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Motion in A Plane question

2011 · Shift 0 · Q77
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Motion in A Plane question

2011 · Shift 0 · Q77

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
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 :
  1. A
    πv4g2\pi {{{v^4}} \over {{g^2}}}πg2v4​
  2. B
    π2v4g2{\pi \over 2}{{{v^4}} \over {{g^2}}}2π​g2v4​
  3. C
    πv2g2\pi {{{v^2}} \over {{g^2}}}πg2v2​
  4. D
    πv2g\pi {{{v^2}} \over g}πgv2​
View written solutionFree

Correct answer: A

  1. Interpret the situation

A fountain at ground level throws water in all directions with the same speed vvv.

For any particular droplet projected at an angle θ\thetaθ with the horizontal, the horizontal range is

R=v2sin⁡2θgR=\frac{v^2\sin 2\theta}{g}R=gv2sin2θ​

  1. Find the maximum horizontal distance reached

The ground gets wet up to the maximum range of the projectile.

Since sin⁡2θ≤1\sin 2\theta \le 1sin2θ≤1, the maximum value of RRR is

Rmax⁡=v2gR_{\max}=\frac{v^2}{g}Rmax​=gv2​

This occurs when

2θ=90∘⇒θ=45∘2\theta=90^\circ \quad \Rightarrow \quad \theta=45^\circ2θ=90∘⇒θ=45∘

  1. Determine the wet region

Because the fountain sprinkles water all around, the wet region on the ground is a circle centered at the fountain with radius

r=Rmax⁡=v2gr=R_{\max}=\frac{v^2}{g}r=Rmax​=gv2​

  1. Compute the area

Area of the circle:

A=πr2A=\pi r^2A=πr2

So,

A=π(v2g)2=πv4g2A=\pi\left(\frac{v^2}{g}\right)^2=\pi\frac{v^4}{g^2}A=π(gv2​)2=πg2v4​

  1. Match with the options

πv4g2\boxed{\pi\frac{v^4}{g^2}}πg2v4​​

This corresponds to Option A.

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