Net gravitational force at the center of a square is found to be F1 when four particles having masses M,2M,3M and 4M are placed at the four corners of the square as shown in figure, and it is F2 when the positions of 3M and 4M are interchanged.
The ratio F2F1=5α.
The value of α is
A
1
B
3
C
25
D
2
Answer
Correct answer:D
Step-by-step solution
Standard Method
Given: Four particles of masses M,2M,3M and 4M are placed at the corners of a square. The net gravitational force at the center is F1 in the first arrangement and F2 after interchanging 3M and 4M.
Find: The value of α if
F2F1=5α
Each corner of the square is at the same distance from the center, so the gravitational force due to each mass at the center is proportional to that mass.
Let the side of the square be a. Distance from center to each corner is
r=2a
Resolve the forces along two perpendicular axes.
Case I — Original configuration:
Horizontal component is proportional to
(3M−M)
Vertical component is proportional to
(4M−2M)
Therefore,
F1∝(2M)2+(2M)2=2M2
Case II — After interchanging 3M and 4M:
Horizontal component is proportional to
(4M−M)
Vertical component is proportional to
(3M−2M)
Therefore,
F2∝(3M)2+(M)2=M10
Now,
F2F1=M102M2=52
Comparing with
F2F1=5α
we get
α=2
Therefore, the correct option is D and the value of α is 2.
Vector Component View
Given: All four masses are at equal distance from the center of the square.
Find: Compare the magnitudes of the resultant gravitational forces in the two arrangements.
Because all corner distances from the center are equal, the common factor in gravitational force,
r2Gm
is the same for every corner contribution. Hence only the mass values matter in forming the resultant.
For the first arrangement, the imbalance along one axis is 2M and along the perpendicular axis is also 2M. So the resultant is proportional to
(2M)2+(2M)2
For the second arrangement, the imbalances become 3M and M. So the resultant is proportional to
(3M)2+(M)2
Thus,
F2F1=10M28M2=108=54=52
Therefore,
α=2
So the correct option is D.
Common mistakes
Taking the forces due to opposite corners as simple scalars instead of vectors is incorrect because gravitational forces at the center act along diagonals. Resolve into perpendicular components first, then use vector addition.
Assuming the force depends only on total mass is wrong because the arrangement of masses changes the component balance. Use the directional distribution of masses, not merely their sum.
Using unequal distances for different corner masses is incorrect because every corner of the square is equally distant from the center. Keep the common distance factor same for all four masses.
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