{"id":143225,"date":"2022-04-15T17:30:57","date_gmt":"2022-04-15T12:00:57","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=143225"},"modified":"2023-04-02T17:25:34","modified_gmt":"2023-04-02T11:55:34","slug":"ncert-physics-revision-notes-for-jee-main","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/ncert-physics-revision-notes-for-jee-main\/","title":{"rendered":"NCERT Physics revision notes for JEE Main 2023"},"content":{"rendered":"

\"jee<\/a>
\nThe curriculum for all schools in the country that follow the Central Board of Secondary Education (CBSE) is set by the National Council of Education Research and Training (NCERT). These <\/span>NCERT Solutions <\/span><\/a>are structured so that students learn to check the marks allotted to a question before correctly solving it.<\/span><\/p>\n

Students would benefit from using <\/span>NCERT Solutions for <\/span>CBSE Class 12<\/b> Physics<\/span><\/a> to improve their problem-solving skills. These NCERT Class 12 Solutions provide detailed, step-by-step explanations of textbook problems. It would also aid students in preparing for undergraduate engineering entrance exams such as VITEEE, <\/span>JEE Main 2022, <\/b>and others. The <\/span>JEE Main<\/b> Physics syllabus <\/span><\/a>is available on the Aakash website for students.<\/span><\/p>\n

Every budding engineer aims to be one of the top <\/span>JEE Main<\/b> test scorers. Due to the vast <\/span>JEE Main <\/b>syllabus, many students find it difficult to cover concepts from all three disciplines. Physics is a tough subject for many students due to the numerical and theoretical concepts. However, it is a high-scoring subject.<\/span><\/p>\n

Important Solved Questions for JEE Main 2022 Physics Examination<\/h2>\n

Q1. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement is<\/b><\/p>\n

Options:<\/span><\/p>\n

    \n
  1. The\u00a0 minimum\u00a0 initial\u00a0 velocity\u00a0 of\u00a0 the\u00a0 mass\u00a0 m\u00a0 to\u00a0 escape\u00a0 the gravitational\u00a0 field\u00a0 of the two bodies\u00a0 is 4<\/span>\u221a<\/span>GM<\/span>L<\/span><\/li>\n
  2. The\u00a0 minimum\u00a0 initial\u00a0 velocity\u00a0 of\u00a0 the\u00a0 mass\u00a0 m\u00a0 to\u00a0 escape\u00a0 the\u00a0 gravitational\u00a0 field\u00a0 of the two bodies\u00a0 is 2<\/span>\u221a<\/span>GM<\/span>L<\/span><\/li>\n
  3. The\u00a0 minimum\u00a0 initial\u00a0 velocity\u00a0 of\u00a0 the\u00a0 mass\u00a0 m\u00a0 to\u00a0 escape\u00a0 the\u00a0 gravitational\u00a0 field\u00a0 of the two bodies\u00a0 is <\/span>\u221a<\/span>2GM<\/span>L<\/span><\/li>\n
  4. The energy of the mass m is not constant.<\/span><\/li>\n<\/ol>\n

    Ans.<\/span><\/p>\n

    Explanation:<\/span><\/p>\n

    Let v is the minimum velocity.<\/span><\/p>\n

    From the energy conservation rule:<\/span><\/p>\n

    \u00a0<\/span>-2GMm<\/span>L<\/span>+<\/span>1<\/span>2<\/span>mv<\/span>2<\/span>= 0<\/span><\/p>\n

    v = 2<\/span>\u221a<\/span>GM<\/span>L<\/span><\/p>\n

    Hence, option (b) is correct.\u00a0<\/span><\/p>\n\n\n\n
    With the JEE Main score, students would be able to apply for admission to 31 NITs, 25 IIITs, and 28 GFTIs.<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Q2. When a 10V\u00a0 potential difference is applied across a wire of length 0.2 m, the drift\u00a0 speed of electrons is 5 \u00d710<\/b>\u22124<\/b> ms<\/b>\u22121<\/b>.\u00a0 If the electron density in the wire is 8 \u00d7 10<\/b>28<\/b>m<\/b>\u22123<\/b>, the resistivity of the material is close to :<\/b><\/p>\n

