Question

Measure of two quantities along with the precision of respective measuring instrument is $A=2.5m{s}^{-1}\pm 0.5m{s}^{-1}$, $B-0.10s\pm 0.01s$ The value of $AB$ will be

• A. $(0.25\pm 0.08)m$
• B. $(0.25\pm 0.5)m$
• C. $(0.25\pm 0.05)m$
• D. $(0.25\pm 0.135)m$

Answer 1

Answer: (A)

Here, $A=2.5m{s}^{-1}\pm 0.5m{s}^{-1},B=0.10s\pm 0.01s$
$AB=\left(2.5m{s}^{-1}\right)(0.10s)=0.25m$
$\frac{\Delta AB}{AB}=\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)=\left(\frac{0.5}{2.5}+\frac{0.01}{0.10}\right)=0.3$
$\Delta AB=0.3\times 0.25m=0.075m=0.08m$
(Rounded off to two significant figures)
The value of $AB(0.25\pm 0.8)m$