Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m². The tank has an opening B near the bottom whose cross-sectional area is 1cm². A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is v ms^-1.

The value of v, to the nearest integer, is..................
[Take value of g to be 10 ms^-2]

<p>v? ²/2 + P? /ρ + gh? = v? ²/2 + P? /ρ + gh? </p><p>⇒ v? ²/2 + (P? + ρgh + mg/A)/ρ + g × 0 = v? ²/2 + P? /ρ + g × 0</p><p>⇒ v? ²/2 + (ρgh + mg/A)/ρ = v? ²/2 . (1)</p><p><img ></p><p>Using Continuity equation, we can write<br>Av? = av? ⇒ v? = (a/A)v? </p><p>Putting the value of v? in equation (1), we have<br> (a²/A²)v? ²/2 + (ρgh + mg/A)/ρ = v? ²/2 ⇒ (A² - a²)/A² * v? ²/2 = (ρgh + mg/A)/ρ</p><p>⇒ v? = √ [ (A²/ (A²-a²) * 2 (ρgh + mg/A)/ρ] = 3.09 m/s ≈ 3 m/s</p>

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