38 public questions tagged with this topic.
g(h) = g(1+h/RE)2. h = 3RE/4, 1+h/RE = 1+3/4 = 7/4. g(h) = 9.8(7/4)2 = 9.8×1649 = 3.2m/s2. Mass: m = 294/9.8 = 30kg. Weight: W = 30×3.2 = 96N.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
It’s attractive (option 1 correct), follows inverse-square (option 2 correct), and is conservative (option 4 correct). Option 3 is incorrect as it is not limited to Earth. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It acts only near Earth. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
At the center (r = 0), the gravitational forces from all mass elements of a uniform sphere cancel out due to symmetry, as the net force from opposite directions balances to zero.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Force between two masses: F = Gm1m2r2 = 6.67×10−114×432 = 1.185×10−10N. Two forces act at 60°: FR = F2+F2+2F2cos60∘. FR = (1.185×10−10)2(1+1+1) = 1.185×10−103. FR≈2.05×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.2 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
For a satellite in circular orbit, the centripetal force (F = mv2r) is provided by Earth’s gravitational force (F = GMEmr2), which keeps it in orbit. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Gravitational force. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Newton observed that the centripetal acceleration of the Moon (am = V2Rm) was much smaller than the acceleration due to gravity on Earth’s surface (g), and their ratio matched the inverse square of their distances from Earth’s center (gam≈Rm2RE2).
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Outside shell: F = GMmr2. F = 6.67×10−11×700×13112. F = 6.0689×10−8121≈5.02×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Distance to 8kg: 8−3 = 5m. U = −Gm1r1−Gm2r2. U = −6.67×10−11(43+85). U = −6.67×10−11(1.333+1.6) = −1.96×10−10J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.0 × 10⁻¹⁰ J/kg. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
The inverse-square law (F∝1/r2) leads to elliptical orbits as a general solution to the two-body problem, matching Kepler’s first law, unlike a linear force which would produce circular orbits. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Inverse-square gravitational force. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Outside shell: F = GMmr2. F = 6.67×10−11×450×1082. F = 3.0015×10−864≈4.69×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.7 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Force due to A on C: FAC = GmAmCr2 = 6.67×10−115×1022 = 8.34×10−10N. Force due to B on C: FBC = 8.34×10−10N (equal magnitude, 60° apart). Resultant: FR = FAC2+FBC2+2FACFBCcos60∘. FR = (8.34×10−10)2+(8.34×10−10)2+2(8.34×10−10)2(0.5). FR = 8.34×10−101+1+1 = 1.44×10−9N.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Force from adjacent: F1 = Gm2r2 = 6.67×10−113×322 = 1.50×10−10N (2 forces at 90°). Force from diagonal: F2 = Gm2(22)2 = 6.67×10−1198 = 7.50×10−11N. Resultant of two F1: FR = 2×1.50×10−10≈2.12×10−10N. Net force (vector sum with F2): F = (2.12×10−10)2+(7.50×10−11)2≈2.25×10−10N.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
