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A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2 divided by pi revolutions per s e c o n d around the vertical axis through the fixed end as shown in figure, then tension in the string is

Physics-General

  1. 16 blank M L    
  2. 4 blank M L    
  3. M L    
  4. 2 blank M L    

    Answer:The correct answer is: 16 blank M LT sin invisible function application theta equals M omega to the power of 2 end exponent R(i)
    T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta(ii)
    From (i) and (ii)
    T equals M omega to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

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