Tiêu đề 7-gravitation-.pdf ✅

Gravitation 1. (A) Energy required to move a satellite from radius r1 to r2 is given by ΔE = GMm 2 [ 1 r1 − 1 r2 ] = GMm 2 ( 1 2R − 1 3R ) = GMm 12R 2. (B) When gravitational force becomes zero, the centripetal force required cannot be provided. So the satellite will move with the velocity as it has at the instant when gravitational force becomes zero, i.e., it moves tangentially to the original orbit. 3. (A) Escape velocity = √2gRe . So the escape velocity is independent of m. So it depends upon mass as m0 . 4. (C) The minimum kinetic energy required to project a body of mass m from the earth's surface to infinite is know as escape energy. Therefore, KE = GMem R = mgR (∵ gR = GMe R ) 5. (C) T2 T1 = ( r2 r1 ) 3/2 = ( 4r r ) 3/2 = 8 ∴ T2 = 8T1 = 8 × 5 = 40 h 6. (C) If x1 and x2 are the distance covered by the two bodies, then x1 + x2 = 9R Also Mx1 = 5Mx2 ⇒ x2 = x1 5 ∴ x1 + x1 5 = 9R ⇒ x1 = 7.5R 7. (C) Work is done against gravitational force, which is a conservative force. Therefore, the escape velocity is independent of the angle of projection. 8. (B) v0 = √ GM R+x = √ gR2 R+x 9. (A) The time period of the satellite is given by T = 2π√ (R+h) 3 GM Where R + h is the radius of the orbit of the satellite, M is mass of the earth. 10. (B) ΔU = mgh 1+ h R = 1 2 mgR(∵ h = R) 11. (A) mω 2R ∝ 1 Rn ⇒ m ( 4π 2 T2 ) R ∝ 1 Rn ⇒ T 2 ∝ R n+1 ∴ T ∝ R ( n+1 2 ) 12. (C) g = GM R2 = G R2 × 4 3 πR 3ρ = 4 3 πRρG ⇒ ρ = g 13. (C) At height h above the surface of the earth. g ′ = g (1 − 2h R ) ⇒ Δg1 = 2h R g At depth d below the surface of the earth, g ′ = g (1 − d R ) ⇒ Δg2 = d R g Since Δg1 = Δg2 d = 2h 14. (D) Work done = Change in GPE = U∞ − UR W = 0 − ( GMm R ) = GMm R = 6.67 × 10−11 × 100 × 10−2 10−1 = 6.67 × 10−10 J 15. (D) Electronic charge = 1.6 × 10−19C It is the fundamental property of an electron. Its value remains the same on the moon as well as on the earth. ∴ Electronic charge on moon Electronic charge on earth = 1 16. (C) Mass of planet MP = 10Me Where Me is the mass of the earth. Radius of planet RP = Re 10 ; Where Re is the radius of the earth. Using formula vplanet = √ 2GMP RP = √ 2G(10Me ) (Re/10) = 10vearth 17. (A) g ′ = GM (R+h) 2 , acceleration due to gravity at height h ⇒ g 9 = GM R2 R 2 (R + h) 2 = g ( R R + h ) 2 ⇒ 1 9 = ( R R + h ) 2 ⇒ R R + h = 1 3 ⇒ 3R = R + h ⇒ 2R = h Using formula ve = √ 2GM R v = 10 18. (C) For solving this problem let us use the concept of electric flux. Let A be the Gaussian surface enclosing a spherical charge Q. Flux enclosed by Gaussian surface
φ = ∮ E⃗ ⋅ d s = Qenclose ε0 ⇒ E. 4πr 2 = Q ε0 Hence electric field at a distance a from charge is E = Q 4πε0r 2 Flux φ = E⃗ . 4πr 2 = Q ε0 Every line passing through A, has to pass through B, whether B is a cube or any surface. It is only for Gaussian surface, the lines of field should be normal. Assuming the mass is a point mass. Gravitational field g = − GM r 2 Flux φg = |g ⋅ 4πr 2 | = 4πr 2 ⋅GM r 2 = 4πGM Hence B is a cube. As explained earlier, whatever be the shape, all the lines passing through A are passing through B, although all the lines are not normal. Statement 2 is correct because when the shape of the earth is spherical, are of the Gaussian surface is 4πr 2 . This ensures inverse square law. 19. (D) Let x be the distance of the point P from the mass m where gravitational field is zero. ∴ Gm x 2 = G(4m) (r−x) 2 or ( x (r−x ) 2 = 1 2 Or x = r 3 Gravitational potential at a point P is : V = − Gm x − G(4m) (r − x) V = − Gm ( r 3 ) − G(4m) (r− r 3 ) = −9 Gm r (Using (i)) 20. (D) W = 0 − (− GMm R ) = GMm R = m ( GM R2 )R = mgR| = 1000 × 10 × 6400 × 103 = 64 × 109 J = 6.4 × 1010 21. (D) Energy of the satellite on the surface of the planet is E1 = KE + PE = 0 + (− GMn R ) = − GMm R = 1 2 GMm 3R − GMm 3R = − GMm 6R ∴ Minimum energy required to launch the satellite is : ΔE = Ef − Ei = − GMm 6R − (− GMm R ) = − GMm 6R + GMm R = 5GMm 6R 22. (B) Net force on any one particle = GM2 (2R) 2 + GM2 (R√2) 2 cos 45∘ + GM2 (R√2) 2 cos 45∘ = GM2 R2 [ 1 4 + 1 √2 ] This force will be equal to centripetal force so Mu 2 R = GM2 R2 [ 1 + 2√2 4 ] u = √ GM R [1 + 2√2] = 1 2 √ GM R (2√2 + 1)
