Tiêu đề DPP-1 SOLUTION.pdf ✅

Class : XIth SUBJECT : PHYSICS Date : DPP No. : 1 1 (c) It is self-evident that the orbit of the comet is elliptic with sun begin at one of the focus. Now, as for elliptic orbits, according to kepler’s third law, T 2 = 4π 2a 3 GM ⇒ a = ( T 2GM 4π 2 ) 1/3 a = [ (76 × 3.14 × 107) × 6.67 × 10−11 × 2 × 1010 4π 2 ] 1/3 But in case of ellipse, 2a = rmin + rmax ∴ rmax = 2a − rmin = 2 × 2.7 × 1012 − 8.9 × 1010 ≅ 5.3 × 1012m 2 (b) Acceleration due to gravity g = GM R2 , M = ( 4 3 πR 3) ρ ∴ g = 4G 3 πR 3 R2 ρ ⟹ g = ( 4GπR 3 ) ρ (ρ = average density) ⟹ g ∝ ρ or ρ ∝ g 3 (c) g = GM R2 and K = L 2 2I If mass of the earth and its angular momentum remains constant then g ∝ 1 R2 and K ∝ 1 R2 i. e., if radius of earth decreases by 2% then g and K both increases by 4% 4 (a) Acceleration due to gravity at a height above the earth surface Topic :- GRAVITATION Solutions OM COACHING CLASSES
g ′ = g ( R R + h ) 2 g g′ = ( R + h R ) 2 g g′ = ( R + nR R ) 2 g g′ = (1 + n) 2 5 (c) Gravitational potential V = GM ( 1 r1 + 1 r2 + 1 r3 + ⋯ ) = G × 1 ( 1 1 + 1 2 + 1 4 + 1 8 + 1 16 + ⋯ . ) = G ( 1 1−1/2 ) (∴ sum of GP = a 1−r ) = 2G 6 (b) ge gm = Reρe Rmρm = 2 3 × 4 1 = 6 or gm = ge 6 For motion on earth, using the relation, s = ut + 1 2 at 2 We have, 1 2 = 0 + 1 2 × 9.8r 2 or t = 1/√9.8s For motion on moon, 3 = 0 + 1 2 (9.8/6)t1 2 or t1 = 6√9.8s ∴ t1 t = 6 or t1 = 6t 7 (c) Escape velocity, vascape = √ 2 GM R = R√ 8 3 πGρ ∴ ve ∝ R if ρ = constant. Since the planet is having double radius in comparision to earth, therefore escape velocity becomes twice ie, 22 kms−1 .
8 (a) vp ve = √ Mp Me × Re Rp = √6 × 1 2 = √3 ∴ vp = √3ve 12 (c) ve vm = √ ge gm Re Rm = √6 × 10 = √60 = 8 (nearly) 13 (c) Gravitational potential energy of a body in the gravitational field, E = −GM m r . When r decreases negative value of E increase ie, E decreases 14 (b) Actually gravitational force provides the centripetal force 15 (c) The earth moves around the sun is elliptical path, so by using the properties of ellipse r1 = (1 + e)a and r2 = (1 − e)a ⇒ a = r1+r2 2 and r1r2 = (1 − e 2)a 2 Where a = semi major axis b = semi minor axis e = eccentricity Now required distance = semi latusrectum = b 2 a = a 2 (1 − e 2 ) a = (r1r2) (r1 + r2)/2 = 2r1r2 r1 + r2 16 (c) At a certain velocity of projection of the body will go out of the gravitational field of earth and never to return to earth. The initial velocity is called escape velocity ve = √2gR Where g is acceleration due to gravity and R the radius. As is clear from above formula, that escape velocity dose not depends upon mass of body hence, it will be same for a body of 100kg as for 1kg body. 17 (d) Telecommunication satellites are geostationary satellite 18 (b) Weight of body at height above the earth’s surface is w ′ = w (1 + h r ) 2
⟹ 40 = 80 (1 + h r ) 2 ⟹ h = 0.41r 19 (d) As we know gas molecules cannot escape from earth’s atmosphere because their root mean square velocity is less than escape velocity at earth’s surface. If we fill this requirement, then gas molecules can escape from earth’s atmosphere. ie, vrms = ves or √ 3RT M = √2gRe or T = 2MgRe 3R ... . (i) Given, M = 2 × 10−3kg, g = 9.8 ms−2 Re = 6.4 × 106 m, R = 8.31 Jmol−1–K −1 Substituting in Eq. (i), we have T = 2 × 2 × 10−3 × 9.8 × 6.4 × 106 3 × 8.31 = 104 K 20 (d) v0 = √ GM r = √ gR2 R + h ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10