A cylindrical tank of height 3 units and diameter 2 units, resting on its base, is partially
filled with water up to a height h units. Thereafter it is turned horizontal and placed on
a level ground with its lateral surface in contact with the ground. In this position, the
maximum depth of water is
1/2
units. Then h is
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volume of cylinder = πr^2h
r = 1 , h=3
volume of cylinder = π× 1^2 × 3
= 3π
volume of water = πr^2h
r =1
volume of water = πh
now when this container turned horizontal it fills at depth 1/2 units which is half of the radius
it means it fills 1/4 of total volume of cylinder
therefore , 3π /4 = πh
h = 3 /4 = 0.75 units
r = 1 , h=3
volume of cylinder = π× 1^2 × 3
= 3π
volume of water = πr^2h
r =1
volume of water = πh
now when this container turned horizontal it fills at depth 1/2 units which is half of the radius
it means it fills 1/4 of total volume of cylinder
therefore , 3π /4 = πh
h = 3 /4 = 0.75 units
rreed8249:
thanks sir
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