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

4 m / s

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6.25 m / s

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16 m / s

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Solution

The correct option is C $4 m / s$
When the bucket is at the highest position
$m g + N = \frac{m v^{2}}{R}$

$N = \frac{m v^{2}}{R} - m g$

For $N \geq 0$

$\frac{m v^{2}}{R} - m g \geq 0$

$v \geq \sqrt{g R}$
$v_{m i n} = \sqrt{g R} = \sqrt{10 \times 1.6} = 4 m / s$

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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 (Take g = 10m/ $s e c^{2}$ )

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A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with a constant speed. What should be the minimum speed, so that the water from the bucket does not spill out when the bucket is at the highest position during the rotation? (g=10 m/s2)

The correct option is C 4 m/sWhen the bucket is at the highest position mg+N=mv2RN=mv2R−mgFor N≥0mv2R−mg≥0v≥√gRvmin=√gR=√10×1.6=4m/s

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

The correct option is C $4 m / s$
When the bucket is at the highest position
$m g + N = \frac{m v^{2}}{R}$

$N = \frac{m v^{2}}{R} - m g$

For $N \geq 0$

$\frac{m v^{2}}{R} - m g \geq 0$

$v \geq \sqrt{g R}$
$v_{m i n} = \sqrt{g R} = \sqrt{10 \times 1.6} = 4 m / s$