physics : class11 Force Between Point Masses : Gravitational Force : Gravitation MCQ Questions and Answers
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Force Between Point Masses
MCQ questions and answers.
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MCQ Questions
1. Two point masses m and 4m are fixed at a separation d. A third point mass is placed on the line joining them so that the net gravitational force on it is zero. Its distance from the mass m is
A.
d/4
B.
d/3
C.
d/2
D.
2d/3
Correct Answer:
d/3
Explanation:
If the third mass is at distance x from m, its distances from m and 4m are x and d − x. Equating the attractive forces gives GmM/x² = G(4m)M/(d − x)², so d − x = 2x and x = d/3.
2. A point mass m is placed at the origin. Two other point masses, m and 2m, are located at (a, 0) and (0, a), respectively. The magnitude and direction of the net gravitational force on the mass at the origin are
A.
Gm²/a² and tan⁻¹(1/2) from the positive x-axis
B.
√5 Gm²/a² and tan⁻¹(2) from the positive x-axis
C.
3Gm²/a² and 45° from the positive x-axis
D.
√5 Gm²/(2a²) and tan⁻¹(1/2) from the positive x-axis
Correct Answer:
√5 Gm²/a² and tan⁻¹(2) from the positive x-axis
Explanation:
The two perpendicular force components are Gm²/a² along +x and 2Gm²/a² along +y. Therefore, the resultant is √5Gm²/a² and its angle satisfies tan θ = 2.
3. The gravitational force between two point masses is F. If one mass is increased by 50%, the other is reduced to one-third of its original value, and their separation is doubled, the new force is
A.
F/8
B.
F/6
C.
F/4
D.
3F/8
Correct Answer:
F/8
Explanation:
Since F is proportional to m₁m₂/r², the force changes by the factor [(1.5)(1/3)]/2² = 1/8. Hence the new force is F/8.
4. Four identical point masses m are fixed at the four corners of a square of side a. The magnitude of the net gravitational force on any one corner mass due to the other three is
A.
(1 + 1/√2)Gm²/a²
B.
(√2 + 1/2)Gm²/a²
C.
(2 + 1/√2)Gm²/a²
D.
(√2 + 1)Gm²/(2a²)
Correct Answer:
(√2 + 1/2)Gm²/a²
Explanation:
The two adjacent masses each exert Gm²/a² along perpendicular sides, while the opposite corner exerts Gm²/(2a²) along the diagonal. Each component of the diagonal force is Gm²/(2√2a²). Thus the resultant is √2[Gm²/a² + Gm²/(2√2a²)] = (√2 + 1/2)Gm²/a².
5. Two point masses m and 4m are separated by a distance d. At what point on the line joining them is the gravitational field zero, considering only the fields due to these two masses?
A.
At d/3 from m
B.
At d/3 from 4m
C.
At d/5 from m
D.
At the midpoint
Correct Answer:
At d/3 from 4m
Explanation:
For cancellation between the masses, Gm/x² = G(4m)/(d − x)², where x is the distance from m. This gives d − x = 2x, so x = 2d/3; equivalently, the point is d/3 from 4m.
6. Two point masses m₁ and m₂ exert mutual gravitational forces of magnitude F on each other. If their masses are changed to 2m₁ and 3m₂ while their separation remains unchanged, the magnitude of the force on each mass becomes
A.
2F
B.
3F
C.
5F
D.
6F
Correct Answer:
6F
Explanation:
The gravitational force is proportional to the product of the two masses. The product changes from m₁m₂ to (2m₁)(3m₂) = 6m₁m₂, so the mutual force becomes 6F.
7. Three identical point masses m are placed at the vertices of an equilateral triangle of side a. The magnitude of the net gravitational force on any one mass due to the other two is
A.
Gm²/a²
B.
√2 Gm²/a²
C.
√3 Gm²/a²
D.
2Gm²/a²
Correct Answer:
√3 Gm²/a²
Explanation:
Each of the two forces has magnitude Gm²/a², and the angle between them is 60°. Therefore, the resultant is √(F² + F² + 2F²cos60°) = √3F = √3Gm²/a².
8. Two point masses m and 2m are separated by a distance d. A third point mass m is placed at the midpoint. The magnitude of the net gravitational force on the third mass is
A.
2Gm²/d²
B.
4Gm²/d²
C.
8Gm²/d²
D.
16Gm²/d²
Correct Answer:
4Gm²/d²
Explanation:
The distances from the midpoint to both masses are d/2. The forces have magnitudes 4Gm²/d² and 8Gm²/d² in opposite directions. Their net magnitude is therefore 8Gm²/d² − 4Gm²/d² = 4Gm²/d².
9. Two point masses m and 4m are fixed at the ends of a segment of length d. A point mass M is placed on the perpendicular bisector at a distance h from the midpoint. The ratio of the magnitudes of the forces exerted by 4m and m on M is
A.
1/4
B.
1
C.
4
D.
16
Correct Answer:
4
Explanation:
Because M lies on the perpendicular bisector, its distance from both fixed masses is identical. Since gravitational force is proportional to the source mass for fixed distance and test mass, the force ratio is 4m/m = 4.
10. Two point masses m and 3m are fixed at a separation d. A third point mass M is placed on the line joining them, outside the segment on the side of the mass m. If the net gravitational force on M is zero, its distance from m is
A.
d/(√3 − 1)
B.
d/(√3 + 1)
C.
d/2
D.
d/(3 − 1)
Correct Answer:
d/(√3 − 1)
Explanation:
Let the distance of M from m be x; its distance from 3m is x + d. For cancellation, GmM/x² = 3GmM/(x + d)². Hence x + d = √3x, giving x = d/(√3 − 1).
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