Question

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings $5.50mm$, $5.55mm$, $5.45mm$, $5.65mm$. The average of these four readings is $5.5375mm$ and the standard deviation of the data is $0.07395mm$. The average diameter of the pencil should therefore be recorded as:

• A. $(5.54\pm 0.07)mm$
• B. $(5.5375\pm 0.0740)mm$
• C. $(5.5375\pm 0.0739)mm$
• D. $(5.538\pm 0.074)mm$

Answer 1

The correct option is A

$(5.54\pm 0.07)mm$

Step 1. Given Data,

Four reading of Vernier Calipers are $5.50mm$, $5.55mm$, $5.45mm$, $5.65mm$,

The average of these four readings is $5.5375mm$,

Standard Deviation of the data is $0.07395mm$.

Step 2. The average diameter of the pencil should be recorded as,

1. The significant rule says that reading should have the same significant figure as that of the reading given.
2. If we do round off $5.5375$then we will get $5.54$ as $7>5$.
3. If we do round off $0.07395$ then we will get $0.07$ as $3<5$.
4. Average diameter will be $(5.54\pm 0.07)mm$ as $0.07mm$is standard deviation of the data.

Hence, option (A) is correct.