if percentage error in the measurement of a cube l is 3%.find the error in it's volume if the given volume of cube l³
Answers
Answer:
Your question is open to interpretation!
When you say that the error is in the side, do you mean that the error occurs in:
the length of one edge; or
the area of one face
In either case, the resultant object would no longer be a cube. So, perhaps, you mean that the error occurs in the length of all 12 edges or the area of all 6 faces. Further, these errors must all be in the same “direction” - e.g. if you increase the area of some faces by 3% and decrease others by 3%, you would no longer have a cube.
First case - we will assume that the error is in the length of the edges.
If the cube is supposed to have edges of length x , its volume would be x3.
If we increase the length of the edges by 3%, the actual length of each edge would be 1.03x , thus the volume would be (1.03x)3=1.092727x3
Subtracting the volume it is supposed to have, we have an error of 0.092727x3 . Dividing this by the volume it is supposed to have, the error is 0.092727 i.e. 9.2727 .
If the error was a decrease, then each edge would have a length of 0.97x thus the volume would be (0.97x)3=0.912673x3
The decrease in volume = (1−0.912673)x3=0.087327x3 . Dividing this by the volume it is supposed to have, we have an error of 0.087327 , i.e. 8.7327 .
Second case - we will assume that the error is in the area of each face.
If the area of each face increases by 3% then, as each face must remain square, this means than each edge must increase by 1.03−−−−√x≈0.014889x . Thus the volume would be approximately (1.014889x)3≈1.045336x3 . This equates to an approximate increase in volume of 4.5336 .
If the area of each face decreases by 3%, then each edge must have a length of 0.97−−−−√x≈0.984886x . Thus the volume would be approximately (0.984886x)3≈0.955339x3 . This equates to an approximate decease of 4.4661 .
Step-by-step explanation:
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