Question

The error in the measurement of force acting normally on a square plate is 3%. If the error in the measurement of the side of the plate is 1%, then the error in the determination of the pressure acting on the plate is

• A. 4%
• B. 3%
• C. 5%
• D. 6%

Hint:

When calculating the maximum percentage error for a formula involving products and divisions, simply add the individual percentage errors, making sure to multiply each by the magnitude of its corresponding exponent in the formula.

Answer 1

The Correct Option is C

Step 1: Write the formula for pressure.
Pressure ($P$) is defined as the normal force ($F$) acting per unit area ($A$). \[ P = \frac{F}{A}. \] For a square plate with side length $L$, the area is $A = L^2$. So, the formula for pressure is: \[ P = \frac{F}{L^2}. \]

Step 2: Apply the formula for propagation of errors.
For a quantity $Z = \frac{X^a}{Y^b}$, the maximum fractional error is given by the sum of the fractional errors of its components, weighted by the magnitude of their exponents: \[ \frac{\Delta Z}{Z} = a \frac{\Delta X}{X} + b \frac{\Delta Y}{Y}. \] In our case, $P = F^1 L^{-2}$. The maximum fractional error in pressure is: \[ \frac{\Delta P}{P} = (1)\frac{\Delta F}{F} + (2)\frac{\Delta L}{L}. \]

Step 3: Convert to percentage errors.
Multiplying the entire equation by 100 gives the relationship for percentage errors: \[ \left(\frac{\Delta P}{P} \times 100%\right) = \left(\frac{\Delta F}{F} \times 100%\right) + 2 \left(\frac{\Delta L}{L} \times 100%\right). \] \[ %\text{Error in } P = %\text{Error in } F + 2 \times (%\text{Error in } L). \]

Step 4: Substitute the given values.
We are given that the percentage error in force is 3% and the percentage error in the side length is 1%. \[ %\text{Error in } P = 3% + 2 \times (1%). \] \[ %\text{Error in } P = 3% + 2% = 5%. \] \[ \boxed{%\text{Error in } P = 5%}. \]