A circular cylinder of height 1.20 m having faces of diameter 0.620 m is immersed in water of density with its axis vertical so that its faces are parallel to the surface of the water.The upper face is 2.50 m below the surface of the water.The net force on this cylinder is observed to be 1120 N downward.Atmospheric pressure is .What net force does the water exert on the lower face of the cylinder? A) 7.40 × 10³ N B) 1.09 × 10⁴ N C) 3.79 × 10⁴ N D) 4.14 × 10⁴ N E) 1.37 × 10⁵ N

Question

Quizplus · Accepted Answer

The net force on the cylinder is equal to the weight of the water displaced by the cylinder plus the weight of the cylinder itself minus the buoyant force acting on it. Since the cylinder is completely submerged, the buoyant force is equal to the weight of the water displaced by it. To find the weight of the water displaced by the cylinder, we can use the formula: V = πr^2h where V is the volume of the cylinder, r is the radius (half of the diameter), and h is the height. V = π(0.310 m)^2 (1.20 m) = 0.458 m^3 The weight of this volume of water is: W = mg = ρVg = (11ea83ea_69e6_2e48_aeff_59165af9800c_TB6394_11)(0.458 m^3)(9.81 m/s^2) = 4.35 × 10³ N Since the cylinder is in equilibrium (not accelerating), the net force on it is equal and opposite to the force exerted by the water on the lower face. Therefore, the force exerted by the water on the lower face is: F = W - 1120 N = (4.35 × 10³ N) - (1120 N) = 4.14 × 10⁴ N Thus, the best choice is D.

A Circular Cylinder of Height 1