Fluid Mechanics

To solve the problem, we need to understand how water drains from a cylindrical vessel under gravity through a small opening at the bottom.

1. Concept of Draining from a Vessel:
The draining rate of a liquid from a hole at the bottom of a vessel is governed by Torricelli’s Law, which states that the speed of outflow is proportional to the square root of the height of the liquid column:

2. Time vs Volume Relation:
As the water level drops, the height of the water column decreases, resulting in a slower flow rate.
- The first 5 litres are removed when the height is highest → fastest flow → least time ( )
- The next 5 litres are removed from a lower height → slower flow → more time ( )
- The last 5 litres are removed from the lowest height → slowest flow → most time ( )

3. Ordering the Times:

Final Answer:
The correct relation between the times is .

To solve the problem, we need to calculate the increase in pressure at the bottom of a vessel when a cylinder of mass and density is lowered into a liquid of density (with ) until fully immersed.

1. Key Concept:
When an object is immersed in a fluid and suspended by a string, the buoyant force acts upward, and the net downward force on the liquid (and thus the vessel) is equal to the tension plus the buoyant force. However, for the pressure increase at the bottom, what matters is the additional force transmitted to the liquid due to the immersion.

2. Displaced Volume:
Let the volume of the cylinder be .
Then, (since )

3. Buoyant Force Acting on the Liquid:
The liquid experiences a buoyant force equal to the weight of the displaced liquid:

4. Pressure Increase at the Bottom of Vessel:
The pressure at the bottom increases due to the net downward force applied by the displaced liquid column over the area :

Final Answer:
The increase in pressure at the bottom of the vessel is .

Step 1: The pressure at the closed end of the tube is given by:

Step 2: When the tube is held vertically, the pressure at the closed end is of mercury.

Step 3: When the tube is held horizontally, the length of the air column changes due to the balance of pressures. The total length of the column is .

Step 1: Understanding the Power-Flow Rate Relationship
- Power required to maintain the flow of an incompressible fluid is given by: where is the volumetric flow rate.

Step 2: Effect of Doubling Flow Rate
- If the flow rate is doubled: - New power required:

Step 3: Increase in Power

Step 4: Conclusion
Since the power increase is 700\%, Option (3) is correct.

Step 1: Understanding the Relation between Pressure and Radius For a soap bubble, the excess pressure inside is given by: where: - is the excess pressure, - is the surface tension, - is the radius of the bubble. Given the ratio of excess pressures: Using the formula:

Step 2: Find the Ratio of Volumes Volume of a sphere is: So the ratio of volumes: Thus, the ratio of volumes is:

Step 1: Understanding the Given Conditions - For an isothermal process, the pressure-volume relation is given by Boyle’s Law: Thus, if volume is doubled, the new pressure is: - For an adiabatic process, the pressure-volume relation is given by Poisson’s equation: where is the adiabatic index, given as 1.5.

Step 2: Applying the Adiabatic Equation For an adiabatic process: Rearranging: Substituting : Since , we get: \

Step 3: Equating and Given that , we equate the expressions: Solving for the pressure ratio: Thus, the correct ratio is:

Step 1: The terminal velocity of a spherical object falling through a fluid is given by: where is the radius of the sphere.

Step 2: Given the ratio of the radii , the ratio of the terminal velocities is:

Step 3: Thus, the ratio of the terminal velocities is .