January 27, 2021

# jee main 2020 question paper 9 january shift 1

Question 1. mui^+m(u2i^+u2j^)=2m(v1i^+v2j^)mu\hat{i} + m(\frac{u}{2}\hat{i}+\frac{u}{2}\hat{j}) = 2m(v_{1}\hat{i}+v_{2}\hat{j})mui^+m(2ui^+2uj^)=2m(v1i^+v2j^), Initial K.E = (mv2/2) + (m/2)×(u/√2)2 = 3mu2/4, Change in K.E = (3mu2/4) - (5mu2/8) = mu2/8. question will be treated as wrong response and marked up for wrong response will be deducted accordingly as ... Paper - 1 Test Date: 9th January 2020 (Second Shift) JEE-MAIN-2020 (9th Jan-Second Shift)-PCM-2 FIITJEE Ltd., FIITJEE House, 29 -A, Kalu Sarai, Sarvapriya Vihar, New Delhi 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. Two plane electromagnetic waves are moving in vacuum in whose electric field vectors are given by, E1⃗=E0j^cos(kx−ωt)\vec{E_{1}} = E_{0}\hat{j} \cos (kx - \omega t)E1=E0j^cos(kx−ωt) and E2⃗=E0k^cos(ky−ωt)\vec{E_{2}} = E_{0}\hat{k} \cos (ky - \omega t)E2=E0k^cos(ky−ωt). The question paper for JEE Main 2020 is uploaded as pdf, along with the answer key. Shift-Wise JEE Main 2019 Question Paper, Answer Keys and Solutions - January. JEE Main 2020 9th Jan shift 1 solved Physics paper consists of accurate solutions, prepared by our subject experts. JEE Main Paper 1 2020 (January 7) - National Testing Agency (NTA) conducted the first two shifts of Paper 1 (B.E./B.Tech) today i.e., January 7. Watch Todays' jee main paper analysis (9 Jan, Shift 2 & 1). Today JEE Mains Question Paper January 2020 Exam Analysis. Three identical solid spheres each have mass ‘m’ and diameter ‘d’ are touching each other as shown in the figure. Determine the work done on the particle by F⃗\vec{F}F in moving the particle from point A to point B (all quantities are in SI units), ds⃗=(dx i^+dy j^)d\vec{s} = (dx \; \hat{i} + dy\; \hat{j})ds=(dxi^+dyj^), =(−xi^+yj^). The velocities at P and Q are respectively vi^v\hat{i}vi^ and −2vj^-2v\hat{j}−2vj^. PART – A (PHYSICS) 1. JEE Main is being conducted by National test Agency (NTA) from 1st of April to 6th of April in two shifts (SHIFT 1: 9.30 AM to 12.30 PM and SHIFT 2: 2.30 PM. C. Rate of work done by both fields at Q is zero. On increasing its K.E by Δ, it's new de–Broglie wavelength becomes λ/2. Copyright © 2020 Entrancei. Learn questions asked in Physics, Chemistry, Maths The forenoon session was held from 9:30 AM to 12:30 PM and the afternoon session was conducted from 2:30 PM to 5:30 PM. Candidates can check the detailed analysis of JEE Main 2020 Paper 1 January 7 in this article. Get JEE Main 2020 (Paper 1 & 2). For process 2-3; pressure is constant , therefore V = kT, Question 6.An electric dipole of moment p⃗=(−i^−3j^+2k^)×10−29Cm\vec{p} = (-\hat{i}-3\hat{j}+2\hat{k})\times 10^{-29}Cmp=(−i^−3j^+2k^)×10−29Cm is at the origin (0,0,0). Then which of the following statements (A, B, C, D) are correct? N⃗ = 0, this means field is along tangential direction and dipole is also perpendicular to radius vector. Candidates can check the detailed analysis of JEE Main 2020 Paper 1 January 7 in this article. JEE Main 2020 (Sep) – Question Paper, Key & Solutions 2nd September 2020 (Morning Shift) 2nd September 2020 (Evening Shift) 3rd September 2020 (Morning Shift) 3rd September 2020 (Evening Shift) 4th September 2020 (Morning Shift) 4th September 2020 (Evening Shift) 5th September 2020 (Morning Shift) 5th September 2020 (Afternoon Shift) 6th September 2020 (Morning Shift) … A charged particle of mass ‘m’ and charge ‘q’ is moving under the influence of uniform electric field Ei^E\hat{i}Ei^ and a uniform magnetic field Bk^B\hat{k}Bk^ follow a trajectory from P to Q as shown in figure. Question 22.A wire of length l = 0.3 m and area of cross section 10–2 cm2 and breaking stress 4.8×107 N/m2 is attached with block of mass 10 kg. Meanwhile, JEE Main 2020 Paper 1 was conducted from 7 – 9 January in online mode and in two shifts – morning shift from 09:30 AM to 12:30 PM and afternoon shift from 02:30 PM to 05:30 PM. The apparent depth of inner surface of the bottom of the vessel will be; Apparent height as seen from liquid 1 (having refractive index 1 = √2 ) to liquid 2 (refractive index 2 = 2√2), Now, actual height perceived from air, h + ℎ/2 = 3ℎ/2, Therefore, apparent depth of bottom surface of the container (apparent depth as seen from air (having refractive index 0 = 1) to liquid 1(having refractive index 1= √2 ). Download JEE Main 2020 