A cylindrical vessel of radius 40 cm is completely filled with water and has a capacity of 528 dm³. The vessel is placed on a solid block whose height is exactly equal to the vessel height. A small hole is made in the side wall at a point 70 cm below the top water level. What is the horizontal range of the water jet when it first reaches the ground?
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Question

A cylindrical vessel of radius 40 cm is completely filled with water and has a capacity of 528 dm³. The vessel is placed on a solid block whose height is exactly equal to the vessel height. A small hole is made in the side wall at a point 70 cm below the top water level. What is the horizontal range of the water jet when it first reaches the ground?

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G1 Physics · Accepted Answer

First determine the height of the cylindrical vessel. Volume: V = 528 dm³ = 0.528 m³. Using V = πr²h with r = 0.40 m, h = 0.528 ÷ [π(0.40)²] ≈ 1.05 m = 105 cm. The hole is 70 cm below the water surface, so the efflux speed is u = √(2g×0.70). The hole is 105-70 = 35 cm above the bottom of the vessel. Since the supporting block is also 105 cm high, the hole is H = 105+35 = 140 cm = 1.40 m above the ground. Time of fall: t = √(2H ÷ g) = √(2×1.40 ÷ 10). Horizontal range: R = ut = √(2g×0.70) × √(2×1…