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If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]
- a)2.5 m
- b)1 m
- c)1.5 m
- d)2 m
Correct answer is option 'D'. Can you explain this answer?
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If the linear density (mass per unit length) of a rod of length 3m is ...
Given, linear density of the rod is proportional to x, where x is the distance from one end of the rod.
Let the linear density of the rod at distance x from one end be ρ(x).
We know that the mass of an element of length dx at a distance x from one end is given by dm = ρ(x) dx.
The total mass of the rod is given by integrating the above expression from 0 to 3m:
M = ∫₀³ ρ(x) dx
We also know that the position of the centre of gravity of the rod is given by:
x_bar = (1/M) ∫₀³ x ρ(x) dx
Substituting dm = ρ(x) dx in the above expression, we get:
x_bar = (1/M) ∫₀³ x dm
x_bar = (1/M) ∫₀³ x ρ(x) dx
x_bar = (1/M) ∫₀³ x (kx) dx (since ρ(x) is proportional to x)
x_bar = (k/M) ∫₀³ x² dx
x_bar = (k/M) [x³/3]₀³
x_bar = (k/M) [3³/3]
x_bar = 3k/M
But we know that M = ∫₀³ ρ(x) dx
M = k ∫₀³ x dx
M = k [x²/2]₀³
M = k (9/2)
Therefore, x_bar = 3k/M = 3k/(k(9/2)) = 2/3 * 3 = 2m
Hence, the distance of the centre of gravity of the rod from one end is 2m, which is option D.
Let the linear density of the rod at distance x from one end be ρ(x).
We know that the mass of an element of length dx at a distance x from one end is given by dm = ρ(x) dx.
The total mass of the rod is given by integrating the above expression from 0 to 3m:
M = ∫₀³ ρ(x) dx
We also know that the position of the centre of gravity of the rod is given by:
x_bar = (1/M) ∫₀³ x ρ(x) dx
Substituting dm = ρ(x) dx in the above expression, we get:
x_bar = (1/M) ∫₀³ x dm
x_bar = (1/M) ∫₀³ x ρ(x) dx
x_bar = (1/M) ∫₀³ x (kx) dx (since ρ(x) is proportional to x)
x_bar = (k/M) ∫₀³ x² dx
x_bar = (k/M) [x³/3]₀³
x_bar = (k/M) [3³/3]
x_bar = 3k/M
But we know that M = ∫₀³ ρ(x) dx
M = k ∫₀³ x dx
M = k [x²/2]₀³
M = k (9/2)
Therefore, x_bar = 3k/M = 3k/(k(9/2)) = 2/3 * 3 = 2m
Hence, the distance of the centre of gravity of the rod from one end is 2m, which is option D.
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If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer?
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If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer?.
If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is[2002]a)2.5 mb)1 mc)1.5 md)2 mCorrect answer is option 'D'. Can you explain this answer?.
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