A rectangular water tank measures $15m\times 6m$ at top and is 10m deep. It is full of water initially. If water is drawn out, lowering the level by 1m, how much water has been drawn out?
[a] 90,000 litres
[b] 45,000 litres
[c] 4,500 litres
[d] 900 litres
[e] None of these
Answer
629.1k+ views
Hint: The amount of water drawn out = initial volume of the water inside the tank – the final volume of water inside the tank. Since water takes the shape of the container, it also will form a cuboidal block. Determine the dimensions of the cuboidal block of water initially and finally. Use volume of a cuboid of length l, breadth b and height h = lbh.
Complete step-by-step answer:
Since the water tank is initially full, the dimensions of the cuboidal block of water inside the tank initially are:
Length (l) = 15m, breadth (b) = 6m and height (h) = 10m.
Using the volume of cuboid = lbh, we get
The volume of water present inside the tank initially $=15\times 6\times 10=900$ cubic metres.
Now after drawing out of some water, the level of water lowers by 1m.
Hence, the new dimensions of the block of water are
Length (l) = 15m, breadth (b) = 6m and height (h) = 10-1 = 9m.
Hence, the volume of water inside the tank finally $=15\times 6\times 9=810$ cubic metres.
Hence the volume of water drawn out $=900-810=90$ cubic metres.
Now, we know that 1 cubic metre = 1000 litres.
Hence 90 cubic metres – 90,000 litres.
Hence the volume of water drawn out of the tank = 90,000 litres.
Hence option [a] is correct.
Note: Let H be the initial height and let d be the depth of lowering of water.
Hence final height = H-d. Let l and b be the length and bread of the tank.
The volume of water initially inside the tank = lbH.
The volume of water finally inside the tank = lb(H-d)
Hence volume of water drawn out of the tank = lbH-lb(H-d) = lbH-lbH+lbd = lbd.
This is equivalent to a cuboidal block of dimensions $l\times b\times d$ .
As can be observed the whole process is as if a cuboidal block of dimensions $l\times b\times d$ is removed from the original block. Hence the decrease in Volume = lbd.
Using d = 1m, l = 15m and b = 6m, we get
The volume of water drawn out of the tank $=15\times 6\times 1=90$ cubic metres, which is the same as obtained above.
It can be noted that the initial height of the water is not necessary to be provided to solve the question.
Complete step-by-step answer:
Since the water tank is initially full, the dimensions of the cuboidal block of water inside the tank initially are:
Length (l) = 15m, breadth (b) = 6m and height (h) = 10m.
Using the volume of cuboid = lbh, we get
The volume of water present inside the tank initially $=15\times 6\times 10=900$ cubic metres.
Now after drawing out of some water, the level of water lowers by 1m.
Hence, the new dimensions of the block of water are
Length (l) = 15m, breadth (b) = 6m and height (h) = 10-1 = 9m.
Hence, the volume of water inside the tank finally $=15\times 6\times 9=810$ cubic metres.
Hence the volume of water drawn out $=900-810=90$ cubic metres.
Now, we know that 1 cubic metre = 1000 litres.
Hence 90 cubic metres – 90,000 litres.
Hence the volume of water drawn out of the tank = 90,000 litres.
Hence option [a] is correct.
Note: Let H be the initial height and let d be the depth of lowering of water.
Hence final height = H-d. Let l and b be the length and bread of the tank.
The volume of water initially inside the tank = lbH.
The volume of water finally inside the tank = lb(H-d)
Hence volume of water drawn out of the tank = lbH-lb(H-d) = lbH-lbH+lbd = lbd.
This is equivalent to a cuboidal block of dimensions $l\times b\times d$ .
As can be observed the whole process is as if a cuboidal block of dimensions $l\times b\times d$ is removed from the original block. Hence the decrease in Volume = lbd.
Using d = 1m, l = 15m and b = 6m, we get
The volume of water drawn out of the tank $=15\times 6\times 1=90$ cubic metres, which is the same as obtained above.
It can be noted that the initial height of the water is not necessary to be provided to solve the question.
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