Consider two identical tanks with a bottom hole of diameter \( d \). One tank is filled with water and the other tank is filled with engine oil. The height of the fluid column \( h \) is the same in both cases. The fluid exit velocity in the two tanks are \( V_1 \) and \( V_2 \). Neglecting all losses, which one of the following options is correct?

The fluid exit velocity from the hole in a tank can be found using Torricelli’s Law, which is derived from the principle of conservation of energy. According to Torricelli’s Law, the exit velocity of a fluid is given by: \[ V = \sqrt{2gh} \] where:
\( V \) is the exit velocity of the fluid,
\( g \) is the acceleration due to gravity, and
\( h \) is the height of the fluid column.
Step 1: Understanding the equation.
The velocity of the fluid at the exit depends on the height \( h \) of the fluid column and the gravitational acceleration \( g \). From the equation, we can observe that the velocity is independent of the type of fluid, but only depends on the height of the column and gravitational acceleration.
Step 2: Considering fluid properties.
While the velocity depends on the height of the column and gravity, the density of the fluid plays a role in the fluid's behavior inside the tank, but does not affect the exit velocity directly in this case (since no losses are considered).
Step 3: Comparing water and engine oil.
Both water and engine oil have different densities, but for fluids flowing through an open hole under the same height, the density does not impact the exit velocity because the velocity is purely determined by the height \( h \) of the fluid and gravity, according to Torricelli’s Law.
Step 4: Conclusion.
Since both tanks have the same height of fluid and the same hole diameter, the exit velocity for both water and engine oil will be the same because the exit velocity is determined by the same height and gravitational force for both fluids. Thus, the correct answer is (B) \( V_2 = V_1 \).
During a welding operation, thermal power of 2500 W is incident normally on a metallic surface. As shown in the figure below (figure is NOT to scale), the heated area is circular. Out of the incident power, 85% of the power is absorbed within a circle of radius 5 mm while 65% is absorbed within an inner concentric circle of radius 3 mm. The power density in the shaded area is _________ W mm^-2 (rounded off to 2 decimal places).

Let a spherical block of ice at -7 °C be exposed to atmospheric air at 30 °C with the gravitational direction as shown in the figure below. What will be the overall direction of air flow in this situation?

A pitot tube connected to a U-tube mercury manometer measures the speed of air flowing in the wind tunnel as shown in the figure below. The density of air is 1.23 kg m\(^{-3}\) while the density of water is 1000 kg m\(^{-3}\). For the manometer reading of \( h = 30 \) mm of mercury, the speed of air in the wind tunnel is _________ m s\(^{-1}\) (rounded off to 1 decimal place).

Consider a velocity field \( \vec{V} = 3z \hat{i} + 0 \hat{j} + Cx \hat{k} \), where \( C \) is a constant. If the flow is irrotational, the value of \( C \) is (rounded off to 1 decimal place).
A pitot tube connected to a U-tube mercury manometer measures the speed of air flowing in the wind tunnel as shown in the figure below. The density of air is 1.23 kg m\(^{-3}\) while the density of water is 1000 kg m\(^{-3}\). For the manometer reading of \( h = 30 \) mm of mercury, the speed of air in the wind tunnel is _________ m s\(^{-1}\) (rounded off to 1 decimal place). 