When the screw and stud touch each other, the edges of a certain screw gauge is on left of the \(\mathrm{O}\) mark on the main scale and the $96^{\text {th }}$ division of the circular scale coincides with the circular line of graduation then what is the value of zero error? (A) zero error \(=+0.96 \mathrm{~mm}\) (B) zero error \(=-0.96 \mathrm{~mm}\) (C) zero error \(=+0.04 \mathrm{~mm}\) (D) zero error \(=-0.04 \mathrm{~mm}\)

To determine whether the zero error is positive or negative, we observe the location of the O mark on the main scale when the screw gauge is closed without any object. In this problem, the O mark is on the left side of the circular graduation line, meaning that the reading is greater than zero. This indicates a positive zero error.

We know that the 96th division on the circular scale coincides with the circular graduation line. To calculate the zero error value, we need to multiply the 96 divisions by the least count of the screw gauge. The least count (LC) of a screw gauge can be calculated as follows: LC = pitch / number of divisions on circular scale In this problem, the pitch and number of divisions on the circular scale are not given. However, we can deduce that the least count is 0.01 mm because it is a standard in most screw gauge problems. So, the zero error value can be calculated as: Zero error value = positive zero error × least count = 96 divisions × 0.01 mm/division = 0.96 mm This means the correct answer is: (A) zero error = +0.96 mm