    Options:<\/span><\/p>\n

    (a) 0.8\u00d710<\/span>\u22128<\/span> \u03a9 m<\/span><\/p>\n

    (b) 0.8\u00d710<\/span>\u22127<\/span> \u03a9 m<\/span><\/p>\n

    (c) 0.8\u00d710<\/span>\u22126<\/span> \u03a9 m<\/span><\/p>\n

    (d) 0.8\u00d710<\/span>\u22125<\/span> \u03a9 m<\/span><\/p>\n

    Ans.<\/span><\/p>\n

    Option (d) is correct<\/span><\/p>\n

    Explanation:<\/span><\/p>\n

    The current density across a wire is<\/span><\/p>\n

    J = nev<\/span>d<\/span><\/p>\n

    I<\/span>A<\/span>= nev<\/span>d<\/span><\/p>\n

    I = neAv<\/span>d<\/span><\/p>\n

    where,<\/span><\/p>\n

    v<\/span>d<\/span> is the drift speed<\/span><\/p>\n

    and R =<\/span>\u03c1l<\/span>A<\/span><\/p>\n

    By Ohm\u2019s law<\/span><\/p>\n

    \u03c1<\/span> = <\/span>V<\/span>ne<\/span>l<\/span>v<\/span>d<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

    \u00a0\u00a0\u00a0<\/span>10<\/span>8 \u00d7 10<\/span> 28<\/span> \u00d7 1.6 \u00d7 10<\/span> -19<\/span> \u00d7 0.2\u00a0 \u00d7 5\u00a0 \u00d7 1<\/span>0<\/span>-4<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

    = 0.8 <\/span> \u00d7<\/span> 10<\/span>-5<\/span> \u03a9 m<\/span><\/p>\n

    Hence, option (d) is correct.\u00a0<\/span><\/p>\n

    Q3. Two identical capacitors each of capacitance C are charged to the same potential V and are connected in two circuits (i) and (ii) at t = 0 as shown. The charged on the capacitor at t = CR are<\/b><\/p>\n

      \n
    1. C<\/span>V<\/span>e<\/span>, <\/span>C<\/span>V<\/span>e<\/span><\/li>\n
    2. C<\/span>V<\/span> , <\/span>C<\/span>V<\/span><\/li>\n
    3. VC<\/span>e<\/span>, <\/span>VC<\/span><\/li>\n
    4. VC<\/span>,<\/span>VC<\/span>e<\/span><\/li>\n<\/ol>\n

      In\u00a0 Fig.\u00a0 (i)\u00a0 the p-n junction\u00a0 diode\u00a0 is\u00a0 forward\u00a0 biassed\u00a0 and\u00a0 represents\u00a0 a\u00a0 very low\u00a0 resistance,\u00a0 the capacitor, therefore discharges itself through resistor R according to relation.<\/span><\/p>\n

      q = q<\/span>0<\/span>e <\/span>-t\/CR<\/span><\/p>\n

      and q0 = CV at t = CR<\/span><\/p>\n

      Therefore,\u00a0<\/span><\/p>\n

      q = q<\/span>0<\/span>e<\/span>-1 <\/span>=<\/span>C<\/span>V<\/span>e<\/span><\/p>\n

      In Fig. (ii), the p-n junction diode is reverse biassed, the capacitor. Therefore, hold the charge intact.<\/span><\/p>\n

      q = q<\/span>0 <\/span>=<\/span>CV<\/span><\/p>\n

      Hence, option (c) is correct.\u00a0<\/span><\/p>\n

      Q4. A particle of mass 0.5 kg is moving in one dimension under a force that delivers a constant power 1 W to the particle.\u00a0 If the initial speed (in m\/s) of the particle is zero, the speed (in m\/s) after 9 s is:\u00a0<\/b><\/p>\n

      Options:<\/span><\/p>\n

      (a) 8<\/span><\/p>\n

      (b) 7<\/span><\/p>\n

      (c) 6<\/span><\/p>\n

      (d) 5<\/span><\/p>\n

      Answer: Power =\u00a0 <\/span>dW<\/span>dt<\/span><\/p>\n

      Total work done in 9 s is<\/span><\/p>\n

      W = 1 x 9 = 9 = KE<\/span>f<\/span> – KE<\/span>i<\/span><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Since the initial speed of the particle is zero, its initial kinetic energy will also be zero.<\/span><\/p>\n

