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Consider two unequal masses m₂ > m₁ connected by a string which passes over a frictionless and massless pulley. The tension T in string is
- a)T = 2m₁+m₂/m₁+m₂ g
- b)T = 2m₁ m₂/m₁+m₂ g
- c)T = (m₁-m₂)/m₁+m₂ g
- d)T = 2(m₁+m₂) g
Correct answer is option 'B'. Can you explain this answer?
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To analyze the motion of two unequal masses, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's assume that the two masses are denoted by m1 and m2, and m1 is greater than m2. Let's also assume that there is a force acting on both masses, denoted by F. The acceleration of each mass can be calculated using Newton's second law:
For mass m1:
F = m1 * a1
For mass m2:
F = m2 * a2
Since the force acting on both masses is the same, we can equate the two equations:
m1 * a1 = m2 * a2
We can rearrange this equation to solve for the acceleration of m2 in terms of m1 and a1:
a2 = (m1/m2) * a1
This equation shows that the acceleration of m2 is directly proportional to the acceleration of m1, with the ratio of their masses as the proportionality constant. This means that if m1 accelerates twice as much as m2, m2 will accelerate half as much as m1.
In summary, when two unequal masses are subjected to the same force, the acceleration of the smaller mass will be greater than the acceleration of the larger mass, and the ratio of their accelerations will be inversely proportional to the ratio of their masses.
Let's assume that the two masses are denoted by m1 and m2, and m1 is greater than m2. Let's also assume that there is a force acting on both masses, denoted by F. The acceleration of each mass can be calculated using Newton's second law:
For mass m1:
F = m1 * a1
For mass m2:
F = m2 * a2
Since the force acting on both masses is the same, we can equate the two equations:
m1 * a1 = m2 * a2
We can rearrange this equation to solve for the acceleration of m2 in terms of m1 and a1:
a2 = (m1/m2) * a1
This equation shows that the acceleration of m2 is directly proportional to the acceleration of m1, with the ratio of their masses as the proportionality constant. This means that if m1 accelerates twice as much as m2, m2 will accelerate half as much as m1.
In summary, when two unequal masses are subjected to the same force, the acceleration of the smaller mass will be greater than the acceleration of the larger mass, and the ratio of their accelerations will be inversely proportional to the ratio of their masses.
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Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. Can you explain this answer?
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Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. 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 Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. Can you explain this answer?.
Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. 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 Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider two unequal masses m > m connected by a string which passes over a frictionless and massless pulley. The tension T in string isa)T = 2m+m/m+m gb)T = 2m m/m+m gc)T = (m-m)/m+m gd)T = 2(m+m) gCorrect answer is option 'B'. Can you explain this answer?.
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