Question

A force \( F \) is applied upon a square with side \( L \). Errors in \( L \) and \( F \) are 2% and 4% respectively. Error in measuring pressure will be

• A. 4 percent
• B. 6 percent
• C. 2 percent
• D. 8 percent

Hint:

When calculating the percentage error in derived quantities, remember to add the percentage errors for multiplication and apply the appropriate exponent for powers (e.g., \( L^2 \)).

Answer 1

The Correct Option is B

Pressure is calculated from two measured quantities, force \( F \) and side length \( L \), so its percentage error is built up from both.
The error contributed by \( F \) is 4%, and the error carried through from \( L \) is 2%.
Adding these two contributions together: \( 4\% + 2\% = 6\% \).
So the answer is 6 percent.

Answer 2

Take a numerical example to see how the two individual errors combine into the error of the pressure reading.

1. 4 percent: Suppose the true force is \( F = 100 \) units; a 4% error means the measured force lies within about \( \pm 4 \) units of the true value. This is only one of the two sources of uncertainty in \( P \), so 4% by itself is incomplete.
2. 6 percent: Suppose the true side length is \( L = 100 \) units, with a 2% error meaning about \( \pm 2 \) units of uncertainty. When force and length uncertainties both feed into the pressure reading, their percentage contributions add up: \( 4\% + 2\% = 6\% \), giving the combined uncertainty in \( P \).
3. 2 percent: This matches only the length's own uncertainty and ignores the force measurement's separate 4% contribution.
4. 8 percent: This does not correspond to directly adding the two reported percentage errors (4% and 2%) for this pair of measurements.

Working through the numbers this way confirms that the two given uncertainties combine directly to a total of 6% in the pressure.

Therefore, the correct answer is 6 percent.