Question

A tank of height 'A' is completely filled with water. A hole is made on one side of the wall at a depth 'b' below the water surface. The distance from the bottom of the wall at which the emerging stream of water strikes the surface of floor is

• A. \(5\sqrt{b(A-b)}\)
• B. \(4\sqrt{b(A-b)}\)
• C. \(3\sqrt{b(A-b)}\)
• D. \(2\sqrt{b(A-b)}\)

Hint:

Use Torricelli's speed for the exit and projectile motion for the fall.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Water leaves the hole horizontally with speed \(v=\sqrt{2gb}\) (Torricelli's theorem).

Step 2: Time of fall
The hole is at height \((A-b)\) above the floor, so \(t=\sqrt{\dfrac{2(A-b)}{g}}\).

Step 3: Range
\(x=vt=\sqrt{2gb}\times\sqrt{\dfrac{2(A-b)}{g}}=2\sqrt{b(A-b)}\)

Final Answer:
The stream strikes the floor at a distance 2 root of b(A - b). \[ \boxed{\text{(D) }2\sqrt{b(A-b)}} \]