First, find the total volume of the godown, which is shaped like a cuboid. The volume of a cuboid is calculated by multiplying its length, width, and height.
Given dimensions:
The volume of the godown ($V_{godown}$) is:
$ V_{godown} = L \times W \times H $
$ V_{godown} = 60 \text{ m} \times 40 \text{ m} \times 30 \text{ m} $
$ V_{godown} = 72000 \text{ m}^3 $
Next, determine how many cuboidal boxes can fit inside the godown. This is found by dividing the total volume of the godown by the volume of a single box.
Volume of one box ($V_{box}$) = $0.8 \text{ m}^3$.
The number of boxes ($N$) that can be stored is:
$ N = \frac{V_{godown}}{V_{box}} $
$ N = \frac{72000 \text{ m}^3}{0.8 \text{ m}^3} $
$ N = \frac{720000}{8} $
$ N = 90000 $
Therefore, 90,000 boxes can be stored in the godown.
The diameter of the base and slant height of a right circular cone are 30 cm and 113 cm, respectively. Find the volume (in cm³) of the given cone.
(Use $\pi = \frac{22}{7}$)
The length and breadth of a cuboid are 50 cm and 75 cm, respectively. If the length of the diagonal is $17\sqrt{29}$ cm, then what is the volume (in $cm^{3}$) of the cuboid?