Question

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be \( \frac{x}{100} \), where \( x \) is:

Hint:

When dealing with measurements involving areas, remember that errors in both dimensions contribute to the total error.

Answer 1

Correct Answer: 3

Step 1: Given Data

• Length of the sheet, \( L = 5 \) mm
• Breadth of the sheet, \( B = 2.5 \) mm
• Pitch of screw gauge = 0.75 mm
• Total divisions in the circular scale = 15
• Least count of the screw gauge = \( \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.75}{15} = 0.05 \) mm

Step 2: Error Calculation

The absolute error in measurement using the screw gauge is the least count, i.e., \( 0.05 \) mm.

Fractional error in length measurement: \[ \frac{\Delta L}{L} = \frac{0.05}{5} = 0.01 \]

Fractional error in breadth measurement: \[ \frac{\Delta B}{B} = \frac{0.05}{2.5} = 0.02 \]

Step 3: Maximum Fractional Error in Area Calculation

Area, \( A = L \times B \)

Maximum fractional error in area: \[ \left( \frac{\Delta A}{A} \right)_{\text{max}} = \left( \frac{\Delta L}{L} + \frac{\Delta B}{B} \right) \] \[ = 0.01 + 0.02 = 0.03 \]

Step 4: Converting to Given Form

The fractional error is given as \( \frac{x}{100} \). \[ 0.03 = \frac{x}{100} \] \[ x = 3 \]

Final Answer:

\[ \boldsymbol{x = 3} \]

Answer 2

Step 1: Understand the given data.
Length of sheet \( L = 5 \, \text{mm} \)
Breadth of sheet \( B = 2.5 \, \text{mm} \)
Pitch of screw gauge \( = 0.75 \, \text{mm} \)
Number of circular scale divisions \( = 15 \)

Step 2: Calculate least count (L.C.) of the screw gauge.
\[ \text{Least count} = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{0.75}{15} = 0.05 \, \text{mm} \] Hence, the least count (smallest measurable length) is \( 0.05 \, \text{mm} \).

Step 3: Find fractional error in the measurement of length and breadth.
Fractional error in a measured quantity is given by: \[ \frac{\Delta L}{L} = \frac{\text{Least count}}{\text{Measured value}} \] So for each dimension: \[ \frac{\Delta L}{L} = \frac{0.05}{5} = 0.01 \] \[ \frac{\Delta B}{B} = \frac{0.05}{2.5} = 0.02 \]

Step 4: Find fractional error in area.
The area \( A = L \times B \).
The fractional error in area is the sum of fractional errors in \( L \) and \( B \): \[ \frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B} \] \[ \frac{\Delta A}{A} = 0.01 + 0.02 = 0.03 \] or \( \frac{3}{100} \).

Step 5: Conclusion.
The maximum fractional error in the measurement of area is \( \frac{x}{100} \), where \( x = 3 \).

Final Answer:
\[ \boxed{x = 3} \]