physics : class11 Zero-Force Points : Gravitational Force : Gravitation MCQ Questions and Answers
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Zero-Force Points
MCQ questions and answers.
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MCQ Questions
1. Two point masses 4M and M are separated by a distance d. At what distance from the mass 4M is the gravitational force on a small test mass zero?
A.
d/3
B.
d/2
C.
2d/3
D.
3d/4
Correct Answer:
2d/3
Explanation:
The zero-force point lies between the masses. If its distance from 4M is x, its distance from M is d − x. Equating the magnitudes of the two attractions, 4M/x² = M/(d − x)², giving x = 2d/3.
2. Two masses M and 9M are fixed at a separation d. The zero-force point on the line joining them is at what distance from the mass 9M?
A.
d/4
B.
d/2
C.
3d/4
D.
d/9
Correct Answer:
3d/4
Explanation:
The zero-force point lies between the masses. If its distance from M is x, then its distance from 9M is d − x. From M/x² = 9M/(d − x)², we obtain d − x = 3x, so the distance from 9M is 3d/4.
3. Two positive point masses are separated by a finite distance. Which statement about points where the gravitational force on a test mass is zero is correct?
A.
There is always one zero-force point outside the heavier mass.
B.
There is always one zero-force point outside the lighter mass.
C.
There is exactly one zero-force point between the masses and none outside them.
D.
There are two zero-force points, one on each outer side.
Correct Answer:
There is exactly one zero-force point between the masses and none outside them.
Explanation:
Between the masses, the gravitational forces act in opposite directions and can balance at one location. Outside the two masses, both gravitational forces point in the same direction, so cancellation is impossible.
4. Masses 16m and m are separated by a distance L. A test particle is placed between them. How far from the mass 16m must it be placed for the net gravitational force to vanish?
A.
L/5
B.
L/4
C.
3L/4
D.
4L/5
Correct Answer:
4L/5
Explanation:
Let the distance from 16m be x; the distance from m is L − x. Balancing the attractions gives 16m/x² = m/(L − x)². Thus 4/x = 1/(L − x), leading to x = 4L/5.
5. The Earth and Moon have masses in the ratio 81:1 and their centers are separated by distance D. Neglecting all other bodies, where is the zero-force point located from Earth's center?
A.
D/81
B.
D/9
C.
D/10
D.
9D/10
Correct Answer:
D/10
Explanation:
The point lies between Earth and Moon. If its distance from Earth is x, its distance from the Moon is D − x. Equating the fields, 81/x² = 1/(D − x)², so 9(D − x) = x and x = D/10.
6. Two equal masses are fixed at x = −a and x = +a on the x-axis. A test particle is placed at the midpoint. If it is displaced slightly in the positive x-direction, what is the direction of the resulting gravitational force?
A.
Positive x-direction, toward the nearer mass
B.
Negative x-direction, toward the nearer mass
C.
Positive x-direction, away from both masses
D.
The force remains zero for every small displacement
Correct Answer:
Positive x-direction, toward the nearer mass
Explanation:
At the midpoint the two forces cancel. After a positive displacement, the mass at x = +a becomes nearer, so its attraction is stronger than that of the mass at x = −a. The net force is therefore in the positive x-direction, making the zero-force point unstable.
7. Two positive masses are fixed at the ends of a line segment. Can the gravitational force on a test mass vanish at a point outside the segment on the same line?
A.
Yes, only on the side of the larger mass
B.
Yes, only on the side of the smaller mass
C.
Yes, on both outer sides
D.
No, because both gravitational forces have the same direction there
Correct Answer:
No, because both gravitational forces have the same direction there
Explanation:
At any point outside the two masses, both masses attract the test particle toward the same general side. Their force vectors are parallel and point in the same direction, so they cannot cancel.
8. Masses m and 4m are located at x = 0 and x = d, respectively. On the x-axis, where is the zero-force point located?
A.
x = d/4
B.
x = d/3
C.
x = d/2
D.
x = 2d/3
Correct Answer:
x = d/3
Explanation:
The zero-force point lies between the masses. At x, the opposing attractions satisfy m/x² = 4m/(d − x)². Therefore d − x = 2x, giving x = d/3.
9. Point masses M and 4M are placed at coordinates (0, 0) and (a, 0), respectively. What are the coordinates of the unique zero-force point in the plane?
A.
(a/3, 0)
B.
(a/2, 0)
C.
(0, a/3)
D.
(a/3, a/3)
Correct Answer:
(a/3, 0)
Explanation:
For the two gravitational force vectors to cancel, they must be equal and opposite, so the test point must lie on the line joining the masses. Along that line, balancing M/x² with 4M/(a − x)² gives x = a/3. Hence the point is (a/3, 0).
10. Two point masses M1 and M2 are separated by distance d, with M2 > M1. The zero-force point lies between them. Its distance from M1 is given by which expression?
A.
d√M2/(√M1 + √M2)
B.
d√M1/(√M1 + √M2)
C.
dM1/(M1 + M2)
D.
dM2/(M1 + M2)
Correct Answer:
d√M1/(√M1 + √M2)
Explanation:
Let x be the distance from M1; then the distance from M2 is d − x. Equating the attractive forces gives M1/x² = M2/(d − x)², so (d − x)/x = √(M2/M1). Solving yields x = d√M1/(√M1 + √M2).
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