A ball of mass, $m$ is being swung on a string of length, $l$ in a horizontal circle as depicted above. When the ball is swung with velocity, $v$ . The tension in the string is equal to, $F$ .
If the length of the string is doubled and the ball is swung at the same angle with the same velocity, in terms of the original force, $F$ , what is the new tension in the string?

F^{'} = \frac{F}{2}

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F^{'} = \frac{F}{4}

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F^{'} = F

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F^{'} = 2 F

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F^{'} = 4 F

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Solution

The correct option is B $F^{'} = \frac{F}{2}$
Let the angle between vertical and string be $θ$
The centripetal force experience by the ball away from the center of circle= $\frac{v^{2}}{R}$
Thus $m \frac{v^{2}}{l s i n θ} = F s i n θ$
$⟹ F \propto \frac{1}{l}$
Hence $F^{'} = \frac{F}{2}$

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The correct option is B F′=F2Let the angle between vertical and string be θThe centripetal force experience by the ball away from the center of circle=v2RThus mv2lsinθ=Fsinθ⟹F∝1lHence F′=F2

The correct option is B $F^{'} = \frac{F}{2}$
Let the angle between vertical and string be $θ$
The centripetal force experience by the ball away from the center of circle= $\frac{v^{2}}{R}$
Thus $m \frac{v^{2}}{l s i n θ} = F s i n θ$
$⟹ F \propto \frac{1}{l}$
Hence $F^{'} = \frac{F}{2}$