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

To find the actual diameter of the ball using the screw gauge, we must consider both the readings obtained and the zero error in the gauge.

1. Identifying the Zero Error:
   • The screw gauge shows a zero error if the zero division on the circular scale is above or below the reference line when the gauge is fully closed.
   • According to the question, when nothing is present between the jaws, the zero of the circular scale is above the reference line by 3 units. This indicates a positive zero error of 3 units.
2. Calculate the Measured Diameter:
   • The main scale reading is given as 1 mm.
   • The circular scale reading is given by the 51st division coinciding with the reference line. Since the least count of the screw gauge is \(0.01\) mm, the actual circular reading is: \(51 \times 0.01 = 0.51 \, \text{mm}\).
   • Thus, the measured diameter is: \(1 + 0.51 = 1.51 \, \text{mm}\).
3. Adjust for Zero Error:
   • Since there is a positive zero error of 3 units, this corresponds to \(3 \times 0.01 = 0.03 \, \text{mm}\).
   • To correct for this error, we add the zero error to the measured value: \(1.51 + 0.03 = 1.54 \, \text{mm}\).

Hence, the actual diameter of the ball is 1.54 mm.