Question

Water from a full cylindrical vessel of height \( h \) and of unknown diameter is completely emptied to precisely fill two cylindrical vessels of the same diameter \( d \) and heights \( h \) and \( 3h \). The diameter of the original vessel is ............ \( \times d \) (round off to 2 decimal places)

Hint:

For volumes of cylinders with the same height and diameter, the volume is proportional to the square of the diameter.

Answer 1

Correct Answer: 1.99

Step 1: Volume of the original vessel.
The volume of the original cylindrical vessel is: \[ V_{\text{original}} = \pi r^2 h \] where \( r \) is the radius of the original vessel, and \( h \) is its height. Step 2: Volume of the two new vessels.
The total volume of the two new vessels (with the same diameter as the original) is: \[ V_{\text{new}} = \pi \left( \frac{d}{2} \right)^2 h + \pi \left( \frac{d}{2} \right)^2 3h = \pi \left( \frac{d}{2} \right)^2 (4h) \] This simplifies to: \[ V_{\text{new}} = \pi \frac{d^2}{4} (4h) = \pi d^2 h \] Step 3: Equating the volumes.
Since the total volume of water in both vessels must be the same as the original volume, we equate: \[ \pi r^2 h = \pi d^2 h \] Cancel \( \pi h \) from both sides: \[ r^2 = d^2 \] Thus, \( r = d \), meaning the diameter of the original vessel is \( \boxed{1.26 \, d} \).