Physics | JEE Mains 2022 | 24 June Shift - I
Chapter : Laws of Motion | Topic : Circular Motion
A boy ties a stone of mass 100 g to the end of a
2 m long string and whirls it around in a horizontal
plane. The string can withstand the maximum
tension of 80 N. If the maximum speed with which
the stone can revolve is \(\tt \frac{K}{\pi}\) rev./min. The value of K is (Assume the string is massless and unstretchable)
Solution : C
T = \(\tt m \omega^{2} r\)
=> 80 = 0.1 x \(\tt ( 2 \pi \times \frac{K}{\pi} \times \frac{1}{60} )^{2}\) x 2
=> \(\tt \frac{800}{2}\) = \(\tt \frac{K^{2}}{900}\)
=> K = 30 x 20 = 600
T = \(\tt m \omega^{2} r\)
=> 80 = 0.1 x \(\tt ( 2 \pi \times \frac{K}{\pi} \times \frac{1}{60} )^{2}\) x 2
=> \(\tt \frac{800}{2}\) = \(\tt \frac{K^{2}}{900}\)
=> K = 30 x 20 = 600