      9 = <\/span>M<\/span>2<\/span> ( v<\/span>f<\/span>2<\/span> – v<\/span>i <\/span>2 <\/span>)<\/span><\/p>\n

      9 = <\/span>M<\/span>2<\/span> ( v<\/span>f<\/span>2<\/span> – 0<\/span> )<\/span><\/p>\n

      v<\/span>f<\/span>2 <\/span>= <\/span>9 x 2<\/span>0.5<\/span>= 36<\/span><\/p>\n

      v<\/span>f <\/span>= 6<\/span><\/p>\n

      \u00a0<\/span>Hence, option (c) is correct.\u00a0<\/span><\/p>\n

      Do you need help filling out your <\/span>JEE application form<\/span><\/a>? You can find all the details along with the guidelines for the same on Aakash!<\/span><\/p>\n

      Q5. Bob of mass m, suspended by a string of length l<\/b>1<\/b> is given a minimum velocity required to complete a full circle in the vertical plane. At the highest\u00a0 point, it\u00a0 collides\u00a0 elastically\u00a0 with\u00a0 another\u00a0 bob\u00a0 of\u00a0 mass m suspended by a string of length l<\/b>2<\/b>, which is initially at rest.\u00a0 Both the strings are massless and inextensible.\u00a0 If\u00a0 the\u00a0 second\u00a0 bob,\u00a0 after\u00a0 collision\u00a0 acquires\u00a0 the\u00a0 minimum\u00a0 speed required\u00a0 to\u00a0 complete\u00a0 a\u00a0 full\u00a0 circle\u00a0 in\u00a0 the vertical plane,\u00a0 the ratio l<\/b>1<\/b>\/l<\/b>2 <\/b>is:<\/b><\/p>\n

      (a) 1 : 5<\/span><\/p>\n

      (b) 5 : 1<\/span><\/p>\n

      (c) 1 : 4<\/span><\/p>\n

      (d) 4 : 1<\/span><\/p>\n

      The\u00a0 minimum\u00a0 velocity\u00a0 required\u00a0 to\u00a0 complete\u00a0 a\u00a0 full\u00a0 circle\u00a0 in\u00a0 the\u00a0 vertical\u00a0 plane\u00a0 by\u00a0 a<\/span><\/p>\n

      bob of mass m, suspended by a string of length l = \u221a5gl<\/span><\/p>\n

      Also, the speed at highest point is given as \u221agl<\/span><\/p>\n

      The initial speed of 1st bob (suspended by a string of length l1) is \u221a5gl<\/span>1<\/span><\/p>\n

      The speed of this bob at highest point will be \u221agl<\/span>1<\/span><\/p>\n

      When this bob collides with the other bob there speeds will be interchanged.<\/span><\/p>\n

      \u221agl<\/span>1<\/span> = \u221a5gl<\/span>2\u00a0<\/span><\/p>\n

      l<\/span>1<\/span>\/l<\/span>2<\/span> = 5\/1<\/span><\/p>\n

      Hence, option (b) is correct.\u00a0<\/span><\/p>\n

       <\/p>\n

      Q6.<\/strong> <\/span>A diatomic molecule is formed by two atoms which may be treated as mass points m1 and m2 joined by a massless rod of length r. Then, the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is:<\/b><\/p>\n

      Options:<\/span><\/p>\n

        \n
      1. Zero<\/span><\/li>\n
      2. (m1 + m2) r<\/span>2<\/span><\/li>\n
      3. (<\/span>m1+ m2<\/span>m1m2<\/span>)<\/span> r<\/span>2<\/span><\/li>\n
      4. (<\/span>m1m2<\/span>m1+m2<\/span>)<\/span> r<\/span>2<\/span><\/li>\n<\/ol>\n

        Ans.<\/span><\/p>\n

        Option (d) is correct<\/span><\/p>\n

        Let\u00a0 the\u00a0 centre\u00a0 of\u00a0 mass be situated\u00a0 at\u00a0 distance x from the atom\u00a0 of\u00a0 mass\u00a0 m1 and\u00a0 at\u00a0 distance (r \u2212 x)\u00a0 from mass m2.<\/span><\/p>\n

