Consider a cylindrical tank of radius 1m is filled with water. The top surface of water is at 15 m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5m from the bottom. A force of  $\mathit{5}\mathit{ }\mathit{\times }\mathit{ }{\mathit{10}}^{\mathit{5}}$ N is applied an the top surface of water using a piston. The speed of efflux from the hole will be: (Assume that there is no atmosphere above the piston)

(given atmospheric pressure $P_{\mathit{A}}\mathit{ }\mathit{=}\mathit{ }\mathit{1}\mathit{.}\mathit{01}\mathit{ }\mathit{\times }\mathit{ }{\mathit{10}}^{\mathit{5}}$Pa, density of water ${\rho }_{\mathit{w}}\mathit{ }\mathit{1000}\mathit{ }kg\mathit{/}{m}^{\mathit{3}}$and gravitational acceleration g = 10 m/s² ) 

Consider a cylindrical tank of radius 1m is filled with water. The top surface of the water is at 15m from the bottom of the cylinder. 

In this scenario, we need to calculate the speed of efflux from a hole that is located at a height of 5m from the bottom of the tank. 

The given conditions include a force of 5 × 10⁵ N applied on the top surface of the water using a piston, with no atmospheric pressure above the piston. Let's begin by analyzing the provided data. 

The cylindrical tank has a radius of 1m, and the top surface of the water is at a height of 15m. The hole is situated 5m above the base. The atmospheric pressure (P_A) is 1.01 × 10⁵ Pa, the density of water (ρ) is 1000 kg/m³, and the gravitational acceleration (g) is 10 m/s².

First, we calculate the pressure exerted by the applied force on the top surface of the water. Since the cross-sectional area (A) of the cylindrical tank is calculated using the formula: 
A = πr² = π × (1)² = π m² 
The applied pressure (P_applied) is: 
P_applied = F / A = (5 × 10⁵) / π ≈ 1.59 × 10⁵ Pa

Since there is no atmospheric pressure above the piston, the total pressure at the top surface of the water is equal to the applied pressure: 
P_top = P_applied = 1.59 × 10⁵ Pa

Now, let's calculate the pressure at the height of the hole. The pressure at this point is influenced by the height of the water column between the top of the tank and the hole. This height is: 
H - h = 15m - 5m = 10m 
Using the hydrostatic pressure formula, the pressure due to this water column is: 
P_water = ρgh = 1000 × 10 × 10 = 1 × 10⁵ Pa

The total pressure at the hole is: 
P_hole = P_top + P_water = 1.59 × 10⁵ + 1 × 10⁵ = 2.59 × 10⁵ Pa

To find the speed of efflux from the hole, we apply Torricelli's theorem: 
v = √[2 (P_hole - P_A) / ρ] 
Substituting the known values: 
v = √[2 × (2.59 × 10⁵ - 1.01 × 10⁵) / 1000] 
v = √[2 × 1.58 × 10⁵ / 1000] 
v = √[3.16 × 10⁵ / 1000] 
v = √316 ≈ 17.8 m/s

Therefore, the speed of efflux from the hole is approximately 17.8 m/s. The scenario where we consider a cylindrical tank of radius 1m is filled with water and the top surface of water is at 15m from the bottom showcases how pressure differences affect the speed of water flow. Whether you consider a cylindrical tank of radius 1m or analyze the cylindrical tank of radius 1m filled with water, the calculation demonstrates the importance of understanding pressure dynamics in fluid mechanics.