JEE Main · 2011 · 10 Apr 2011 · Offline · Kinematics

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A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :

(A) $π \frac{v ^{4}}{g ^{2}}$ ✓ Correct
(B) $\frac{π}{2} \frac{v ^{4}}{g ^{2}}$
(C) $π \frac{v ^{2}}{g ^{2}}$
(D) $π \frac{v ^{4}}{g}$

Answer: (A)

Solution

Water leaves the fountain in every direction with speed

v

, and the farthest it lands is the maximum projectile range, reached at a launch angle of

4 5^{\circ}

, namely

R_{m a x} = \frac{v ^{2}}{g}

. The wetted ground is the disc of this radius, so its area is

π R_{m a x}^{2} = π (\frac{v ^{2}}{g})^{2} = \frac{π v ^{4}}{g ^{2}}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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