A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Answers
38,500 cm^3 of the volume of the soup the hospital has to prepare daily to serve 250 patients.
Given,
The diameter of the cylindrical bowl used to feed the patient with soup = 7 cm
The height of the cylindrical bowl filled with the soup = 4 cm
Total number of patients = 250
The volume of the cylinder is given by,
V = π*r^2*h
here, r = d/2 = 7/2 = 3.5 cm
Hence, the volume of soup to be prepared is,
⇒ V = 22/7 × (3.5)^2 × 4
∴ V = 154 cm^3
154 cm^3 is the volume of the soup to be prepared for a single patient.
For a total of 250 patients, the volume of soup to be prepared is,
= 250 × 154
= 38500 cm^3
Given : In a cylindrical bowl of diameter 7 cm and bowl is filled with soup to a height of 4 cm.
Radius of cylindrical bowl (r) = diameter/2 = 7/2 cm
r = 3.5 cm
Height , h = 4 cm
Now,
Volume of soup in 1 bowl ,V = πr²h
V = 22/7 × 3.5² × 4
V = 22/7 × 3.5 × 3.5 × 4
V = 22 × 0.5 × 3.5 × 4
V = 22 × 2 × 3.5
V = 22 × 7
V = 154 cm³
Volume of soup in 1 bowl ,V = 154 cm³
Volume of soup in 250 bowls = (250 x 154)
= 38500 cm³
= 38500/1000
[1 cm³ = 1/1000 litres]
Volume of soup in 1 bowl ,V = 38.5 litres
Hence , hospital has to prepare 38.5 litres of soup daily in order to serve 250 patients.
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