## Question

secondary 3 | Chemistry

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##### Chase Tan

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Physics, vernier caliper

Date Posted: 5 months ago
Views: 60
Eric Nicholas K
5 months ago
To find the reading, these are the steps.

First, we look at the point in which the marking below has not crossed the marking below.

In the case of Q1, we see that the 0 mm marking below has definitely passed the 0.9 cm mark and has passed the 1.0 cm mark on top, but has not passed the 1.1 cm mark yet. This corresponds to a reading of 1.0__ cm where __ is a number to be found.

Next, for the missing digit, we look at the point where one of the lines on top is in line with one of the lines below. Then, we read off the number at the bottom part.

Here, in Q1, it just so happens that the '0' mark below coincides with the line on top. So the missing digit is 0.

Hence, the reading is 1.00 cm.

In the case of Q6, we see that the 0 mm marking below has passed the 6.8 cm mark, but not the 6.9 cm mark. So the reading is 6.8__ cm.

The '5' marking at the bottom coincides with one of the lines on top, so the missing digit is 5.

Hence, the reading is 6.85 cm.

In the case of Q2, we see that the 0 mm marking below has passed the 5.2 cm mark, but not the 5.3 cm mark. So the reading is 5.2__ cm.

The '3' marking at the bottom coincides with one of the lines on top, so the missing digit is 3.

Hence, the reading is 5.23 cm.

All the rest of the questions use similar approaches.

My solutions are

Q1: 1.00 cm
Q2: 5.23 cm
Q3. 2.88 cm
Q4: 9.16 cm
Q5: 6.07 cm
Q6: 6.85 cm
Q7: 11.23 cm
Q8: 8.38 cm
Q9: 3.89 cm
Q10: 2.21 cm

We do not need to adjust for zero error since the zero error is given to be 0.00 cm.