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A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string is
- a)17.88 m/s
- b)16 m/s
- c)20 m/s
- d)19.4 m/s
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A stone of mass 5 kg is attached to a string of 10 m length and is whi...
maximum tension = (mv^2)/ r,
60 = 5*v^2/10
v = 17.88
v = 17.88
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A stone of mass 5 kg is attached to a string of 10 m length and is whi...
Given, mass of stone, m = 5 kg
Length of string, l = 10 m
Maximum tension, T = 160 N
We need to find the maximum velocity of revolution that can be given to the stone without breaking the string.
Maximum velocity of revolution:
We know that the tension in the string, T is given by:
T = (mv²) / r
where m is the mass of the stone, v is the velocity of the stone and r is the radius of the circle.
Finding the radius:
The length of the string, l is equal to the circumference of the circle, so:
l = 2πr
=> r = l / (2π)
=> r = 10 / (2π) = 1.59 m
Finding the maximum velocity:
Substituting the values in the tension equation, we get:
T = (mv²) / r
=> v² = (Tr) / m
=> v = √[(Tr) / m]
Now, substituting the given values, we get:
v = √[(160 × 1.59) / 5]
=> v = 17.88 m/s
Therefore, the maximum velocity of revolution that can be given to the stone without breaking the string is 17.88 m/s.
Length of string, l = 10 m
Maximum tension, T = 160 N
We need to find the maximum velocity of revolution that can be given to the stone without breaking the string.
Maximum velocity of revolution:
We know that the tension in the string, T is given by:
T = (mv²) / r
where m is the mass of the stone, v is the velocity of the stone and r is the radius of the circle.
Finding the radius:
The length of the string, l is equal to the circumference of the circle, so:
l = 2πr
=> r = l / (2π)
=> r = 10 / (2π) = 1.59 m
Finding the maximum velocity:
Substituting the values in the tension equation, we get:
T = (mv²) / r
=> v² = (Tr) / m
=> v = √[(Tr) / m]
Now, substituting the given values, we get:
v = √[(160 × 1.59) / 5]
=> v = 17.88 m/s
Therefore, the maximum velocity of revolution that can be given to the stone without breaking the string is 17.88 m/s.
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A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string isa)17.88 m/sb)16 m/sc)20 m/sd)19.4 m/sCorrect answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string isa)17.88 m/sb)16 m/sc)20 m/sd)19.4 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string isa)17.88 m/sb)16 m/sc)20 m/sd)19.4 m/sCorrect answer is option 'A'. Can you explain this answer?.
A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string isa)17.88 m/sb)16 m/sc)20 m/sd)19.4 m/sCorrect answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string isa)17.88 m/sb)16 m/sc)20 m/sd)19.4 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string isa)17.88 m/sb)16 m/sc)20 m/sd)19.4 m/sCorrect answer is option 'A'. Can you explain this answer?.
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