Question

a large tank is filled with water a small hole is made at a depth 10 m below water surface. the range of water issuing out of the hole is R on ground . what extra pressure must be applied on water surface so that the range becomes 2R.

Answer 1

Extra pressure must be applied on water surface so that the range becomes 2R is about 294300 Pa.

Explanation:

1. Here initial height(h) above hole is 10(m), initial range is R; taking gravitational acceleration(g) is 9.81 ($\frac{m}{s^{2}}$).

2. Velocity comes out from hole is in horizontal direction, so time to reach water jet on ground is independent on speed.

3. Here initial range is R. For range 2 R is possible only when velocity comes out from hole with double of initial speed.

4. For initial case of when range is R with speed(V)

    $\rho gh = \frac{1}{2}\rho V^{2}$        ...a)

5. For final case of when range is 2R with speed(2V)

  $P+ \rho gh = \frac{1}{2}\rho (2V)^{2}$

  $P+ \rho gh = 4\times \frac{1}{2}\rho (V)^{2}$   ...b)

6. On subtracting equation b) - equation a)

    We got  $P= 3\times \frac{1}{2}\rho (V)^{2}$      ...c)

     From equation a) and equation c)

    $P= 3\rho gh$

So P = 3×1000×9.81×10 = 294300 Pa