Question

A cylindrical tank \(0.5\,m\) in radius, rests on a platform \(1.5\,m\) high. Initially the tank is filled with water to a height of \(2.5\,m\). A small plug whose area is \(10^{-4}\,m^2\) is removed from an orifice located on the side of the tank at the bottom. The speed with which the water strikes the ground is: [Assume \(g = 10\,ms^{-2}\)]

• A. \(7.07\,ms^{-1}\)
• B. \(5.1\,ms^{-1}\)
• C. \(5.47\,ms^{-1}\)
• D. \(7.62\,ms^{-1}\)

Hint:

For water coming out horizontally from a side orifice, vertical speed before hitting ground is found using \(v_y = \sqrt{2gh}\).

Answer 1

The Correct Option is C

Step 1: Identify the vertical height from orifice to ground.
The orifice is at the bottom of the tank and the tank is kept on a platform of height \(1.5\,m\).
\[ h = 1.5\,m \]

Step 2: Water comes out horizontally.
Since the hole is on the side of the tank, water initially comes out horizontally.
So, the vertical component of velocity initially is zero.

Step 3: Vertical speed gained while falling.
The vertical speed gained by water before reaching the ground is:
\[ v_y = \sqrt{2gh} \]

Step 4: Substitute values.
\[ v_y = \sqrt{2 \times 10 \times 1.5} \]
\[ v_y = \sqrt{30} \]

Step 5: Calculate speed.
\[ v_y = 5.47\,ms^{-1} \]

Step 6: Final conclusion.
\[ \boxed{5.47\,ms^{-1}} \] Hence, correct answer is option (C).