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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A ball of mass $0.500\;kg$ is attached to the end of a cord $1.50\;m$ long. The ball is whirled in a horizontal circle as shown in fig.If the cord can withstand a maximum tension of $50.0\;N$, What is the maximum speed at which the ball can be whirled before the cord breaks ? Assume that the string remains horizontal during the motion.

$\begin{array}{1 1} 12.2\;m/s \\ 3.5\;m/s \\ 4.8\;m/s\\ 2.9\;m/s\end{array} $

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Answer : $ 12.2\;m/s $
The force causing the centripetal acceleration in this case is the force T exerted by the cord on the ball, Solving for v, we have
$T= m \large\frac{v^2}{r}$
$v=\sqrt { \large\frac{Tr}{m}}$
$v_{max} =\sqrt { \large\frac{T_{max}T}{m}}$
$\qquad= \sqrt { \large\frac{(50.0 \;N)(1.50\;m)}{0.500\;kg}}$
$\qquad= 12.2 \;m/s$
answered Aug 26, 2014 by meena.p

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