        By definition, we have, m1x = m2 (r\u2212x)\u00a0<\/span><\/p>\n

        x = <\/span>(<\/span>m2<\/span>m1+m2<\/span>)<\/span> r<\/span>2<\/span> \u00a0 \u00a0 … 1<\/span><\/p>\n

        Now, moment of inertia about the centre of mass is<\/span><\/p>\n

        l = m1x<\/span>2<\/span> + m2 (r\u2212x) <\/span>2<\/span>\u00a0<\/span><\/p>\n

        Putting the value of x from equation (1)<\/span><\/p>\n

        I = m<\/span>1<\/span> (<\/span>m2r<\/span>m1m2<\/span>)<\/span>2<\/span> + m<\/span>2<\/span> (r-<\/span>m2r<\/span>m1 +m2<\/span>)<\/span> 2<\/span><\/p>\n

        \u00a0\u00a0= m<\/span>1<\/span> (<\/span>m2r<\/span>m1m2<\/span>)<\/span>2<\/span> + m<\/span>2<\/span> (<\/span>m1r<\/span>m1 +m2<\/span>)<\/span> 2<\/span><\/p>\n

        \u00a0\u00a0= <\/span>(<\/span>1<\/span>m1 +m<\/span>2<\/span>2<\/span>)<\/span> (m<\/span>1<\/span>m<\/span>2<\/span>2<\/span>r<\/span>2<\/span> + m<\/span>2<\/span>m<\/span>1<\/span>2<\/span>r<\/span>2<\/span>)<\/span><\/p>\n

        \u00a0\u00a0=\u00a0 <\/span>(<\/span>m<\/span>1<\/span>m<\/span>2<\/span>r<\/span>2<\/span>m1 +m<\/span>2<\/span>2<\/span>)<\/span> (m<\/span>2<\/span> + m<\/span>1<\/span>)<\/span><\/p>\n

        I = <\/span>(<\/span>m<\/span>1<\/span>m<\/span>2<\/span>r<\/span>2<\/span>m1 +m<\/span>2<\/span>)<\/span><\/p>\n

        Hence, option (d) is correct.\u00a0<\/span><\/p>\n

        Conclusion<\/h3>\n

        The Physics syllabus is vital during the preparation and final revision phases. Students should be meticulous in their revision strategy with only a few weeks until the <\/span>JEE Main Examination 2022<\/b>. To avoid wasting their time and efforts, pupils must rigorously adhere to the Physics syllabus. For every entrance test, NCERT books are essential because they cover all the fundamental concepts. JEE experts also recommend the NCERT solutions. You can study from NCERT and reference books to improve your JEE Main preparation and score a high rank!<\/span><\/p>\n

        Students can access<\/span> the previous 14 years’ question papers<\/span><\/a> on the Aakash website and detailed solutions.\u00a0<\/span><\/p>\n

        FAQs<\/h2>\n

        <\/b>1.The picture of an object created by a plano-convex lens at an 8-metre distance behind the lens is actually one-third the size of the item. The light inside the lens has a wavelength that is 2\/3 that of light in open space. The lens’s curved surface has a radius of<\/b><\/p>\n

          \n
        1. 3m<\/span><\/li>\n
        2. 10m<\/span><\/li>\n
        3. 7m\u00a0<\/span><\/li>\n
        4. 9m<\/span><\/li>\n<\/ol>\n
            \n
          1. <\/b> Given that- positive of image, v = +8<\/span><\/li>\n<\/ol>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0magnification, m = 1\/3<\/span><\/p>\n

            \u00a0m = v\/u = 1\/3<\/span><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a08\/u = 1\/3\u00a0<\/span><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0u = 8 * 3<\/span><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0u = -24<\/span><\/p>\n

            \u03bc = <\/b>\u03bb air \/ \u03bb medium<\/span><\/p>\n

            \u03bb air = 1 and \u03bb medium = \u2154<\/span><\/p>\n

            So, <\/span>\u03bc = 3\/2<\/b><\/p>\n

            According to the lens formula, <\/span>1\/<\/b>f<\/b> = 1\/<\/b>v<\/b> + 1\/<\/b>u<\/b>\u00a0<\/b><\/p>\n

            where , <\/span>1\/<\/b>f<\/b> = (u – 1) (1\/R \u2013 1\/<\/b>\u221e)<\/b><\/p>\n

