Question

A screw gauge has a least count of $0.01$ mm. Using this screw gauge a measurement was done. When nothing was present in the jaw, zero of circular scale was above reference line by $3$ units. When a sphere was kept between the jaws, main scale reads $1$ mm and $51^{\text{st}$ division of circular scale coincides with reference line. Find the actual diameter of the ball:}

• A. $1.54$ mm
• B. $1.48$ mm
• C. $1.51$ mm
• D. $1.53$ mm

Hint:

Always check zero error before final calculation. Zero above reference line $\Rightarrow$ negative zero error, which must be added to the observed reading.

Answer 1

The Correct Option is A

Concept: In screw gauge measurements:
Observed reading $=$ Main scale reading $+$ (Circular scale reading $\times$ Least count)
Zero error must be corrected to obtain the true reading
If zero of circular scale is above reference line, the zero error is negative
Step 1: Least count \[ \text{L.C.} = 0.01 \text{ mm} \]
Step 2: Zero error Zero of circular scale is above reference line by $3$ divisions. \[ \text{Zero error} = -3 \times 0.01 = -0.03 \text{ mm} \]
Step 3: Observed reading Main scale reading: \[ \text{MSR} = 1 \text{ mm} \] Circular scale reading: \[ \text{CSR} = 51 \] \[ \text{Observed reading} = 1 + (51 \times 0.01) = 1 + 0.51 = 1.51 \text{ mm} \] Step 4: Apply zero correction \[ \text{True diameter} = \text{Observed reading} - (\text{Zero error}) \] \[ = 1.51 - (-0.03) = 1.54 \text{ mm} \]
Step 5: Hence, the actual diameter of the ball is: \[ \boxed{1.54 \text{ mm}} \]