the figure. Calculate the current for      0-1msec.<\/p>\n\n\n\n

    \"\"<\/a><\/figure>\n\n\n\n
      \n
    1. a) 5A<\/li>\n\n\n\n
    2. b) 1A<\/li>\n\n\n\n
    3. c) -5A<\/li>\n\n\n\n
    4. d) -1A<\/li>\n<\/ol>\n\n\n\n

      Correct Answer<\/strong>:  5A<\/p>\n\n\n\n

      Solution<\/strong>:    For 0 <= t <= 1msec,
      V (t) =mt (y=mx form)
      100= <\/em> 10-3<\/sup>m <\/em>
      <\/em>m= 105<\/sup>
      V (t) = 105<\/sup> t
      Current I (t) = c. d (v (t))\/dt = 50 x10-6<\/sup> x (1 x105<\/sup> )= 5A.<\/p>\n\n\n\n

      Photoelectric effect \u2013 Modern Physics: <\/strong><\/p>\n\n\n\n

      Different types of question on Bohr\u2019s Model and Photoelectric Effect can come in the exam.<\/p>\n\n\n\n

      Question.<\/strong> Let K1<\/sub> be the maximum kinetic energy of photoelectrons emitted by light of wavelength $\\lambda _1$ and K2<\/sub> corresponding to wavelength $\\lambda _2$ If $\\lambda _1=2\\lambda _2$ then,<\/p>\n\n\n\n

        \n
      1. 2K1<\/sub>= K2<\/sub><\/li>\n\n\n\n
      2. K1<\/sub>= 2K2<\/sub><\/li>\n\n\n\n
      3. K1<\/sub>< K2<\/sub>\/2<\/li>\n\n\n\n
      4. K1<\/sub>> 2K2<\/sub><\/li>\n<\/ol>\n\n\n\n

        Correct Answer<\/strong>: Option (C)<\/p>\n\n\n\n

        Solution:   <\/strong><\/p>\n\n\n\n

        As we know, $K_1 = \\frac {hc}{\\lambda_1} \u2013 W$         \u2026… (1)<\/p>\n\n\n\n

        and  $K_2 =\\frac { hc}{\\lambda _2} \u2013 W$                      \u2026… (2)<\/p>\n\n\n\n

        Substituting $\\lambda _1=2\\lambda _2$in equation (1), we get,<\/p>\n\n\n\n

        $K_1 = \\frac {hc}{2 \\lambda _2} \u2013 W = \\frac {1}{2} \\frac {hc}{\\lambda _2} \u2013 W = \\frac {1}{2} (K_2+W) \u2013 W$<\/p>\n\n\n\n

        Thus, K1<\/sub> = K2<\/sub>\/2  \u2013 W\/2 Or, K1<\/sub><K2<\/sub>\/2<\/p>\n\n\n\n

        Therefore from the above observation, we conclude that option (C) is correct.<\/p>\n","protected":false},"excerpt":{"rendered":"

        All students preparing themselves for JEE Main 2025 should have already started their preparation from class 11. The simple mantra for success here is to have the complete knowledge of JEE Main 2025 Syllabus. It\u2019s not easy to cover the entire syllabus within the time for most of the students as most of them start […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[14],"tags":[],"class_list":["post-4597","post","type-post","status-publish","format-standard","hentry","category-physics"],"yoast_head":"\nExpect these 7 Physics Questions in JEE 2026 - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/expect-7-physics-questions-jee\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Expect these 7 Physics Questions in JEE 2026 - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"All students preparing themselves for JEE Main 2025 should have already started their preparation from class 11. 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The simple mantra for success here is to have the complete knowledge of JEE Main 2025 Syllabus. It\u2019s not easy to cover the entire syllabus within the time for most of the students as most of them start…","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/4597","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=4597"}],"version-history":[{"count":6,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/4597\/revisions"}],"predecessor-version":[{"id":9950,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/4597\/revisions\/9950"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=4597"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=4597"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=4597"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}