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A wet open umbrella is held vertical and whirled about the handle at a uniform rate of 21 rev in $44 s$ . If the rim of the umbrella is a circle of $1 m$ in diameter and the height of the rim above the floor is $4.9 m$ . The locus of the drop on floor is a circle of radius

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$\sqrt{2.5} m$

$1 m$

$3 m$

$10 m$

answer is A.

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Detailed Solution

$v = \frac{a}{t_{1}} = \frac{21 \times 2 π \times 0.5}{44} = 1.5 m / s$

$t_{2} = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 4.9}{9.8}} = 1 s$

Horizontal distance travelled by drop $X = v_{2} Xt = 1.5 m$

$r = \sqrt{(1.5)^{2} + (0.5)^{2}}$

$= \sqrt{2.5} m$

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