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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

$20 m / \sec$
$10 \sqrt{3} m / s e c$
$5 \sqrt{2} m / s e c$
$10 m / \sec$

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 $v subscript B end subscript superscript 2 end superscript equals v subscript T end subscript superscript 2 end superscript plus 4 g L$
\ $v subscript T end subscript superscript 2 end superscript plus 4 g L plus g L equals 4 v subscript T end subscript superscript 2 end superscript minus 4 g L$ Þ $3 v subscript T end subscript superscript 2 end superscript equals 9 g L$
\ $v subscript T end subscript superscript 2 end superscript equals 3 cross times g cross times L equals 3 cross times 10 cross times fraction numerator 10 over denominator 3 end fraction$ or $v subscript T end subscript equals 10 m divided by s e c$

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