23. (B) V = V1 − V2 V1 = − GM 2R3 [3R 2 − ( R 2 ) 2 ]; V2 = − 3G ( M 8 ) 2 ( R 2 ) ⇒ V = −GM R INTEGER VALUE TYPE 24. (2) dm = D × dl = D × Sdα cos α Gravitational force, dF = GMdm ( S cos α ) 2 cos α Total force F = ∫−π/2 π/2 MGD S cos αdα = 2MGD S 25. (8) dF = GMdm (3R) 2 ; F = ΣdFcos θ ⇒ F = GMm2√2 27R2 (⬚ as cos θ = 2√2 3 ) 26. (7) V = −G4M 3 2 R − MG 2R = − 19 6 GM R 27. (2) Let v be the speed of the projectile at highest point and rmax its distance from the centre of the earth. Applying conservation of angular momentum and mechanical energy. mv0sin α + mvrmax 1 2 mv0 2 − GMem Re = 1 2 mv 2 − GMem rmax Solving these two equations with given data, we get rmax = 3Re 2 or, the maximum height hmax = rmax − Re = Re 2 28. (6) LA = 2.2Msr1 2ω; LB = 11Msr2 2ω L LB = 2.2Msr1 2 + 11Msr2 2 11Msr2 2 = 1 + 2.2 11 ( r1 r2 ) 2 = 1 + 2.2 11 × ( 11 2.2 ) 2 (∵ m1r1 = m2r2 ) = 6 29. (4) From given information, semi-major axis is equal to 4 units. Let e be the eccentricity of ellipse, then ae = 1 ⇒ e = 1/4 Semi-major axis, b = a√1 − e 2 = 4√1 − 1 16 units = √15 units So required distance = √b 2 + 1 2 = 4 units 30. (2) F1 = DMm (2R) 2 , F2 = F1 − F ′ = GMm (2R) 2 − GMm 4×(R+ R 2 ) 2 Now, F2 F1 = ( 7 9 ) 31. (2) For a particle at a distance r from the centre of earth, force is given by, F = Gm1m2 r 2 Force becomes one-fourth, when r = 2R ( R = radius of earth) Escape velocity, ve = √ 2GM R Using conservation of energy for the given particle, 1 2 mv 2 − GmM R = − GmM 2R This gives, v = √ GM R ve = v√2 ⇒ ve = v√2Hence, N = 2 32. (7)From FBD of middle mass: GMm 9l 2 − Gm2 l 2 − T = ma Since T = 0, GMm 9l 2 − Gm2 l 2 = ma Also, for the right most mass:
GMm 16l 2 + Gm2 l 2 = ma Equating (i) and (ii) GMm 9l 2 − GMm 16l 2 = 2Gm2 l 2 Solving m = 7M 288 33. (9) √3F = mv 2 r ⇒ v = [ √3Fr M ] 1/2 Where F = Gm2 a 2 34. (6) Let the projectile of mass m be fired with minimum velocity, v from the surface of sphere of mass M to reach the surface of mass 4M. Let N be neutral point at a distance r from the centre of the sphere of mass M. At neutral point N. GMm r 2 = G(4M) (6R − r) 2 (6R − r) 2 = 4r 2 6R − r = ±2r or r = 2R or − 6R The point r = −6R does not concern us. Thus, ON = r = 2R It is sufficient to project the projectile with a speed which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy of m at the surface of M is Ei = 1 2 mv 2 − GMm R − G(4M)m 5R At the neutral point N, the speed approaches zero ∴ The mechanical energy at N is EN = − GMm 2R − G(4M)m 4R = − GMm 2R − GMm R According to law of conservation of mechanical energy, Ei = EN 1 2 mv 2 − GMm R − 4GMm 5R = − GMm 2R − GMm R v 2 = 2GM R [ 4 5 − 1 2 ] = 3 5 GM R or v = ( 3 5 GM R ) 1/2 35. (6) Here, mass of the particle = M Mass of the spherical shell = M Radius of the spherical shell = R Let O be centre of spherical shell Gravitational potential at point P due to particle at O is V1 = − GM (R/2) Gravitational potential at point P due to spherical shell is V2 = − GM R Hence, total gravitational potential at the point P is V = V1 + V2 = − GM (R/2) + (− GM R ) = − 2GM R − GM R = − 3GM R 36. (2) Removed mass m = M 4 3 πR3 × 4 3 π ( R 2 ) 2 = M 8 Gravitational potential at centre of cavity is VP = Vsphere P − VmP