Tiêu đề NEET MOCK TEST - 01(Exp).pdf ✅

NEET MOCK TEST PAPERS 16. (3) : Explanation Given, and The orbital velocity of satellite is Now 17. (3) : Explanation 18. (1) : Explanation Conceptual Question 19. (3) : Explanation Conceptual Question 20. (2) : Explanation Conceptual Question 21. (4) : Explanation 22. (4) : Explanation Frequency of oscillations made by a simple pendu‐ lum of length is Since, completes 10 oscillations in Similarly, or 23. (4) : Explanation Effective spring constant . When one spring is removed, becomes new frequency 24. (1) : Explanation . 25. (4) : Explanation 26. (1) : Explanation Electric lines of force are always normal to metal‐ lic body. 27. (3) : Explanation Given: When point charges and placed in medium with dielectric constant and 28. (1) : Explanation Potential difference From the relation 29. (4) : Explanation If there is a component of electric field parallel to the surface then charge will experience force and it starts moving on the surface, so electric field should be always perpendicular to the surface in electrostatic situation. If there is any charge inside a conductor then it should flows to the outer sur‐ face because of repulsion, so no charge resides on outer surface. If the electric field is along - axis then equipotential surfaces will be along - plane. 30. (4) : Explanation Fuse wire has less melting point so when excess current flows, it melts 31. (1) : Explanation The circuit can be redrawn as R1 = Re R2 = Re + = Re Re 2 3 2 vo = √ ⇒ vo ∝ √ GMe R 1 R = √ = √ = √ v1 v2 R2 R1 3Re 2Re 3 2 v2 = √ v1 = √ v0 (∵ v1 = v0) 2 3 2 3 g 1 = r GM R3 = r = gR2 R3 gr R = 9.8×2000 6400 = 3.06 ms−2 CP = ( + 1) R = ( + 1) R = R f 2 5 2 7 2 l f = √ ⇒ f ∝ 1 2π g l 1 √l A 20 s ⇒ fA = = 0.5 Hz 10 20 fB = = 0.8 Hz 8 10 ∴ = = √ fA fB 5 8 lB lA = lA lB 64 25 Keff = 2K f = √ 1 2π 2K m Keff K f ′ = √ = ⇒ f ′ = 1 2π K m f ′ f 1 √2 f √2 Amax = A1 + A2; Amin = A1 − A2 ∴ Amax − Amin = A1 + A2 − A1 + A2 = 2A2 = 0.14 ⇒ T = 4 × 0.14s T 4 ∴ f = = s−1 = 1.79Hz 1 T 1 4 × 0.14 F = (1) q1q2 4πε0r 2 q1 q2 k = 5 r ′ = ,then r 5 F ′ = = = 5 F q1q2 4πε0r ′2k q1q2 4πε0×5( ) 2 r 5 = 10 − (−10) = 20 q = CV C = = = 2F q V 40 20 y xz RAB = 3.12Ω