A bucket tied at the end of 1.6 m long string is whirled in a vertical circle, when the bucket is at the highest point, the minimum speed for the bucket not to spill water from it is (take $g=10{\text{\hspace{0.17em}ms}}^{-\text{2}}$)

Given that a bucket is whirled in vertical circle as shown.

Length of string r = 1.6m

Weight of bucket W = mg

g value $=10{\text{m/s}}^{\text{2}}$

Tension in the string = T

From F.B.D of bucket

At highest point A.

We have for body performing vertical circular motion at any point.

$\text{T}-\text{Wcos}\theta =\frac{{\text{mv}}^{\text{2}}}{\text{R}}$

Here T, W, $\frac{{\text{mv}}^{\text{2}}}{\text{R}}$ all are in downward direction.

$-\text{T}-\text{W}=\frac{-{\text{mv}}^{\text{2}}}{\text{R}}$

To get minimum centripetal force let us consider T = 0.

$-\text{mg}=\frac{-{\text{mv}}^{\text{2}}}{\text{R}}$

$\therefore \text{v}=\sqrt{\text{gr}}$

$\text{v}=\sqrt{10\times 1.6}=\sqrt{16}=4{\text{ms}}^{-1}$

$\therefore \text{v}=4{\text{ms}}^{-1}$