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CBSE Class 11-science Answered

one meter long string can bear maximum of 0.5 kg mass . a mass of 0.05 kg is tied to one end and rotated in a horizontal circle , calculate the maximum number of revolution so that string does not brakes
Asked by pushpakumari291279 | 20 Aug, 2022, 08:15: AM
answered-by-expert Expert Answer
Maximum Tension force Tmax that is bearable by string = M g
 
where M is mass hanging to give tension force in the string and g is acceleration due to gravity
 
Tension force F experienced by string, when it is attached to mass m and is revolved horizontally is given as
 
F = m ω2 L
 
where ω is angular speed of revolution and L is length of string
 
For maximum revolution speed , we have ,  F = Tmax
 
m ω2 L = M g
 
begin mathsize 14px style omega space equals space square root of M over m cross times g over L end root space equals space square root of fraction numerator.5 over denominator.05 end fraction cross times fraction numerator 9.8 over denominator 1 end fraction end root space almost equal to space 9.9 space r a d space divided by s end style
Number of revolution per second = 9.9 / ( 2 π )  ≈ 1.57
 
Number of revolution per minute  = 1.57 × 60  ≈ 94
Answered by Thiyagarajan K | 20 Aug, 2022, 11:31: AM

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CBSE 11-science - Physics
one meter long string can bear maximum of 0.5 kg mass . a mass of 0.05 kg is tied to one end and rotated in a horizontal circle , calculate the maximum number of revolution so that string does not brakes
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