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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

  • Option 1)

    \pi \frac{v^{2}}{g}

  • Option 2)

    \pi \frac{v^{4}}{g^{2}}

  • Option 3)

    \frac{\pi}{2} \frac{v^{4}}{g^{2}}

  • Option 4)

    \pi \frac{v^{2}}{g^{2}}

 

Answers (1)

best_answer

As we learnt in 

Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

R=\frac{u^{2}\sin 2\Theta }{g}

- wherein

Special case of horizontal range

For man horizontal range.

\Theta = 45^{0}

R_{max}=\frac{u^{2}\sin 2 (45) }{g}=\frac{u^{2}\times 1}{g}=\frac{u^{2}}{g}

 

 R_{max}=\frac{v^2_0sin 90^{o} }{g}=\frac{v^2_0}{g}

Area=\Pi (R_{max})^2 \: =\frac{\Pi v^4_0}{g^2}


Option 1)

\pi \frac{v^{2}}{g}

Incorrect

Option 2)

\pi \frac{v^{4}}{g^{2}}

Correct

Option 3)

\frac{\pi}{2} \frac{v^{4}}{g^{2}}

Incorrect

Option 4)

\pi \frac{v^{2}}{g^{2}}

Incorrect

Posted by

divya.saini

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