physics : class11 Integration for Continuous Mass : Centre of Mass of a Uniform Rod : Motion of System of Particles and Rigid Body MCQ Questions and Answers
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Integration for Continuous Mass
MCQ questions and answers.
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MCQ Questions
1. A uniform rod of length L and mass M lies along the x-axis from x = 0 to x = L. What is its linear mass density?
A.
M/L
B.
ML
C.
L/M
D.
M + L
Correct Answer:
M/L
Explanation:
For a uniform rod, the linear mass density is constant and is given by λ = dm/dx = M/L.
2. For a uniform rod extending from x = 0 to x = L, which expression correctly represents the mass element dm?
A.
dm = λ dx
B.
dm = x dλ
C.
dm = λ/x
D.
dm = dx/λ
Correct Answer:
dm = λ dx
Explanation:
For continuous mass distribution along a rod, a small element of mass is dm = λ dx, where λ is the constant linear mass density.
3. A uniform rod of length L extends from x = 0 to x = L. Using integration, its centre of mass is given by which expression?
A.
∫₀ᴸ x dx
B.
(1/L)∫₀ᴸ x dx
C.
L∫₀ᴸ x dx
D.
∫₀ᴸ dx/x
Correct Answer:
(1/L)∫₀ᴸ x dx
Explanation:
Since λ is constant, x_cm = (1/M)∫ x dm = (1/λL)∫₀ᴸ xλ dx = (1/L)∫₀ᴸ x dx.
4. Evaluate the integral ∫₀ᴸ x dx, which appears in calculating the centre of mass of a uniform rod.
A.
L
B.
L²
C.
L²/2
D.
2L²
Correct Answer:
L²/2
Explanation:
The integration gives ∫₀ᴸ x dx = [x²/2]₀ᴸ = L²/2.
5. A uniform rod of mass M and length L lies from x = 0 to x = L. What is the value of ∫ x dm over the rod?
A.
M/L
B.
ML/2
C.
ML
D.
ML²/2
Correct Answer:
ML/2
Explanation:
Using dm = (M/L)dx, ∫ x dm = (M/L)∫₀ᴸ x dx = (M/L)(L²/2) = ML/2.
6. A uniform rod extends from x = 0 to x = L. Which integral gives its total mass?
A.
∫₀ᴸ xλ dx
B.
∫₀ᴸ λ dx
C.
∫₀ᴸ dx/λ
D.
∫₀ᴸ λx² dx
Correct Answer:
∫₀ᴸ λ dx
Explanation:
The total mass is obtained by integrating the mass elements: M = ∫ dm = ∫₀ᴸ λ dx.
7. A uniform rod lies along the x-axis from x = −L/2 to x = L/2. Which integral correctly gives its centre of mass coordinate?
A.
(1/M)∫₋L/2ᴸ/2 λ dx
B.
(1/M)∫₋L/2ᴸ/2 xλ dx
C.
(1/M)∫₋L/2ᴸ/2 x²λ dx
D.
M∫₋L/2ᴸ/2 xλ dx
Correct Answer:
(1/M)∫₋L/2ᴸ/2 xλ dx
Explanation:
For a continuous mass distribution, x_cm = (1/M)∫ x dm. Since dm = λ dx, this becomes (1/M)∫ xλ dx.
8. For a uniform rod extending from x = −L/2 to x = L/2, what result is obtained for the integral ∫ xλ dx?
A.
0
B.
λL
C.
λL²/2
D.
λL²
Correct Answer:
0
Explanation:
The integrand xλ is an odd function and the limits are symmetric about zero, so ∫₋L/2ᴸ/2 xλ dx = 0.
9. A uniform rod extends from x = a to x = b, with constant linear density λ. Which expression represents its centre of mass?
A.
(1/(b−a))∫ₐᵇ x dx
B.
(1/(b+a))∫ₐᵇ x dx
C.
(b−a)∫ₐᵇ x dx
D.
∫ₐᵇ dx/x
Correct Answer:
(1/(b−a))∫ₐᵇ x dx
Explanation:
The total mass is λ(b−a), so x_cm = [∫ₐᵇ xλ dx]/[λ(b−a)] = (1/(b−a))∫ₐᵇ x dx.
10. A uniform rod occupies the interval x = 2 cm to x = 6 cm. Using integration for its continuous mass distribution, what is its centre of mass coordinate?
A.
2 cm
B.
3 cm
C.
4 cm
D.
6 cm
Correct Answer:
4 cm
Explanation:
For a uniform rod, x_cm = [1/(6−2)]∫₂⁶ x dx = (1/4)[x²/2]₂⁶ = 4 cm.