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A frictionless wire $\mathrm{AB}$ is fixed on a sphere of radius $\mathrm{R}$. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to $\mathrm{B}$ is
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The correct answer is:
$2 \sqrt{\frac{R}{g}}$
Acceleration of body along $\mathrm{AB}$ is g $\cos \theta$ Distance travelled in time $\mathrm{t}$ sec $=$
$A B=\frac{1}{2}(g \cos \theta) t^{2}$
From $\Delta \mathrm{ABC}, \mathrm{AB}=2 \mathrm{R} \cos \theta$
Thus, $2 \mathrm{R} \cos \theta=\frac{1}{2} g \cos \theta t^{2}$
$\Rightarrow t^{2}=\frac{4 R}{g} \Rightarrow t=2 \sqrt{\frac{R}{g}}$
$A B=\frac{1}{2}(g \cos \theta) t^{2}$
From $\Delta \mathrm{ABC}, \mathrm{AB}=2 \mathrm{R} \cos \theta$
Thus, $2 \mathrm{R} \cos \theta=\frac{1}{2} g \cos \theta t^{2}$
$\Rightarrow t^{2}=\frac{4 R}{g} \Rightarrow t=2 \sqrt{\frac{R}{g}}$
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