Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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 $g = 10 \text{ m/s}^2$ )

4 m/s

6.25 m/s

16 m/s

None of the above

Step-by-Step Solution

Concept: For water not to spill from the bucket at the highest point of a vertical circle, the weight of the water must be just sufficient to provide the necessary centripetal force. This corresponds to the condition where the normal reaction (or tension) becomes zero [Source 85].
Critical Speed Formula: The minimum speed ( $v_{min}$ ) at the highest point is given by equating the gravitational force to the centripetal force: $mg = \frac{mv^2}{R}$ $v_{min} = \sqrt{Rg}$ [Source 85]
Calculation:

Radius ( $R$ ) = Length of string = $1.6 \text{ m}$
Gravity ( $g$ ) = $10 \text{ m/s}^2$ Substitute the values: $v_{min} = \sqrt{1.6 \times 10}$ $v_{min} = \sqrt{16}$ $v_{min} = 4 \text{ m/s}$

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