Question
Question: A cylinder containing water stands on a table of height H. A small hole is punched in the side of th...
A cylinder containing water stands on a table of height H. A small hole is punched in the side of the cylinder at its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of water in the cylinder is:
& \left( A \right)H \\\ & \left( B \right)R \\\ & \left( C \right)\sqrt{RH} \\\ & \left( D \right)\dfrac{{{R}^{2}}}{4H} \\\ \end{aligned}$$Solution
These types of problems are quite easy to solve once we understand the underlying concepts behind the sums. This particular problem requires some prerequisite knowledge of physics as well some algebraic concepts of mathematics. We need to be through with the chapter of kinematics to be able to solve the problem efficiently. Here in this problem we need to assume that the height of the cylinder is equal to ‘h’, and we need to recall from the concepts of fluid dynamics that the velocity of the stream flow from the bottom of a cylinder of height ‘h’ is given by,
v=2gh . Using this relation we apply the concept that the time required by the stream of water to cover the horizontal distance and the vertical distance will be equal.
Complete step-by-step solution:
Now we start off with the solution to the problem by writing that, the time required for the stream of water to travel the horizontal distance ‘R’ is given by,