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Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation is
- a)2ma2
- b)4ma2
- c)2m(a2 + b2)
- d)2m(a2 – b2)
Correct answer is option 'A'. Can you explain this answer?
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Four small objects each of mass m are fixed at the corners of a rectan...
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Four small objects each of mass m are fixed at the corners of a rectan...
Let's label the four corners of the rectangular wire-frame as A, B, C, and D.
To find the moment of inertia of the system, we need to consider the contribution from each individual object and then sum them up.
The moment of inertia of each object around its center of mass is given by the formula:
I = (1/12)mL^2
where I is the moment of inertia, m is the mass, and L is the length of the object.
In this case, the length of each object is a/2 and b/2. So, the moment of inertia of each object is:
I = (1/12)m(a/2)^2 + (1/12)m(b/2)^2
Simplifying this equation, we get:
I = (1/48)m(a^2 + b^2)
Now, we need to consider the distance of each object from the axis of rotation. Let's assume the axis of rotation passes through the center of the rectangle.
The distance of each object from the center of the rectangle is a/2 and b/2. So, the total moment of inertia of the system is:
I_total = 4 * (m * (a/2)^2 + m * (b/2)^2)
Simplifying this equation, we get:
I_total = (1/12)m(a^2 + b^2)
Therefore, the moment of inertia of the system is equal to (1/12)m(a^2 + b^2).
To find the moment of inertia of the system, we need to consider the contribution from each individual object and then sum them up.
The moment of inertia of each object around its center of mass is given by the formula:
I = (1/12)mL^2
where I is the moment of inertia, m is the mass, and L is the length of the object.
In this case, the length of each object is a/2 and b/2. So, the moment of inertia of each object is:
I = (1/12)m(a/2)^2 + (1/12)m(b/2)^2
Simplifying this equation, we get:
I = (1/48)m(a^2 + b^2)
Now, we need to consider the distance of each object from the axis of rotation. Let's assume the axis of rotation passes through the center of the rectangle.
The distance of each object from the center of the rectangle is a/2 and b/2. So, the total moment of inertia of the system is:
I_total = 4 * (m * (a/2)^2 + m * (b/2)^2)
Simplifying this equation, we get:
I_total = (1/12)m(a^2 + b^2)
Therefore, the moment of inertia of the system is equal to (1/12)m(a^2 + b^2).
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Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer?.
Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer?.
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Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation isa)2ma2b)4ma2c)2m(a2 + b2)d)2m(a2 – b2)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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