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Grade: 12


A string breaks on suspending a 50KG mass.One Kg mass is tied to one end of the 10m long string and revolved in a horizontal plane. find the miximum number of rotations per minute so that the string may not break. 1)67 2)57 3)47 4)97

3 years ago

Answers : (2)

35 Points
we known critical mass is 500 N 
            mw²r= 500
 we want to have no. of rotations per minute  I.e.  
no. of 2py radians per 60 sec 
       let no. of rotation = n
    { n× 2 py/60  }²= 50 on calculating n comes to 67.52
        =67 rotations approximately 
   I don't known whether I did wrong or right as I did it in my first attempt
  only the last line was tricky rest easy one
    thank you
3 years ago
24741 Points
Length r= 10 m
Mass m= 1 kg
Centripetal force F = 50 kg wt = 50* 9.8 N
Let n be the maximum number of revolutions made per
second without breaking string.
Then F = m*r*w= m r (2\pi n)2
50 * 9.8 = 1*10 (44/7 *  n)2
490 = 395.10 n2
n = 1.11 revolution per second
Hence revolution per minute = 1.11* 60 = 66.66 = 67
3 years ago
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