Tiêu đề 08. GRAVITATION(MOBILE).pdf ✅

SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 08. GRAVITATION #QID# 40010 (1.) Halley’s comet has a period of 76, had distance of closest approach to the sun equal to 8.9 × 1010m. the comet’s farthest distance from the sun if the mass of sun is 2 × 1030 kg and G = 6.67 × 1011 in MKS units is (a.) 2 × 1012m (b.) 2.7 × 1013m (c.) 5.3 × 1012m (d.) 5.3 × 1013m ANSWER: c EXPLANATION: (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, 1/3 2 3 2 2 2 4 4 a T GM T a GM     =  =     ( ) 1/3 7 11 10 2 76 3.14 10 6.67 10 2 10 4 a  −             =         But in case of ellipse, 2a = rmin + rmax ∴ rmax = 2a − rmin = 2 × 2.7 × 1012 − 8.9 × 1010 ≅ 5.3 × 1012m #QID# 40011 (2.) Average density of the earth (a.) does not depend on g (b.) is a complex function of g
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (c.) is directly proportional to g (d.) is inversely proportional g ANSWER: b EXPLANATION: (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 #QID# 40012 (3.) Let g be the acceleration due to gravity at earth’s surface and K be the rotational kinetic energy of the earth. Suppose the earth’s radius decreases by 2% keeping all other quantities same, then (a.) g decreases by 2% and K decreases by 4% (b.) g decreases by 4% and K increases by 2% (c.) g increases by 4% and K increases by 4% (d.) g decreases by 4% and K increases by 4% ANSWER: c EXPLANATION: (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%
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 #QID# 40013 (4.) A body is taken to a height of nR from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is (a.) (n + 1) 2 (b.) (n + 1) −2 (c.) (n + 1) −1 (d.) (n + 1) ANSWER: a EXPLANATION: (a) Acceleration due to gravity at a height above the earth surface g ′ = g ( R R + h ) 2 g g′ = ( R + h R ) 2 g g′ = ( R + nR R ) 2 g g′ = (1 + n) 2 #QID# 40014 (5.) Infinite number of masses, each 1 kg, are placed along the x-axis at x = ±1m, ±2m,±4m, ±8m,±16m ..... The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x = 0) is (a.) G/2
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (b.) G (c.) 2G (d.) 4G ANSWER: c EXPLANATION: (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 #QID# 40015 (6.) In the above problem, the ratio of the time duration of his jump on the moon to that of his jump on the earth is (a.) 1 : 6 (b.) 6 : 1 (c.) √6 ∶ 1 (d.) 1 ∶ √6 ANSWER: b EXPLANATION: (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