physics : class11 Net Force in Symmetric Mass Configurations : Gravitational Force : Gravitation MCQ Questions and Answers
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Net Force in Symmetric Mass Configurations
MCQ questions and answers.
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MCQ Questions
1. Four identical point masses m are fixed at the vertices of a square of side a. What is the magnitude of the net gravitational force on the mass at one vertex due to the other three masses?
A.
Gm²/a²
B.
(√2 + 1/2)Gm²/a²
C.
(2 + 1/√2)Gm²/a²
D.
2Gm²/a²
Correct Answer:
(√2 + 1/2)Gm²/a²
Explanation:
The two adjacent masses exert perpendicular forces, each of magnitude Gm²/a². The diagonally opposite mass exerts Gm²/(2a²) along the square diagonal. The two equal components of the resultant are Gm²/a² + Gm²/(2√2a²), giving a magnitude √2 times this value, namely (√2 + 1/2)Gm²/a².
2. Four masses are placed at the corners of a square of side a. Three have mass m and one has mass 2m. A test mass M is placed at the square's center. What is the magnitude and direction of the net gravitational force on M?
A.
2GMm/a², directed toward the 2m mass
B.
4GMm/a², directed away from the 2m mass
C.
√2GMm/a², directed toward the 2m mass
D.
Zero, because the masses are at the corners of a square
Correct Answer:
2GMm/a², directed toward the 2m mass
Explanation:
If all four corner masses were m, symmetry would make the force at the center zero. Replacing one mass by 2m leaves an excess mass m at that corner. Its distance from the center is a/√2, so the remaining force is G(M)(m)/(a²/2) = 2GMm/a² toward that corner.
3. Three masses are fixed at the vertices of an equilateral triangle of circumradius R. Two masses are m and the third is 2m. What is the gravitational force on a test mass M at the triangle's center?
A.
Zero
B.
GMm/R² directed away from the 2m mass
C.
GMm/R² directed toward the 2m mass
D.
2GMm/R² directed toward the 2m mass
Correct Answer:
GMm/R² directed toward the 2m mass
Explanation:
Three equal masses would produce zero force at the center because their radial force vectors cancel. The 2m mass can be regarded as an ordinary m mass plus an excess m. Therefore, only the excess mass contributes, producing GMm/R² toward the vertex containing 2m.
4. Six masses are placed at the vertices of a regular hexagon, with masses m and 2m alternating around the hexagon. A test mass M is at the center. What is the net gravitational force on M?
A.
Zero
B.
GMm/R² toward the nearest m mass
C.
GMm/R² toward the nearest 2m mass
D.
2GMm/R² toward the nearest 2m mass
Correct Answer:
Zero
Explanation:
In a regular hexagon, each vertex has an opposite vertex. The alternating arrangement gives equal masses at every pair of opposite vertices: m is opposite m and 2m is opposite 2m. Each opposite pair produces equal and opposite forces at the center, so the net force is zero.
5. A test mass M is at the center of a square of side a. The two vertices on the right side each contain mass 2m, while the two vertices on the left side each contain mass m. What is the magnitude and direction of the net force on M?
A.
2√2GMm/a² toward the right
B.
4GMm/a² toward the right
C.
√2GMm/a² toward the left
D.
Zero
Correct Answer:
2√2GMm/a² toward the right
Explanation:
The vertical components cancel pairwise. Each corner is at distance a/√2 from the center, so the force due to a corner mass m has magnitude 2GMm/a². The two right-side masses contribute rightward components totaling 4√2GMm/a², while the two left-side masses contribute leftward components totaling 2√2GMm/a². The resultant is 2√2GMm/a² toward the right.
6. Eight identical masses m are placed at the vertices of a cube of side a. A test mass M is at the cube's center. If one corner mass is removed, what is the magnitude and direction of the resulting force on M?
A.
GMm/(3a²), toward the missing corner
B.
GMm/(3a²), away from the missing corner
C.
√3GMm/a², away from the missing corner
D.
Zero
Correct Answer:
GMm/(3a²), away from the missing corner
Explanation:
With all eight equal corner masses present, cubic symmetry gives zero net force at the center. Removing one mass is equivalent to adding a force opposite to the force that the removed mass would have exerted. The center-to-corner distance is √3a, so the magnitude is GMm/(3a²), directed away from the missing corner.
7. Five equal masses m are placed at the vertices of a regular pentagon of circumradius R. A test mass M is at the center. One of the five masses is replaced by 3m. What is the net force on M?
A.
2GMm/R² toward the 3m mass
B.
GMm/R² toward the 3m mass
C.
2GMm/R² away from the 3m mass
D.
Zero
Correct Answer:
2GMm/R² toward the 3m mass
Explanation:
The five equal masses would produce zero force at the center by rotational symmetry. Replacing one m by 3m adds an excess mass 2m at that vertex. Since its distance from the center is R, the net force is G(M)(2m)/R² = 2GMm/R² toward that vertex.
8. Four masses are fixed on a straight line at positions x = -2a, -a, +a, and +2a relative to a test mass M at x = 0. The masses at ±a are each 2m, and those at ±2a are each m. What is the net gravitational force on M?
A.
Zero
B.
2GMm/a² toward positive x
C.
GMm/(2a²) toward negative x
D.
3GMm/a² toward positive x
Correct Answer:
Zero
Explanation:
Each mass has an identical mass at the opposite position, at the same distance from M. The forces from the pair at ±a cancel, and the forces from the pair at ±2a also cancel independently. Therefore, the total net force is zero.
9. Four masses are placed at the vertices of a square of side a, with masses m, 2m, m, and 2m in cyclic order. A test mass M is at the center. What is the net gravitational force?
A.
Zero
B.
2√2GMm/a² along a diagonal joining the two 2m vertices
C.
4GMm/a² toward one of the 2m vertices
D.
√2GMm/a² along a diagonal joining the two m vertices
Correct Answer:
Zero
Explanation:
The two 2m masses occupy opposite vertices, so their forces at the center cancel. The two m masses also occupy opposite vertices and cancel. Thus, despite unequal masses, the opposite-pair symmetry makes the net force zero.
10. A regular hexagon has circumradius R. Five vertices contain mass m and one vertex contains mass 4m. A test mass M is at the center. What is the magnitude and direction of the net gravitational force on M?
A.
3GMm/R² toward the 4m vertex
B.
4GMm/R² away from the 4m vertex
C.
GMm/R² toward the 4m vertex
D.
Zero
Correct Answer:
3GMm/R² toward the 4m vertex
Explanation:
If all six vertices had mass m, the net force at the center would be zero. The 4m vertex therefore represents an excess mass of 3m over the symmetric arrangement. Its force on M has magnitude 3GMm/R² and points toward that vertex.
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