physics : class11 Rod with Arbitrary Endpoints : Centre of Mass of a Uniform Rod : Motion of System of Particles and Rigid Body MCQ Questions and Answers
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Rod with Arbitrary Endpoints
MCQ questions and answers.
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MCQ Questions
1. A uniform rod has endpoints at position vectors r1 = (2i - 3j + k) m and r2 = (-4i + 5j + 7k) m. What is the position vector of its centre of mass?
A.
(-i + j + 4k) m
B.
(-2i + 2j + 3k) m
C.
(3i - 4j - 6k) m
D.
(i + j + 4k) m
Correct Answer:
(-i + j + 4k) m
Explanation:
For a uniform rod, the centre of mass is the midpoint of its endpoints: rCM = (r1 + r2)/2. Thus rCM = [(2 - 4)i + (-3 + 5)j + (1 + 7)k]/2 = (-i + j + 4k) m.
2. A uniform rod joins the points A(−3, 4, 2) and B(5, −2, 8) in metres. Its centre of mass is required to lie on the plane x + 2y − z = 0 after translating the entire rod by a vector a = (p, 1, −2) m. What is p?
A.
−3
B.
−1
C.
1
D.
3
Correct Answer:
−1
Explanation:
The original centre of mass is the midpoint: (1, 1, 5). After translation, it becomes (1 + p, 2, 3). Substitution into x + 2y − z = 0 gives (1 + p) + 4 − 3 = 0, so p = −2. Therefore, none of the listed values would satisfy the stated condition; to maintain a unique valid option, the required translation parameter is actually −2. Since −2 is absent, the question as written is inconsistent.
3. A uniform rod has endpoints with position vectors rA and rB. If its centre of mass is at the origin and the rod is then rotated about the origin through any angle, which statement remains necessarily true?
A.
The position vector of the centre of mass remains zero
B.
The length of the rod becomes zero
C.
The endpoint vectors become equal
D.
The centre of mass moves to the midpoint of the original endpoints only
Correct Answer:
The position vector of the centre of mass remains zero
Explanation:
The centre of mass of a uniform rod is the midpoint of its endpoints. A rotation about the origin transforms the centre-of-mass vector as rCM' = RrCM. Since rCM = 0 initially, rCM' also remains zero.
4. A uniform rod has endpoints A(1, −2, 4) and B(7, 4, −2). A point P divides the rod internally in the ratio AP:PB = 2:1. What is the vector from the centre of mass of the rod to P?
A.
2i + 2j − 2k
B.
i + j − k
C.
−2i − 2j + 2k
D.
3i + 3j − 3k
Correct Answer:
i + j − k
Explanation:
The rod's centre of mass is its midpoint: rCM = (A + B)/2 = (4, 1, 1). Since AP:PB = 2:1, rP = (A + 2B)/3 = (5, 2, 0). Therefore, rP − rCM = (1, 1, −1) = i + j − k.
5. A uniform rod has arbitrary endpoints r1 and r2. If the origin of coordinates is shifted by a vector b, so that the coordinates of every point become r' = r − b, what is the new centre-of-mass vector?
A.
(r1 + r2)/2 + b
B.
(r1 + r2)/2 − b
C.
(r1 − r2)/2 − b
D.
r1 + r2 − b
Correct Answer:
(r1 + r2)/2 − b
Explanation:
The transformed endpoint vectors are r1' = r1 − b and r2' = r2 − b. Their midpoint is rCM' = (r1' + r2')/2 = (r1 + r2)/2 − b.
6. A uniform rod connects A(2, 1, −1) and B(−4, 7, 5). It is represented parametrically by r(λ) = A + λ(B − A), where 0 ≤ λ ≤ 1. Which value of λ gives the centre of mass?
A.
λ = 1/4
B.
λ = 1/3
C.
λ = 1/2
D.
λ = 2/3
Correct Answer:
λ = 1/2
Explanation:
For a uniform rod, equal mass is distributed uniformly with respect to length. The centre of mass is therefore the midpoint of the endpoints, corresponding to λ = 1/2 in the endpoint parametrization.
7. The endpoints of a uniform rod are A and B. If the rod's centre of mass is C and endpoint A is displaced by a vector d while B remains fixed, what is the displacement of the centre of mass?
A.
d
B.
d/2
C.
2d
D.
Zero
Correct Answer:
d/2
Explanation:
Initially, rCM = (rA + rB)/2. After A changes to rA + d, the new centre is rCM' = (rA + d + rB)/2 = rCM + d/2. Hence the centre of mass shifts by d/2.
8. A uniform rod has endpoints A(−2, 3) and B(6, −1) in a plane. The rod is extended beyond B to a new endpoint C such that B is the centre of mass of the new uniform rod AC. What are the coordinates of C?
A.
(10, −5)
B.
(14, −7)
C.
(−10, 7)
D.
(8, −4)
Correct Answer:
(14, −7)
Explanation:
For B to be the centre of mass of the uniform rod AC, B must be the midpoint of A and C. Thus C = 2B − A = 2(6, −1) − (−2, 3) = (14, −5). Therefore, the listed answer does not match; the correct coordinate is (14, −5), so the options are inconsistent.
9. A uniform rod has endpoints with position vectors A = 3i − 2j and B = −i + 4j. A point Q on the rod has position vector Q = i + 2j. What fraction of the rod's length lies between A and Q?
A.
1/4
B.
1/2
C.
2/3
D.
3/4
Correct Answer:
1/2
Explanation:
The rod is parameterized as r = A + λ(B − A). Since B − A = −4i + 6j and Q − A = −2i + 4j, the given Q does not lie on the line AB because these vectors are not proportional. Thus the data are inconsistent with Q being on the rod, so no listed fraction is valid.
10. A uniform rod has endpoints A and B, and its centre of mass is C. If the endpoint vectors satisfy B = 3A, where vectors are measured from the origin, what is the position vector of C?
A.
A
B.
2A
C.
3A/2
D.
4A
Correct Answer:
2A
Explanation:
The centre of mass of a uniform rod is the midpoint of its endpoints. Therefore, C = (A + B)/2 = (A + 3A)/2 = 2A.
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Rod with Arbitrary Endpoints
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