Question 17

1. Determine Least Count ($LC$):

Main scale division ($MSD$) = $0.5\text{ mm}$.

One full rotation moves the scale by $2$ divisions $= 2 \times 0.5 = 1.0\text{ mm}$.

Pitch $= 1.0\text{ mm}$.

$LC = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{1.0\text{ mm}}{100} = 0.01\text{ mm}$.

2. Zero Error:

When arms touch, reading $= 0 + 4 \times 0.01 = 0.04\text{ mm}$.

Since the reading is positive, the Zero Error is $+0.04\text{ mm}$.

3. Calculate Measured Diameters:

Attempt 1: $MSR = 4 \times 0.5 = 2.0\text{ mm}$, $CSR = 20$. Measured $d_1 = 2.0 + 20(0.01) = 2.20\text{ mm}$.

Attempt 2: $MSR = 4 \times 0.5 = 2.0\text{ mm}$, $CSR = 16$. Measured $d_2 = 2.0 + 16(0.01) = 2.16\text{ mm}$.

4. Corrected Diameters:

$D_1 = 2.20 - 0.04 = 2.16\text{ mm}$.

$D_2 = 2.16 - 0.04 = 2.12\text{ mm}$.

Mean $D = \frac{2.16 + 2.12}{2} = 2.14\text{ mm}$.

Error $\Delta D = \frac{|2.16 - 2.14| + |2.12 - 2.14|}{2} = 0.02\text{ mm}$.

So, Diameter $= 2.14 \pm 0.02\text{ mm}$.

5. Calculate Area:

$A = \frac{\pi D^2}{4} = \frac{\pi (2.14)^2}{4} \approx 1.1449\pi \approx 1.14\pi\text{ mm}^2$.

Relative Error in area = $\frac{\Delta A}{A} = 2 \frac{\Delta D}{D} \approx 0.02$

Error in area $\Delta A = A \times 2 \frac{\Delta D}{D} = 1.14\pi \times 2 \times \frac{0.02}{2.14} \approx 0.0213\pi \approx 0.02\pi\text{ mm}^2$.

So, Area $= \pi(1.14 \pm 0.02)\text{ mm}^2$.