Maths Answer Key 9 Jan Shift 1 by Resonance in PDF Format form aglasem.com JEE MAIN 2020 Paper Analysis for 7th January SHIFT-1 Today on 7 th January, the national level competition exam JEE MAIN was held across the nation and with the completion of the exam, there are many queries in the mind of aspirants. There are three papers: Paper-1 (BTech ), Paper-2 (BArch), and Paper-3 (BPlanning). A telescope of aperture diameter 5 m is used to observe the moon from the earth. Now voltage at E is 12 volt and voltage at H is 4 volt and since, diode between E and H is reversed biased and any difference of voltage is possible across reverse biased. Find the maximum possible value of angular velocity (/) with which block can be moved in a circle with string fixed at one end. Let v1 and v2 be the final velocities after collision in x and y direction respectively. We have to apply nodal analysis on both left and right side and check what can be voltage at E. For nodal analysis, voltage at B, F and G will be 0 volts and voltage at A will be 12.7 volt and voltage at H will be 4 volts. JEE Main 2020 Question Paper – Candidates can download JEE Main question paper for January 07, 08, 09, 2020, for Shift 1 and Shift 2 from this page. NTA JEE Mains 2020 Answer Key Official for 6th,7th,8th,9th January with Shift Wise Question Paper (Paper 1 and 2): The National Testing Agency (NTA) is set to publish the Official JEE Main 2020 Answer Paper 1 & Paper 2 (shift 1 and 2). Find least count of screw gauge? JEE Main 2020 Question Paper – Candidates can download JEE Main question paper for January 07, 08, 09, 2020, for Shift 1 and Shift 2 from this page. NTA JEE Mains 2020 Answer Key PCM Shift 1, Shift 2- National Testing Agency NTA would publish official JEE Mains 2020 Answer Key along with Question Paper solutions at website. Furthermore, other coaching institutes will release JEE Main 2020 answer key. For JEE Main Analysis 2020 of the Joint Entrance Examination held on 6th 7th 8th and 9th January 2020 and JEE Main Question Paper Solution with Answer Sheet Pdf download, Aspirants stya here. JEE Main Official Question Papers and Solutions 2019 . Plan has been released by NTA. Total Number of questions asked in 9th January 1st shift 2020 JEE Main are 75 of which 60 questions are MCQ based having one choice correct and 15 are integer based. So, this case is not possible. Calculate ratio of moment of inertia about the axis perpendicular to plane of paper and passing through point P and B as shown in the figure. Find the loss in kinetic energy. Given P is centroid of the triangle. Practicing JEE Main Previous Year Question Paper 2020 with Answer Keys will help aspirants to score more marks in your IIT JEE Examinations. The liquids are immiscible. JEE Main Question Paper 2020 with Solutions (9th Jan – Morning) – Free PDF. This page consists of questions from JEE Main 2020 from 2nd shift of 8th January. Aspirants who appeared for the JEE Mains exam can check the 7th January, Morning … to 5.30 PM). Information such as difficulty … (Graphs are schematic and not to scale). Particle moves from point to point along the line shown in figure under the action of force F⃗=−xi^+yi^\vec{F} = -x\hat{i}+ y\hat{i}F=−xi^+yi^. The question paper consists of 3 Parts (Part I: Physics, Part II: Chemistry, Part III: Mathematics). Therefore current will also not flow through GH. Then, the combined body. To calculate I.E., we’ll have to put n lower= 1, which isn’t possible here. 3. JEE Main 2020 second session or the April session that was supposed to be conducted has been postponed due to the Covid-19 pandemic. JEE Main Question Paper with Solutions: The National Testing Agency will be releasing the JEE Main 2021 question paper on the official website. As per anticipation NTA may publish JEE Mains 2020 8 January Paper 1 answer key of B.E, B.Tech (Mathematics, Physics, Chemistry within 10 days of conduct of final exam. JEE MAIN QUESTION PAPERS 2019 (Last Update :18/09/2019) 8 January 2019 shift 1 7 April 2019 shift 1; 8 January 2019 shift 2 7 April 2019 shift 2; 9 January 2019 shift 1 8 April 2019 shift 1; 9 January 2019 shift … Calculate ratio … NTA has released the JEE Main official question paper for each shift. The first shift exam was held from 9.30 AM to 12.30 PM. One question cancelled is from Maths from the second shift paper held on