Measure of two quantities along with the precision of respective measuring instrument is A=2.5 m s-1 ± 0.5 m s-1, B=0.10 s ± 0.01 s.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
$(0.25 \pm 0.135) m$

Solution:

Here,

$A = 2.5 m s^{- 1} \pm 0.5 m s^{- 1}, B = 0.10 s \pm 0.01 s AB = (2.5 m s^{- 1}) (0.10 s) = 0.25 m \frac{∆ AB}{AB} = (\frac{∆ A}{A} + \frac{∆ B}{B}) = (\frac{0.5}{2.5} + \frac{0.01}{0.10}) = 0.3 ∆ AB = 0.3 \times 0.25 m = 0.075 m = 0.08 m$

(Rounded off to two significant figures)

The value of $AB$ is $(0.25 \pm 0.08) m$ .

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

Solution:

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