Question

Water leaks out from an open tank through a hole of area 2 mm² in the bottom. Suppose water is filled up to a height of 80 cm and the area of cross section of the tanks is 0.4 m². The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of water coming out of the hole. (b) Find the speed of water coming out when half of water has leaked out. (c) Find the volume of eater leaked out using a time interval dt after the height remained is h. Thus find the decrease in height dh in terms of h and dt.
(d) From the result of park (c) find the time required for half of the water to leak out.

Answer 1

Given:
Area of the hole of the water tank = 2 mm²
Height of filled water, h = 80 cm
Area of the cross-section of the tanks, A = 0.4 m²
Acceleration due to gravity, g = 9.8 ms^{- 2}

Pressure at the open surface and at the hole is equal to atmospheric pressure.

(a) Velocity of water:

V \[= \sqrt{2gh}\]

\[= \sqrt{2 \times 10 \times 0 . 80}\]

\[ = 4 m/\sec\]

(b) Velocity of water when the tank is half-filled and h \[is \frac{80}{2}, i . e . , 40 cm:\]

\[= \sqrt{2 \times 10 \times 0 . 40}\]

\[ = \sqrt{8} m/\sec\]

(c) Volume:

Volume = Ah =A v dt

\[ = A \times \sqrt{2gh} dt\]

\[ = \left( 2 {mm}^2 \right)\sqrt{2gh}dt\]

Volume of the tank = Ah = V (say)

\[i . e . , \frac{dV}{dt} = A\frac{dh}{dt}\]

\[a_1 v_1 = A\frac{dh}{dt}\]

\[ \Rightarrow 2 \times {10}^{- 6} \sqrt{2gh} = 0 . 4\frac{dh}{dt}\]

\[ \Rightarrow dh = 5 \times {10}^{- 6} \sqrt{2gh}dt\]

\[(d) \because dh = 5 \times {10}^{- 6} \sqrt{2gh}\ dt\]

\[ \therefore \frac{dh}{\sqrt{2gh}} = 5 \times {10}^{- 6}\ dt\]

On integrating, we get:

\[5 \times {10}^{- 6} \int_0^t dt = \frac{1}{\sqrt{28}} \int_{0 . 8}^{0 . 4} \frac{dh}{\sqrt{h}}\]

\[ = 5 \times {10}^{- 6} \times t = \frac{1}{\sqrt{28}} \times 2 \left[ h^\frac{1}{2} \right]_{0 . 8}^{0 . 4} \]

\[ = 5 \times {10}^{- 6} \times t = \frac{1}{\sqrt{28}} \times 2 \left[ h^\frac{1}{2} \right]_{0 . 8}^{0 . 4} \]

\[ \Rightarrow t = \frac{1}{\sqrt{20}} \times 2 \times \left[ (0 . 4 )^{1/2} - (0 . 8 )^{1/4} \right] \times \frac{1}{5 \times {10}^{- 6}}\]

\[ \Rightarrow t = \frac{1}{4 . 47} \times 2 \times \frac{2}{3 . 16} \times \frac{1}{5 \times {10}^{- 6}} \times \frac{1}{3600} h\]

\[ = 6 . 51 h\]

Thus, the time required to leak half of the water out is 6.51 hours.