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Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is
- a)2ml 2
- b)√3ml2
- c)3ml 2
- d)ml 2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Four point masses, each of value m, are placed at the corners of a squ...
Inn' = M.I due to the point mass at B + M.I due to the point mass at D + M.I due to the point mass at C.
= ml2 + 2ml2=3ml2
Most Upvoted Answer
Four point masses, each of value m, are placed at the corners of a squ...
To find the moment of inertia of the system about an axis passing through A and parallel to BD, we need to consider the contributions of each point mass.
Let's label the point masses as M1, M2, M3, and M4, with M1 at corner A, M2 at corner B, M3 at corner C, and M4 at corner D.
The moment of inertia of a single point mass about an axis passing through A and parallel to BD can be found using the formula:
I = m * r^2
Where m is the mass of the point mass and r is the distance from the point mass to the axis of rotation (in this case, the distance from A to the point mass).
For M1 at corner A, the distance from A to M1 is 0, so its contribution to the moment of inertia is:
I1 = m * (0)^2 = 0
For M2 at corner B, the distance from A to M2 is l, so its contribution to the moment of inertia is:
I2 = m * (l)^2 = ml^2
For M3 at corner C, the distance from A to M3 is √(l^2 + l^2) = √2l, so its contribution to the moment of inertia is:
I3 = m * (√2l)^2 = 2ml
For M4 at corner D, the distance from A to M4 is l, so its contribution to the moment of inertia is:
I4 = m * (l)^2 = ml^2
Now, we can find the total moment of inertia of the system by summing up the contributions from each point mass:
I_total = I1 + I2 + I3 + I4 = 0 + ml^2 + 2ml + ml^2 = 2ml + 2ml^2
Therefore, the moment of inertia of the system about an axis passing through A and parallel to BD is 2ml + 2ml^2.
Answer: b) 2ml + 2ml^2
Let's label the point masses as M1, M2, M3, and M4, with M1 at corner A, M2 at corner B, M3 at corner C, and M4 at corner D.
The moment of inertia of a single point mass about an axis passing through A and parallel to BD can be found using the formula:
I = m * r^2
Where m is the mass of the point mass and r is the distance from the point mass to the axis of rotation (in this case, the distance from A to the point mass).
For M1 at corner A, the distance from A to M1 is 0, so its contribution to the moment of inertia is:
I1 = m * (0)^2 = 0
For M2 at corner B, the distance from A to M2 is l, so its contribution to the moment of inertia is:
I2 = m * (l)^2 = ml^2
For M3 at corner C, the distance from A to M3 is √(l^2 + l^2) = √2l, so its contribution to the moment of inertia is:
I3 = m * (√2l)^2 = 2ml
For M4 at corner D, the distance from A to M4 is l, so its contribution to the moment of inertia is:
I4 = m * (l)^2 = ml^2
Now, we can find the total moment of inertia of the system by summing up the contributions from each point mass:
I_total = I1 + I2 + I3 + I4 = 0 + ml^2 + 2ml + ml^2 = 2ml + 2ml^2
Therefore, the moment of inertia of the system about an axis passing through A and parallel to BD is 2ml + 2ml^2.
Answer: b) 2ml + 2ml^2
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Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer?
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Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer?.
Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD isa)2ml 2b)√3ml2c)3ml 2d)ml 2Correct answer is option 'C'. Can you explain this answer?.
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