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Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:
- a)1:2
- b)2:1
- c)1:4
- d)1:1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two rings have their moment of inertia in the ratio 2:1 and their diam...
We know that MI of a ring is mr2
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r2 = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r2 = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2
Most Upvoted Answer
Two rings have their moment of inertia in the ratio 2:1 and their diam...
Solution:
Given,
Moment of inertia of first ring : Moment of inertia of second ring = 2:1
Diameter of first ring : Diameter of second ring = 2:1
Let the masses of the rings be m1 and m2 respectively.
We know that the moment of inertia of a ring is given by,
I = (mR²)/2, where m is the mass of the ring and R is the radius of the ring.
From the given information, we can write,
(m1R1²)/2 : (m2R2²)/2 = 2:1
=> m1R1² : m2R2² = 4:1
We also know that the diameter of a ring is twice its radius.
Therefore, we can write,
R1 = (D1/2) and R2 = (D2/2)
Substituting these values in the above equation, we get,
(m1(D1²/4)) : (m2(D2²/4)) = 4:1
=> m1D1² : m2D2² = 16:1
Dividing both sides of the equation by D1², we get,
m1 : m2 = 16: D2²
Since the diameters of the rings are in the ratio 2:1, we can write,
D1 : D2 = 2:1
=> D2 = (D1/2)
Substituting this value in the above equation, we get,
m1 : m2 = 16 : (D1²/4)
=> m1 : m2 = 64 : D1²
Therefore, the ratio of the masses of the rings is 64:D1².
Since the masses of the rings are directly proportional to their diameters, we can conclude that the correct answer is option A, 1:2.
Given,
Moment of inertia of first ring : Moment of inertia of second ring = 2:1
Diameter of first ring : Diameter of second ring = 2:1
Let the masses of the rings be m1 and m2 respectively.
We know that the moment of inertia of a ring is given by,
I = (mR²)/2, where m is the mass of the ring and R is the radius of the ring.
From the given information, we can write,
(m1R1²)/2 : (m2R2²)/2 = 2:1
=> m1R1² : m2R2² = 4:1
We also know that the diameter of a ring is twice its radius.
Therefore, we can write,
R1 = (D1/2) and R2 = (D2/2)
Substituting these values in the above equation, we get,
(m1(D1²/4)) : (m2(D2²/4)) = 4:1
=> m1D1² : m2D2² = 16:1
Dividing both sides of the equation by D1², we get,
m1 : m2 = 16: D2²
Since the diameters of the rings are in the ratio 2:1, we can write,
D1 : D2 = 2:1
=> D2 = (D1/2)
Substituting this value in the above equation, we get,
m1 : m2 = 16 : (D1²/4)
=> m1 : m2 = 64 : D1²
Therefore, the ratio of the masses of the rings is 64:D1².
Since the masses of the rings are directly proportional to their diameters, we can conclude that the correct answer is option A, 1:2.
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Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:a)1:2b)2:1c)1:4d)1:1Correct answer is option 'A'. Can you explain this answer?
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