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A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s)
- a)9
- b)18
- c)27
- d)36
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A ball of mass 0.5 kg is attached to the end of a string having length...
T sin θ = mrω2; r = l sin θ
Most Upvoted Answer
A ball of mass 0.5 kg is attached to the end of a string having length...
T = mw²r
324 = 0.5 ×0.5×w²
w² = 324÷0.25
w = √1296
w = 36 rad/s
324 = 0.5 ×0.5×w²
w² = 324÷0.25
w = √1296
w = 36 rad/s
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A ball of mass 0.5 kg is attached to the end of a string having length...
Given:
- Mass of ball, m = 0.5 kg
- Length of string, l = 0.5 m
- Maximum tension in string, T = 324 N
To find:
- Maximum angular velocity, ω in rad/s
Approach:
- We can use the equation for centripetal force, Fc = mv^2/r, where m is mass of the ball, v is velocity of the ball, and r is radius of circular path.
- The tension in the string provides the centripetal force required to keep the ball in circular motion.
- At maximum tension, T = Fc, so we can equate the two equations and solve for v.
- Then we can use the formula for angular velocity, ω = v/r, where r is the length of the string.
Solution:
1. Equating T and Fc:
- T = mv^2/r
- Substituting the given values: 324 = 0.5 * v^2 / 0.5
- Solving for v: v = sqrt(324*2) = 36 m/s
2. Finding angular velocity:
- ω = v/r
- Substituting the given value of r: ω = 36/0.5 = 72 rad/s
3. Checking if the answer is within bounds:
- We know that the tension in the string cannot exceed the maximum value, T = 324 N.
- The centripetal force required to keep the ball in circular motion is Fc = mv^2/r.
- At maximum tension, T = Fc, so we can set the two equations equal to each other and solve for v.
- If the calculated value of v exceeds the speed of light or any other physically impossible value, then the answer is incorrect.
- In this case, the value of v = 36 m/s is within reasonable bounds, so the answer is correct.
Therefore, the correct answer is option 'D', 36 rad/s.
- Mass of ball, m = 0.5 kg
- Length of string, l = 0.5 m
- Maximum tension in string, T = 324 N
To find:
- Maximum angular velocity, ω in rad/s
Approach:
- We can use the equation for centripetal force, Fc = mv^2/r, where m is mass of the ball, v is velocity of the ball, and r is radius of circular path.
- The tension in the string provides the centripetal force required to keep the ball in circular motion.
- At maximum tension, T = Fc, so we can equate the two equations and solve for v.
- Then we can use the formula for angular velocity, ω = v/r, where r is the length of the string.
Solution:
1. Equating T and Fc:
- T = mv^2/r
- Substituting the given values: 324 = 0.5 * v^2 / 0.5
- Solving for v: v = sqrt(324*2) = 36 m/s
2. Finding angular velocity:
- ω = v/r
- Substituting the given value of r: ω = 36/0.5 = 72 rad/s
3. Checking if the answer is within bounds:
- We know that the tension in the string cannot exceed the maximum value, T = 324 N.
- The centripetal force required to keep the ball in circular motion is Fc = mv^2/r.
- At maximum tension, T = Fc, so we can set the two equations equal to each other and solve for v.
- If the calculated value of v exceeds the speed of light or any other physically impossible value, then the answer is incorrect.
- In this case, the value of v = 36 m/s is within reasonable bounds, so the answer is correct.
Therefore, the correct answer is option 'D', 36 rad/s.
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A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s) a)9 b)18 c)27 d)36Correct answer is option 'D'. Can you explain this answer?
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A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s) a)9 b)18 c)27 d)36Correct answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s) a)9 b)18 c)27 d)36Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s) a)9 b)18 c)27 d)36Correct answer is option 'D'. Can you explain this answer?.
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