A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7 m. Find the height of the platform.
Answers
Answer:
The height of the platform is 80 cm.
Step-by-step explanation:
Given :
Height of deep well which is in the form of cylinder, H = 16 m
Diameter of the deep well = 3.5 m
Radius (r) of the deep well = 3.5/2 m = 1.75 m
Length of the platform (l) = 27.5 m
Breadth of the platform (b) = 7 m
Let the height of the platform be ‘h’ m.
Here, the earth obtained from digging the well of cylindrical shape is used to make a platform of cuboidal shaped. So the volume of Earth will be equal to the volume of cylindrical well and it will be equal to volume of cuboidal platform.
Volume of the deep well (cylinder) = Volume of cuboidal platform.
πr²H = lbh
π × (1.75)² ×16 = 27.5 × 7 × h
22/7 × 1.75 × 1.75 × 16 = 27.5 × 7 × h
h = (22× 1.75 × 1.75 × 16) / (27.5 × 7 × 7)
h = (22 × 0.25 × 0.25 × 16) / 27.5
h = 22 / 27.5
h = 0.8 m = 0.8 × 100 = 80 cm
[1 m = 100 cm ]
height of the platform = 80 cm
Hence, the height of the platform is 80 cm.
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Answer:
Volume of the Platform = Volume of mud dug out
Volume of Mud dug out can be calculated by calculating the volume of the cylindrical column dug.
Given that, Height of the mud dug = 16 m
Diameter = 3.5 m which implies r = 1.75 m
Volume of mud dug out = πr²h = 22/7 × 1.75 × 1.75 × 16 = 154 m³
Volume of Platform = 27.5 m × 7 m × h
⇒ 27.5 m × 7 m × h = 154 m³
⇒ 192.5 m² × h = 154 m³
⇒ h = 154 m³ / 192.5 m² = 0.8 m
Hence the height of the platform is 0.8 m or 80 cm.