Question

A cylindrical vessel of \(40\,\text{cm}\) radius is completely filled with water and its capacity is \(528\,\text{dm}^3\) (\(\text{dm}=\text{decimetre}\)). The vessel is placed on a solid block of same height as vessel. If a small hole is made at \(70\,\text{cm}\) below the top of water level, then the horizontal range of water falling on the ground in the beginning is ______ cm.

• A. \(120\sqrt2\)
• B. \(140\sqrt2\)
• C. \(140\sqrt3\)
• D. \(120\sqrt3\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to calculate the initial horizontal range of water escaping from a hole in a cylindrical tank that is placed on an elevated platform (block).
Step 2: Key Formula or Approach:
1. Find the height of the cylinder (\(H\)) using volume \(V = \pi r^2 H\).
2. Horizontal range \(R = 2\sqrt{h \cdot y}\), where \(h\) is the depth of the hole from the free surface and \(y\) is the height of the hole from the ground.
Step 3: Detailed Explanation:
Conversion: \(1 \text{ dm}^3 = 1000 \text{ cm}^3\).
Volume \(V = 528 \text{ dm}^3 = 528000 \text{ cm}^3\).
Radius \(r = 40 \text{ cm}\).
\[ V = \pi r^2 H \Rightarrow 528000 = \frac{22}{7} \times (40)^2 \times H \]
\[ 528000 = \frac{22}{7} \times 1600 \times H \]
\[ 330 = \frac{22}{7} H \Rightarrow H = 105 \text{ cm} \]
The height of the cylinder is \(105 \text{ cm}\). The block has the same height, \(H_{block} = 105 \text{ cm}\).
The hole is at a depth \(h = 70 \text{ cm}\) from the top.
Height of the hole from the base of the cylinder \(= 105 - 70 = 35 \text{ cm}\).
Total height of the hole from the ground \(y = 35 + 105 = 140 \text{ cm}\).
Range \(R = 2\sqrt{h \cdot y} = 2\sqrt{70 \times 140}\)
\[ R = 2\sqrt{2 \times 35 \times 4 \times 35} = 2 \times 35 \times 2\sqrt{2} = 140\sqrt{2} \text{ cm} \]
Step 4: Final Answer:
The horizontal range is \(140\sqrt{2} \text{ cm}\).