Question

Measure of two quantities along with the precision of respective measuring instrument is $A=2.5m/s\underset{–}{+}0.5m/s$
$B=0.10s\underset{–}{+}0.01s$. The value of AB will be

• A. $\left(0.25\underset{–}{+}0.5\right)m$
• B. $\left(0.25\underset{–}{+}0.05\right)m$
• C. $\left(0.25\underset{–}{+}0.08\right)m$
• D. $\left(0.25\underset{–}{+}0.135\right)m$

Answer 1

The correct option is C $\left(0.25\underset{–}{+}0.08\right)m$

Step:1 Calculate the value of AB.

Given, $A=2.5m/s\underset{–}{+}0.5m/s$
$B=0.10s\underset{–}{+}0.01s$
by multiplying the first term of A and B,
AB = (2.5) (0.10)
AB = 0.25 m

Step:2 Find the error in the value AB.

Formula used: $\frac{\Delta AB}{AB}=\frac{\Delta A}{A}+\frac{\Delta B}{B}$
Error in A, $\Delta A$ = 0.5 m/s
Error in B, $\Delta B$ = 0.01 s
So, error in AB,
$\frac{\Delta AB}{AB}=\frac{\Delta A}{A}+\frac{\Delta B}{B}$

$\frac{\Delta AB}{0.25}=\frac{0.5}{2.5}+\frac{0.01}{0.1}$

$\Delta AB=0.075$
Rounding off to two significant figures, $\Delta AB=0.08$ m
So, AB = $\left(0.25\underset{–}{+}0.08\right)m$

Final answer: (a) $\left(0.25\underset{–}{+}0.08\right)m$