January 9 while two questions in Physics … – 2nd Step: Click on the link to download the answer key. JEE Main January 2020 Answer Key & Paper Solutions are Available here. Radiation with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The question paper consists of 3 Parts (Part I: Physics, Part II: Chemistry, Part III: Mathematics). In the given circuit diagram, a wire is joining point B & C. Find the current in this wire; Since resistance 1 Ω and 4 Ω are in parallel, Similarly we can find equivalent resistance (′′) for resistances 2 Ω and 3 Ω, So total current flowing in the circuit ‘’ can be given as. What Students say about JEE Main 2020 January - 6 January Shift 1. Therefore, Molar heat capacity of B at constant volume (CV)B = 7R/2, Ratio of molar specific heat of A and B = (CV)A / (CV)B = 5/7, Question 20. Each section consisted of 25 questions … The forenoon session was held from 9:30 AM to 12:30 PM and the afternoon session was conducted from 2:30 PM to 5:30 PM. The final answer key of JEE Main 2020 Paper 1 shows three questions as cancelled by NTA. Information Bulletin JEE(Main) January-2020; Public Notice DASA; PRESS RELEASE:NTA Declares JEE(Main) Paper-2 (B.Arch./B.Planning) Results; SYLLABUS FOR JEE (Main) -2020 ; Other Links. So current in the branch BC will be = (4/5) - (3/ 5), Question 18. 2. JEE Main 2020 (6th, 7th, 8th, 9th January) Paper analysis & review. Students can easily access these … Question 12. In the given circuit both diodes are ideal having zero forward resistance and built-in potential of 0.7 V. Find the potential of point E in volts. If, we apply Nodal from right side, voltage at E will be 12 volt (diode between A and E will be forward biased). The official JEE Main 2020 answer key, along with question papers will be released by the NTA on the official website of JEE Main 2020 in the second week of September. Candidates can check and download the question paper from the official website – jeemain.nta.nic.in.JEE Main exam will be conducted in 4 sessions, starting from February, March, April and May 2021. JEE Main 2020 Answer Key and Question Paper PDF for September 1, 2 & 3 exam is now available. Get Shift wise JEE Main September 2020 answer key & question paper PDF for paper 1, 2 & 3 here. JEE Main 2020 Answer Key and Question Paper PDF for September 1, 2 & 3 exam is now available. 2. The electrons are made to enter a uniform magnetic field of 3×10-4 T. If the radius of largest circular path followed by electron is 10 mm, the work function of metal is close to; Radius of charged particle moving in a magnetic field is given by, r=2evm×meB=1B2mver = \frac{\frac{\sqrt{2ev}}{m}\times m}{eB}= \frac{1}{B}\sqrt{\frac{2mv}{e}}r=eBm2ev×m=B1e2mv, Question 14. JEE Main Sample Paper contains q uestions as per the most recent exam pattern of the Joint Entrance Examination. The final answer key of JEE Main 2020 Paper 1 shows three questions as cancelled by NTA. D. The difference between the magnitude of angular momentum of the particle at P and Q is 2mva. – 1st Step: Vist the official website of Resonance at resonance.ac.in. Three identical solid spheres each have mass ‘m’ and diameter ‘d’ are touching each other as shown in the figure. So, it is not possible to calculate I.E. Home; Exams. NTA conducted JEE Main 2020 on 6th (Paper 1), 7th, 8th, 9th January (Paper 2). Distance of centroid from (point P) centre of sphere = (2/3)×(√3d/2) = d/√3, Moment of Inertia about P = 3[ (2/5)M(d/2)2 + M(d/√3)2] = (13/10) Md2, Moment of Inertia about B = 2[ (2/5)M(d/2)2 + M(d)2] + (2/5)M(d/2)2 = (23/10)Md2, Question 2. Answer Key, Jee Main. Question 11. Now voltage at E is 3.3 volt and voltage at A is 12 volt and since, diode between E and A is forward biased and in forward biased difference of voltage of 0.7 volt is allowable. If a sphere of radius R/2 is carved out of it as shown in the figure. There are three papers: Paper-1 (BTech ), Paper-2 (BArch), and Paper-3 (BPlanning). Download free JEE Main 2020 Question Paper Solutions to ace your exams on the 7th January Morning covering Physics, Chemistry and Mathematics. Rate of work done at P = Power of electric force, So, dw/dt = 0 for both forces dwdt=q(E⃗+v⃗×B⃗).v⃗\frac{dw}{dt}= q(\vec{E}+\vec{v}\times \vec{B}).