Ncert Exercises & Solutions

You measure two quantities as A = 1.0 m $\pm$ 0.2 m, B = 2.0 m $\pm$ 0.2 m. We should report the correct value for $\sqrt{AB}$ as

1. 1.4 m $\pm$ 0.4 m

2. 1.41 m $\pm$ 0.15 m

3. 1.4 m $\pm$ 0.3 m

4. 1.4 m $\pm$ 0.2 m

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

Step 1: Find the value of $\sqrt{AB}$ .

$\begin{array}{ll} Given, & A = 1 .0 m \pm 0 .2 m, B = 2 .0 m \pm 0 .2 m \\ Let, & Y = \sqrt{AB} = \sqrt{(1 .0) (2 .0)} = 1 .414 m \end{array}$

Rounding off to two significant digit Y 1.4 m.

Step 2: Find the relative error in $\sqrt{AB}$ .

$\begin{array}{l} \frac{ΔY}{Y} = \frac{1}{2} [\frac{ΔA}{A} + \frac{ΔB}{B}] = \frac{1}{2} [\frac{0 .2}{1 .0} + \frac{0 .2}{2 .0}] = \frac{0 .6}{2 \times 2 .0} \\ \Rightarrow ΔY = \frac{0 .6 Y}{2 \times 2 .0} = \frac{0 .6 \times 1 .4}{2 \times 2 .0} = 0 .212 \end{array}$

Rounding off to one significant digit $∆$ Y = 0.2m
Thus, the correct value for $\sqrt{AB} = r + Δr = 1 .4 \pm 0 .2 m$

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