            Substituting all the values we get <\/span>R = 3m. <\/b>Therefore the correct answer is (a) R = 3m.<\/span><\/p>\n

            2. The volume of a gas becomes four times if\u00a0<\/b><\/p>\n

              \n
            1. Temperature becomes four times at constant pressure<\/span><\/li>\n
            2. Temperature becomes one fourth at constant pressure<\/span><\/li>\n
            3. Temperature becomes two times at constant pressure<\/span><\/li>\n
            4. Temperature becomes half at constant pressur<\/span>e<\/b><\/li>\n<\/ol>\n
                \n
              1. \u00a0<\/b>V\u221dT (as constant pressure)<\/span><\/li>\n<\/ol>\n

                Hence, the correct answer is (a) Temperature becomes four times at constant pressure.<\/span><\/p>\n

                3. A wire’s resistance at 20\u00b0C room temperature is found to be 20W. Now, in order to increase the resistance by 20%, the wire must be heated to?<\/b><\/p>\n

                [The temperature coefficient of resistance of the wire material is 0.002 per \u00b0C].<\/span><\/p>\n

                (a) 176\u00b0C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                (b) 124\u00b0C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                (c) 142\u00b0C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                (d) 133\u00b0C<\/span><\/p>\n

                  \n
                1. <\/b> R1 = R0 (1 + <\/span>\u03b1t)<\/span><\/li>\n<\/ol>\n

                  At the beginning R0 (1 + 20\u03b1) = 20\u03a9<\/span><\/p>\n

                  Finally, <\/span>R0 (1 + <\/span>\u03b1t) = 24\u00a0<\/span><\/p>\n

                  At the beginning R0 (1 + 20\u03b1) = 20\u03a9<\/span><\/p>\n

                  Finally, <\/span>R0 (1 + <\/span>\u03b1t) = 24<\/span><\/p>\n

                  Now, 6\/5 = <\/span>(1 + <\/span>\u03b1t) \/ (1 + 20\u03b1)\u00a0<\/span><\/p>\n

                  Substituting all the values we get:\u00a0<\/span><\/p>\n

                  5 + (5 x 0.002 x t) = 6 + 120 x 0.002<\/span><\/p>\n

                  0.01t = 1.24<\/span><\/p>\n

                  t = 1.24\/0.01 = 124\u00b0C\u00a0<\/span><\/p>\n

                  Hence, (b) 124\u00b0C is the correct answer.\u00a0<\/span><\/p>\n

                  4. Can students from other boards utilise the NCERT Class 12 Physics Solutions?<\/b><\/p>\n

                  Answer: CBSE and NCERT guidelines are followed by NCERT Solutions for <\/span>CBSE Class 12<\/b> Physics. Although NCERT books may not follow the same criteria and standards as other boards, students can still get information on the ideas and concepts included in the textbooks. The NCERT Solutions for <\/span>CBSE Class 12 <\/b>Physics might be useful to students from all boards because the majority of the syllabus in Physics is the same.<\/span><\/p>\n

                  Are you looking for a coaching institute near you for revision before your <\/span>JEE Main 2022 <\/b>examination? Find an <\/span>Aakash coaching institute<\/span><\/a> near you today!\u00a0<\/span><\/p>\n

                  5. How much time should you spend studying for JEE Physics?<\/b><\/p>\n

                  The subject of physics covers a wide range of topics. It would be beneficial if you had enough time to cover all the issues completely. If you begin your JEE Main 2022 <\/b>preparation early, you must devote approximately 5 hours every day. However, increasing the preparation time over time would be beneficial. You should devote about 7-8 hours to studying Physics throughout the final revision period.<\/p>\n

                  Is it your dream to study at an IIT? Then make sure you go through the <\/span>JEE Main<\/b> <\/a>and <\/span>Advanced eligibility criteria<\/span><\/a> before sitting for the exam.\u00a0<\/span><\/p>\n

                  6. How can I score 360 in JEE Main 2022 after studying the important Physics topics?<\/b><\/p>\n

                  Here are some tips to help you achieve a good rank in JEE Main 2022:<\/b><\/p>\n