Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h (see figure).  Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :

 

  • Option 1)

    x = r\left ( \frac{H}{H + h} \right )

  • Option 2)

    x = r\left ( \frac{H}{H + h} \right )^{\frac{1}{2}}

  • Option 3)

    x = r\left ( \frac{H}{H + h} \right )^{\frac{1}{4}}

  • Option 4)

    x = r \left ( \frac{H}{H + h} \right )^{2}

 

Answers (1)

As we learnt in

Equation of Continuity -

Mass of the liquid entering per second at A = mass of the liquid leaving per second at B. 

a1 v1 = a2 v2 

- wherein

a1  and abe the area of cross section.

 

 A1v1 = A2v2

\pi r^{2}\sqrt{2gH}=\pi x^{2}\sqrt{2g(H+h)}

\therefore\ \;x=r\left(\frac{H}{H+h} \right )^{1/4}

Correct option is 3.


Option 1)

x = r\left ( \frac{H}{H + h} \right )

This is an incorrect option.

Option 2)

x = r\left ( \frac{H}{H + h} \right )^{\frac{1}{2}}

This is an incorrect option.

Option 3)

x = r\left ( \frac{H}{H + h} \right )^{\frac{1}{4}}

This is the correct option.

Option 4)

x = r \left ( \frac{H}{H + h} \right )^{2}

This is an incorrect option.

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