physics : class11 Mass Element of a Rod : Centre of Mass of a Uniform Rod : Motion of System of Particles and Rigid Body MCQ Questions and Answers
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Mass Element of a Rod
MCQ questions and answers.
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MCQ Questions
1. A uniform rod of length L and total mass M lies along the x-axis from x = 0 to x = L. The mass contained in an infinitesimal element between x and x + dx is:
A.
(M/L) dx
B.
(M/x) dx
C.
(M/L^2) dx
D.
M dx
Correct Answer:
(M/L) dx
Explanation:
For a uniform rod, the linear mass density is constant and equals lambda = M/L. Hence the mass element is dm = lambda dx = (M/L) dx.
2. A uniform rod has linear mass density lambda and extends from x = -L/2 to x = L/2. Which expression gives the mass of the element located between x and x + dx?
A.
lambda x dx
B.
lambda dx
C.
lambda L dx
D.
(lambda/x) dx
Correct Answer:
lambda dx
Explanation:
Since the rod is uniform, its linear mass density is independent of position. Therefore, an element of length dx has mass dm = lambda dx.
3. A uniform rod of mass M and length L is described using the coordinate x measured from its left end. If a small segment has length 2 dx, its mass, to first order in dx, is:
A.
(M/L) dx
B.
(2M/L) dx
C.
(M/L^2) 2 dx
D.
2ML dx
Correct Answer:
(2M/L) dx
Explanation:
The mass per unit length is M/L. A segment of length 2 dx therefore has mass dm = (M/L)(2 dx) = (2M/L) dx.
4. A uniform rod of length 4a and mass M extends from x = -a to x = 3a. The mass element corresponding to a coordinate interval dx is:
A.
(M/a) dx
B.
(M/4a) dx
C.
(4Ma) dx
D.
(M/16a^2) dx
Correct Answer:
(M/4a) dx
Explanation:
The rod's total length is 3a - (-a) = 4a. Its uniform linear density is therefore lambda = M/(4a), so dm = (M/4a) dx.
5. For a uniform rod of total mass M and length L, an element is specified by its distance r from the centre, where the element extends from r to r + dr on one side of the centre. Its mass is:
A.
(M/L) dr
B.
(2M/L) dr
C.
(M/r) dr
D.
(M/L^2) dr
Correct Answer:
(M/L) dr
Explanation:
Distance from the centre is only a coordinate choice. The element still has physical length dr, and the uniform linear density is M/L. Thus dm = (M/L) dr.
6. A uniform rod of mass M and length L is divided into infinitesimal elements using the angular variable theta defined by x = L sin(theta), where theta lies between -pi/6 and pi/6. What is the mass element in terms of dtheta?
A.
(M/L) cos(theta) dtheta
B.
M cos(theta) dtheta
C.
M sin(theta) dtheta
D.
(M/L) sin(theta) dtheta
Correct Answer:
M cos(theta) dtheta
Explanation:
For a uniform rod, dm = (M/L) dx. Since x = L sin(theta), dx = L cos(theta) dtheta. Therefore dm = (M/L)(L cos(theta)dtheta) = M cos(theta)dtheta.
7. A uniform rod of mass M and length L is parametrized by x = L u^2, with u ranging from 0 to 1. Which expression represents the mass element associated with du?
A.
(M/L) du
B.
2Mu du
C.
(2M/L)u du
D.
MLu^2 du
Correct Answer:
2Mu du
Explanation:
The uniform-rod mass element is dm = (M/L)dx. With x = Lu^2, dx = 2Lu du. Hence dm = (M/L)(2Lu du) = 2Mu du.
8. A uniform rod has mass density per unit length lambda. Its position is described by x = a t^3, where t is a parameter. What is the mass element corresponding to an increment dt?
A.
lambda a dt
B.
3lambda a t^2 dt
C.
lambda t^3 dt
D.
(lambda/a) t^2 dt
Correct Answer:
3lambda a t^2 dt
Explanation:
The mass element is dm = lambda dx. Differentiating x = a t^3 gives dx = 3a t^2 dt, so dm = 3lambda a t^2 dt.
9. A uniform rod extends from x = 0 to x = L. It is divided into elements using y = sqrt(x/L). Which differential mass expression is correct?
A.
dm = (M/L) dy
B.
dm = 2My dy
C.
dm = (M/2y) dy
D.
dm = 2MLy dy
Correct Answer:
dm = 2My dy
Explanation:
From y = sqrt(x/L), x = Ly^2 and dx = 2Ly dy. Since dm = (M/L)dx for a uniform rod, dm = (M/L)(2Ly dy) = 2My dy.
10. A uniform rod of length L and mass M is described by the coordinate q = 2x/L - 1, so that q ranges from -1 to 1. What is the mass element in terms of dq?
A.
(M/L) dq
B.
(M/2) dq
C.
2M dq
D.
(2M/L) dq
Correct Answer:
(M/2) dq
Explanation:
Solving for x gives x = L(q + 1)/2, so dx = (L/2)dq. Using dm = (M/L)dx yields dm = (M/L)(L/2)dq = (M/2)dq.
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