A bucket tied at the end of a 1.6, m long str | vedclass

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Question

$\begin{aligned} \frac{\mathrm{mV}^{2}}{\mathrm{r}}=\mathrm{mg} & & \mathrm{V}=\sqrt{\mathrm{rg}}=\sqrt{1.6 	imes 10} \ & \mathrm{V}=4 \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$\begin{aligned} \frac{\mathrm{mV}^{2}}{\mathrm{r}}=\mathrm{mg} & & \mathrm{V}=\sqrt{\mathrm{rg}}=\sqrt{1.6 	imes 10} \ & \mathrm{V}=4 \mathrm{m} / \mathrm{s} \end{aligned}$

A bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position ............ $m/sec$ (Take $g = 10\, m/sec^2$)

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