Physics-
General
Easy

Question

A stone of mass 1 kg tied to a light inextensible string of length L equals fraction numerator 10 over denominator 3 end fraction m is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m divided by sec to the power of 2 end exponent invisible function application blank, the speed of the stone at the highest point of the circle is

  1.  20 straight m divided by sec    
  2. 10 square root of 3 m divided by s e c    
  3. 5 square root of 2 m divided by s e c    
  4.  10 straight m divided by sec    

The correct answer is: 10 straight m divided by sec


    Since the maximum tension T subscript B end subscript in the string moving in the vertical circle is at the bottom and minimum tension T subscript T end subscript is at the top.
    \ T subscript B end subscript equals fraction numerator m v subscript B end subscript superscript 2 end superscript over denominator L end fraction plus m g and T subscript T end subscript equals fraction numerator m v subscript T end subscript superscript 2 end superscript over denominator L end fraction minus m g
    \ fraction numerator T subscript B end subscript over denominator T subscript T end subscript end fraction equals fraction numerator fraction numerator m v subscript B end subscript superscript 2 end superscript over denominator L end fraction plus m g over denominator fraction numerator m v subscript T end subscript superscript 2 end superscript over denominator L end fraction minus m g end fraction equals fraction numerator 4 over denominator 1 end fraction or fraction numerator v subscript B end subscript superscript 2 end superscript plus g L over denominator v subscript T end subscript superscript 2 end superscript minus g L end fraction equals fraction numerator 4 over denominator 1 end fraction
    or v subscript B end subscript superscript 2 end superscript plus g L equals 4 v subscript T end subscript superscript 2 end superscript minus 4 g L but v subscript B end subscript superscript 2 end superscript equals v subscript T end subscript superscript 2 end superscript plus 4 g L
    \ v subscript T end subscript superscript 2 end superscript plus 4 g L plus g L equals 4 v subscript T end subscript superscript 2 end superscript minus 4 g LÞ 3 v subscript T end subscript superscript 2 end superscript equals 9 g L
    \ v subscript T end subscript superscript 2 end superscript equals 3 cross times g cross times L equals 3 cross times 10 cross times fraction numerator 10 over denominator 3 end fraction or v subscript T end subscript equals 10 m divided by s e c

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