Ncert Exercises & Solutions

The measure of two quantities along with the precision of respective measuring instrument is $A = 2 .5 {ms}^{- 1} \pm 0 .5 {ms}^{- 1}, B = 0 .10 s \pm 0 .01 s$ . The value of AB will be:

1. (0.25 $\pm$ 0.08) m

2. (0.25 $\pm$ 0.5) m

3. (0.25 $\pm$ 0.05) m

4. (0.25 $\pm$ 0.135) m

(1) Hint: The relative errors in the multiplication of quantities get added.

Step 1: Find the value of AB.

Given,

$\begin{array}{l} A = 2 .5 {ms}^{- 1} \pm 0 .5 {ms}^{- 1}, B = 0 .10 s \pm 0 .01 s \\ x = AB = (2 .5) (0 .10) = 0 .25 m \\ \frac{Δx}{x} = \frac{ΔA}{A} + \frac{ΔB}{B} \\ = \frac{0 .5}{2 .5} + \frac{0 .01}{0 .10} = \frac{0 .05 + 0 .025}{0 .25} = \frac{0 .075}{0 .25} \end{array}$
$\begin{array}{l} Δx = 0 .075 = 0 .08 m, rounding off to two significant figures. \\ AB = (0 .25 \pm 0 .08) m \end{array}$

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