Tiêu đề NEET MOCK TEST - 02(EXP).pdf ✅

NEET MOCK TEST PAPERS NEET MOCK TEST - 02 NEET MOCK TEST PAPERS Date: March 18, 2025 Dura on: 3:00:00 Total Marks: 720 Important Instructions: 1. On the Answer Sheet, fill in the particulars on Side-1 and Side-2 carefully with blue/black ball point pen only. 2. The test is of 3.20 hours duration and this Test Booklet contains 180 questions. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be deducted from the total scores. The maximum marks are 720. 3. In case of more than one option correct in any question, the best correct option will be considered as the answer. 4. Use Blue/Black Ball Point Pen only for writing particulars on this page/marking responses. 5. Rough work is to be done on the space provided for this purpose in the Test Booklet only. 6. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator before leav‐ ing the Room/Hall. The candidates are allowed to take away this Test Booklet with them. 7. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on the Answer Sheet. Do not write your Form No. anywhere else except in the specified space in the Test Booklet/Answer Sheet. 8. Use of white fluid for correction is not permissible on the Answer Sheet. PHYSICS 1. () : Explana on Linear scale reading Circular scale reading Total reading 2. () : Explana on Conceptual Ques on 3. () : Explana on Displacement is given by 4. () : Explana on According to the defini on, displacement equals velocity mul plied by me. As displace‐ ment is a vector quan ty, its value corresponds to the signed sum of the area under the veloc‐ ity- me graph. In this context, areas above the -axis are considered posi ve, while areas be‐ low the -axis are considered nega ve. So cor‐ rect op on is (2). 5. () : Explana on Due to earth axial rota on, the speed of the trains rela ve to the earth will be different and hence the centripetal forces on them as given by and will be different. So, they will exert different pressure on the rails. Also, the magnitude of accelera on at every point of a circular path is same but its direc on changes con nuously. L. C = = 0.01mm 1 100 = 7 × pitch = 7mm = 45 × 0.01 = 0.45 mm ∴ = (7 + 0.45) = 7.45 mm = √2r ∴ d = √2 × 2 = √8 km x x (mg − ) mv 2 r (mg + ) mv 2 r

NEET MOCK TEST PAPERS 21. () : Explana on If volume is doubled at constant pressure, then absolute temperature of the gas is doubled. So, average transla onal kine c energy is also doubled. 22. () : Explana on In one complete oscilla on par cle comes to its star ng posi on, so displacement is zero. 23. () : Explana on 24. () : Explana on Phase difference between any two points present in two adjacent loops is always equal to . 25. () : Explana on 26. () : Explana on Here, number of electrons removed number of atoms in 1 27. () : Explana on Electric field intensity at a point near and out‐ side the surface of any charged conductor is given by (1) where, is charge density (charge per unit area). Electric field intensity due to uniformly charged infinite thin plane sheet is given by (2) From eqs.(1) and (2) we have 28. () : Explana on Air cra s move on 'runway ground'. The asser‐ on is true that aircra res are slightly con‐ duc ng. So correct op on is (1). 29. () : Explana on We know that, 30. () : Explana on Let be the current through the circuit Current in circuit Terminal voltage 31. () : Explana on The rela on between balancing lengths & emf is 32. () : Explana on As 33. () : Explana on Force due to magne c field, here and are same for both electron and proton. So force on both will be same in magni‐ tude but different in direc on. 34. () : Explana on for turn For rewinding the coil in three turns, new ra‐ dius /3, number of turns New magne c fields V = ω√A2 − x 2 ⇒ = √ = √ = √ V1 V2 A2 − x 2 1 A2 − x 2 2 A2 − A2/9 A2 − 4A2/9 8 5 π Iav = 1 2 B2 0 c μ0 = × 1 2 (12×10 −8)×3×10 8 4π×10 −7 = 1.71 Wm−2 . = g ⇒ n = = 4 × 10 4×10 17 20 10 3 q = ne = 4 × 10 17 × 1.6 × 10 −19C = 6.4 × 10 −2C E1 = σ ε0 σ E2 = σ 2ε0 E1 = 2E2 qE = mg ⇒ = mg qQ ε0A ⇒ q = [ 8.85 × 10 −12 × 2× 10 −2 × 2.5 × 10 −7 × 10 ] 5 × 10 −7 = 8.85 × 10 −13C i i = = 2 A 10 4+1 = ε − iR = 10 − 2 × 1 = 8 V = ⇒ = E1 l1 E2 l2 1.5 36 2.5 l2 ⇒ l2 = 36 ( ) = 36 × = 60 cm 2.5 1.5 5 3 I = neAvd = neAv v = = I neA 1.5 9×10 28×1.6×10 −19×5×10 −6 = 0.02 × 10 −3m/s = 0.02 mm/s F = Bqv B, |q| v B0 = = = μ0NI 2a μ0I × I 2a μ0I 2a I a (N ′) = 3 = − = 9 B0 μ0I×3 2×(a/3) 9μ0I 2a