Question

Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter $d$ are given. To obtain the value of $d$, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm. The table shows the readings of LS and CS for the measurements. The value of $d$ (in µm) is:

Hint:

Always remember: True Value = Observed Value - Zero Error. If the observed value of a standard object is greater than its actual value, the instrument has a positive zero error, which must be subtracted from all subsequent readings.

Answer 1

Correct Answer: 1915

Step 1: Understanding the Question:
The problem involves determining the diameter of a wire using a screw gauge. We need to account for the Least Count of the instrument and determine the zero error using the known diameter of Wire-1.

Step 2: Key Formula or Approach:

• Least Count (LC) $= \frac{\text{Pitch}}{\text{Number of circular scale divisions}}$.

• Observed Reading $= \text{LS reading} + (\text{CS reading} \times \text{LC})$.

• True Reading $= \text{Observed Reading} - \text{Zero Error}$.

Step 3: Detailed Explanation:

• Calculate Least Count (LC):
Given Pitch $= 0.5$ mm and total divisions $= 100$.
$LC = \frac{0.5}{100} = 0.005$ mm.

• Determine Zero Error using Wire-1:
Observed diameter of Wire-1 $= LS + (CS \times LC) = 0.5 + (42 \times 0.005) = 0.5 + 0.210 = 0.710$ mm.
The true diameter is given as $0.650$ mm.
Zero Error $= \text{Observed} - \text{True} = 0.710 - 0.650 = +0.060$ mm.

• Calculate diameter $d$ of Wire-2:
Observed diameter of Wire-2 $= LS + (CS \times LC) = 1.5 + (95 \times 0.005) = 1.5 + 0.475 = 1.975$ mm.
True diameter $d = \text{Observed} - \text{Zero Error} = 1.975 - 0.060 = 1.915$ mm.

• Convert to µm:
$d = 1.915 \times 1000 = 1915$ µm.

Step 4: Final Answer:
The value of $d$ is 1915.