physics : class11 Newton's Law of Universal Gravitation : Gravitational Force : Gravitation MCQ Questions and Answers
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Newton's Law of Universal Gravitation
MCQ questions and answers.
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MCQ Questions
1. Three identical point masses, each of mass m, are fixed at the other three vertices of a square of side a. What is the magnitude of the net gravitational force on a mass m placed at the fourth vertex?
A.
Gm²/a²
B.
(1 + √2/2)Gm²/a²
C.
(√2 + 1/2)Gm²/a²
D.
2Gm²/a²
Correct Answer:
(√2 + 1/2)Gm²/a²
Explanation:
The two adjacent masses each exert a force Gm²/a² along perpendicular sides. The diagonally opposite mass exerts Gm²/(2a²). The resultant of the adjacent forces is √2Gm²/a², while the diagonal force has equal components along both axes. Therefore, the total magnitude is Gm²/a²(√2 + 1/2).
2. The gravitational force between two masses is F. If one mass is tripled, the other is doubled, and the separation between them is increased to four times its original value, the new force is:
A.
3F/16
B.
3F/8
C.
3F/4
D.
8F/3
Correct Answer:
3F/8
Explanation:
According to Newton's law of universal gravitation, F = Gm₁m₂/r². Thus, the new force divided by the original force is (3)(2)/(4²) = 6/16 = 3/8.
3. Two point masses m and 4m are fixed 10 m apart. At what distance from the mass m, between the two masses, should a third point mass be placed so that the net gravitational force on it is zero?
A.
2 m
B.
10/3 m
C.
5 m
D.
20/3 m
Correct Answer:
10/3 m
Explanation:
For zero net force, the attractions toward the two masses must have equal magnitudes: GmM/x² = G(4m)M/(10 − x)². Hence, (10 − x) = 2x, giving x = 10/3 m from the mass m.
4. Two equal point masses M are separated by a distance d. A test mass m is placed on the line joining them at a distance d/6 from their midpoint, toward one of the masses. What is the magnitude of the net gravitational force on m?
A.
9G Mm/(4d²)
B.
27G Mm/(4d²)
C.
36G Mm/(5d²)
D.
G Mm/d²
Correct Answer:
27G Mm/(4d²)
Explanation:
The distances from the test mass to the nearer and farther masses are d/3 and 2d/3. The forces oppose one another, so the net magnitude is G Mm[(3/d)² − (3/(2d))²] = G Mm(9/d² − 9/(4d²)) = 27G Mm/(4d²).
5. A mass m is placed at the point (0, a), while masses m and 2m are fixed at (0, 0) and (a, 0), respectively. What is the magnitude of the net gravitational force on the mass at (0, a)?
A.
(1 + √2)Gm²/a²
B.
√(2 + √2)Gm²/a²
C.
√(3 + √2)Gm²/a²
D.
2Gm²/a²
Correct Answer:
√(2 + √2)Gm²/a²
Explanation:
The force from the mass at (0, 0) has magnitude Gm²/a² downward. The force from the mass 2m is directed along (−i + j) and has magnitude Gm²/a², giving components (−F/√2, F/√2), where F = Gm²/a². The resultant magnitude is F√[1/2 + (1 + 1/√2)²] = F√(2 + √2).
6. Two point masses m₁ and m₂ exert a force F on each other when separated by r. If both masses are increased by 50% and the separation is decreased by 20%, the new force is:
A.
1.5625F
B.
2.25F
C.
2.34375F
D.
3.125F
Correct Answer:
2.34375F
Explanation:
The masses become 1.5m₁ and 1.5m₂, while the separation becomes 0.8r. Therefore, F'/F = (1.5 × 1.5)/(0.8)² = 2.25/0.64 = 2.34375.
7. Four identical point masses m occupy the vertices of a square of side a. A test mass M is placed at the center, and one of the four vertex masses is removed. What is the magnitude of the resulting gravitational force on M?
A.
GmM/a²
B.
2GmM/a²
C.
4GmM/a²
D.
GmM/(2a²)
Correct Answer:
2GmM/a²
Explanation:
With all four masses present, the net force at the center is zero by symmetry. Removing one mass leaves a resultant equal in magnitude and opposite in direction to the force that the removed mass would have exerted. The center-to-vertex distance is a/√2, so the magnitude is GmM/(a²/2) = 2GmM/a².
8. A mass m is divided into n equal point masses, and another mass M is divided into p equal point masses. If all resulting small masses are placed at the original locations of their respective bodies, with the separation unchanged, the total gravitational force between the two groups is:
A.
F/(np)
B.
F/n
C.
F/p
D.
F
Correct Answer:
F
Explanation:
Each pair of small masses has force G(m/n)(M/p)/r² = F/(np). There are np such pairs, so the total force is np × F/(np) = F. This agrees with the dependence of gravitational force on the total masses.
9. Two point masses M and 9M are separated by 4a. A third mass is placed between them so that the magnitudes of the gravitational forces due to M and 9M are equal. Its distance from M is:
A.
a/2
B.
a
C.
4a/3
D.
2a
Correct Answer:
a
Explanation:
Let the distance from M be x; its distance from 9M is 4a − x. Equality of force magnitudes gives GMm/x² = 9GMm/(4a − x)². Therefore, 4a − x = 3x, so x = a.
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