Question

Measure of two quantities along with the precision of respective measuring instrument is $A = 2.5m{s^{ - 1}} \pm 0.5m{s^{ - 1}}$ and $B = 0.10s \pm 0.01s$ the value of AB will be
A. $\left( {0.25 \pm 0.08} \right)m$
B. $\left( {0.25 \pm 0.8} \right)m$
C. $\left( {0.25 \pm 0.05} \right)m$
D. $\left( {0.25 \pm 0.135} \right)m$

Answer 1

Hint: In the above question, we are provided with two quantities along with the precision of the respective instrument and we have to find the product of those quantities. The first term is calculated by simply multiplying both the first term of the two quantities and the error is calculated by the formula $\dfrac{{\Delta AB}}{{AB}} = \dfrac{{\Delta A}}{A} + \dfrac{{\Delta B}}{B}$.

Complete step by step answer:
Here in the question, we are given that $A = 2.5m{s^{ - 1}} \pm 0.5m{s^{ - 1}}$ and $B = 0.10s \pm 0.01s$. We have to find the AB. These two quantities are given in the form of precision which means the closeness of two or more measurements to each other
Firstly, multiplying the first terms,
$AB = \left( {2.5} \right)\left( {0.10} \right) = 0.25m$

Now, calculating the error by the formula
$\dfrac{{\Delta AB}}{{AB}} = \dfrac{{\Delta A}}{A} + \dfrac{{\Delta B}}{B}$
Substituting the values,
$\dfrac{{\Delta AB}}{{AB}} = \dfrac{{0.5}}{{2.5}} + \dfrac{{0.01}}{{0.1}} \\
\Rightarrow\dfrac{{\Delta AB}}{{AB}} = 0.2 + 0.1 \\
\Rightarrow\dfrac{{\Delta AB}}{{AB}} = 0.3$
Whereas, we had already found the AB,
$\Delta AB = \left( {0.3} \right) \times 0.25 \\
\therefore\Delta AB= 0.075 \approx 0.08m$
Hence, combining $AB = \left( {0.25 \pm 0.08} \right)m$

So, the correct option is A.

Note: Precision for an instrument is defined as the spread in the measurements. The multiplying of the error is a small quantity. We can neglect those quantities for our convenience. High precision instruments are the ideal instruments as the lower the values show the productive results.