Question

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Answer 1

In the vertical plane maximum tension (Tmax) in the string occurs when the stone is at the bottom of the circular and minimum tension (Tmin) when the stone is at the top of the circular motion.

Given: length of the string (l) = 0.5 m        

           mass of the stone = 5 gm

            Tmax  : Tmin = 13 : 1

$Maximum \ tension \ (Tmax) = \frac{mv^{2} }{R} + mg$

$Tmax = \frac{0.005v^{2} }{0.5} + 0.005*g$

$Minimum \ tension \ (Tmin) = \frac{mv^{2} }{R} - mg$

$Tmin = \frac{0.005v^{2} }{0.5} - 0.005*g$

$\frac{Tmax}{Tmin} = \frac{\frac{0.005v^{2} }{0.5} + 0.005*g}{\frac{0.005v^{2} }{0.5} - 0.005*g} = \frac{13}{1}$

⇒ v = 0.9 m/s

Hope this helps you !!