\vec{v}dtdw=q(E+v×B).v. In a fluorescent lamp choke (a small transformer) 100 V of reversible voltage is produced when choke changes current in from 0.25 A to 0 A in 0.025 ms. Download FREE PDF for JEE Mains 2020 Question Paper solved by experts. c) Escape from the Planet’s Gravitational field, d) Start moving in an elliptical orbit around the planet. Total 11,18,673 candidates registered for JEE Main 2020 in which included 9,21,261 for B.E./B.Tech, 1,38,409 for B.Arch. There are 50 divisions on circular scale. Only desired c andidates and also those who applied for the exam should solve the practice papers before the exam. Kota’s most experiences top IIT JEE Faculty Team design a best JEE Main 2020 Paper solutions and Answer key. Students can find below the links to download all sets of JEE Main question papers. (dxi^+dyj^), =∫10−x dx+∫01y dy=\int_{1}^{0}-x\: dx+ \int_{0}^{1}y\: dy=∫10−xdx+∫01ydy, Question 5. to 5.30 PM). Escape velocity will be √2V and at velocity less than escape velocity but greater than orbital velocity (V), the path will be elliptical. By using faraday law to write induced emf. Two particles of equal mass m have respective initial velocities u1⃗=ui^\vec{u_{1}} = u\hat{i}u1=ui^ and u2⃗=u2i^+u2j^\vec{u_{2}} = \frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}u2=2ui^+2uj^ . Students can easily access these solutions from our website and download them in PDF format for free. A solid sphere having a radius R and uniform charge density ρ. They collide completely inelastically. A screw gauge advances by 3 mm on main scale in 6 rotations. A rod of length 1 m pivoted at one end is released from rest when it makes 30° from the horizontal as shown in the figure below. The forenoon session was held from 9:30 AM to 12:30 PM and the afternoon session was conducted from 2:30 PM to 5:30 PM. If pressure difference between A & B is 700 N/m2, then volume flow rate is (density of water = 1000 kgm−3). NTA conducted JEE Main 2020 on 6th (Paper 1), 7th, 8th, 9th January (Paper 2). (Trajectory shown is schematic and not to scale), A. Since E⃗×B⃗\vec{E}\times \vec{B}E×B gives direction of propagation and E0/B0 = c and variation of magnetic field will be same to magnetic field. A vessel of depth 2h is half filled with a liquid of refractive index √2 in upper half and with a liquid of refractive index 2√2 in lower half. Let, Net work done by magnetic field be WB and net work done by electric field be WE. Only desired c andidates and also those who applied for the exam should solve the practice papers before the exam. For process 1 - 2, PVγ Constant , and PV = nRT therefore TVγ-1 = Constant ; therefore as V increases T decreases and also relation is non linear, so curve will not be a straight line. The Question paper is divided into 3 sections: Chemistry, Physics, and Mathematics. Candidates looking for NTA JEE Mains 2020 Paper 1 Online CBT/ Offline OMR sheet B.E/ B.Tech should download access to latest answer keys from National Testing Agency. Get Shift wise JEE Main September 2020 answer key & question paper PDF for paper 1, 2 … Question 21. Dimension of f is that of; Question 16. As per anticipation NTA may publish JEE Mains 2020 … Furthermore, other coaching institutes will release JEE Main 2020 answer key. Available Soon. Since, the variation of vectors E1 and E2 are along x and y respectively. Download JEE Main 2020 Question Paper (7th January – Morning) with Solutions for Physics, Chemistry and Mathematics in PDF format for free on Mathongo.com. Change in magnitude of angular momentum of the particle at P and Q about origin, ΔL⃗=ΔLP⃗−ΔLQ⃗\Delta \vec{L} = \Delta \vec{L_{P}}-\Delta \vec{L_{Q}}ΔL=ΔLP−ΔLQ, ΔLQ⃗=m(2v)(2a)\Delta \vec{L_{Q}}= m (2v)(2a)ΔLQ=m(2v)(2a), ΔLP⃗=m(v)(a)\Delta \vec{L_{P}}= m (v)(a)ΔLP=m(v)(a). JEE Main Paper 1 2020 (January 9) - National Testing Agency (NTA) conducted the first two shifts of Paper 1 (B.E./B.Tech) today i.e., January 9. A quantity is given by f = √(hc5/G), where c is speed of light, G is universal gravitational constant and h is the Planck’s constant. The forenoon session was held from 9:30 AM to 12:30 PM and the afternoon session was conducted from 2:30 PM to 5:30 PM. We have curated a list of memory based question and these will aid you in preparing and checking the marks efficiently for the exam held in a very short span of time. Examsnet. Specially how I just fucked up in chemistry. Hence, VE = 12 V. 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