Horizontal Range of Water Falling From Cylindrical Vessel Hole

To find the horizontal range of the water falling on the ground in the beginning, we first determine the height of the cylindrical vessel.

Step 1: Calculate the height of the cylindrical vessel ($H$) Given:

• Radius of the vessel, $r = 40\text{ cm} = 4\text{ dm}$
• Volume (capacity) of the vessel, $V = 528\text{ dm}^3$

The volume of a cylinder is given by: $V = \pi r^2 H$

Taking $\pi = \frac{22}{7}$: $528 = \frac{22}{7} \times 4^2 \times H$ $528 = \frac{352}{7} \times H$ $H = \frac{528 \times 7}{352} = 10.5\text{ dm} = 105\text{ cm}$

Step 2: Determine the heights relative to the ground

• The vessel is placed on a solid block of height equal to the height of the vessel, so: $H_{\text{block}} = 105\text{ cm}$
• The hole is made at a depth $h = 70\text{ cm}$ below the top water level.
• The height of the hole above the bottom of the vessel is: $h_{\text{vessel}} = H - h = 105\text{ cm} - 70\text{ cm} = 35\text{ cm}$
• Therefore, the total height of the hole above the ground ($y$) is: $y = H_{\text{block}} + h_{\text{vessel}} = 105\text{ cm} + 35\text{ cm} = 140\text{ cm}$

Step 3: Calculate the horizontal velocity of efflux and time of flight By Torricelli's Law, the velocity of water exiting the hole is: $v = \sqrt{2gh}$

The time taken $t$ for the water to fall from height $y$ to the ground is: $t = \sqrt{\frac{2y}{g}}$

Step 4: Calculate the horizontal range ($R$) The horizontal range $R$ of the water jet is: $R = v \cdot t = \sqrt{2gh} \times \sqrt{\frac{2y}{g}} = 2\sqrt{h \cdot y}$

Substituting $h = 70\text{ cm}$ and $y = 140\text{ cm}$: $R = 2\sqrt{70 \times 140} = 2\sqrt{70 \times 70 \times 2} = 2 \times 70\sqrt{2} = 140\sqrt{2}\text{ cm}$

Thus, the horizontal range of the water falling on the ground is $140\sqrt{2}\text{ cm}$.

Correct Answer: B ($140\sqrt{2}$)