Given the following:
Main scale reading: 6.4 cm
Vernier scale reading: [tex]4 \times 0.01 cm=0.04 cm[/tex]
Vernier constant = 0.1 mm
[tex]\begin{aligned}
\text { TS } & =\text { MSR }+ \text { USR } \\
\text { TS } & =0.04 cm+6.4 cm \times \\
\text { TS } & =6.4 cm+0.04 \times 0.1 \\
& =6.4 cm+
\end{aligned}[/tex]
What is the precise depth of the beaker, assuming there is no zero error?
Answer (2)
The precise depth of the beaker is 6.44 cm, calculated by adding the main scale reading of 6.4 cm to the Vernier scale reading of 0.04 cm. The total is given by the formula TS = MSR + VSR. Thus, TS = 6.4 cm + 0.04 cm = 6.44 cm.
;
Identify the main scale reading (MSR) as 6.4 c m and the Vernier scale reading (VSR) as 0.04 c m .
Apply the formula TS = MSR + V SR to find the total reading.
Calculate the total reading: TS = 6.4 c m + 0.04 c m = 6.44 c m .
State the final answer: 6.44 c m .
Explanation
Understanding the Problem We are given the main scale reading (MSR) as 6.4 cm and the Vernier scale reading (VSR) as 4 × 0.01 c m = 0.04 c m . We need to find the total reading (TS), which represents the precise depth of the beaker. The formula for total reading is TS = MSR + VSR.
Calculating the Total Reading Now, we will calculate the total reading by adding the main scale reading and the Vernier scale reading: TS = MSR + V SR TS = 6.4 c m + 0.04 c m
Final Calculation Adding the two values, we get: TS = 6.44 c m The Vernier constant is not needed in this calculation since the Vernier scale reading is already provided.
Conclusion Therefore, the precise depth of the beaker is 6.44 cm.
Examples
Imagine you're measuring the thickness of a textbook using a ruler. The main scale gives you a reading of 2 cm, and the Vernier scale helps you get a more precise reading of 0.05 cm. By adding these two readings, you find the textbook's thickness to be 2.05 cm. This principle is used in various precise measurements, such as in engineering, manufacturing, and scientific research, where accuracy is crucial.
