Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • You measure two quantities as A = 1.0 \; m \pm 0.2 \; m, B = 2.0 \; m \pm 0.2 \; m. We should report correct value for √ AB as: a) 1.4 \; m \pm 0.4 \; m b) 1.41 \; m \pm 0.15\; m c) 1.4 \; m \pm 0.3 \; m d) 1.4 \; m \pm 0.2\; m

You measure two quantities as A = 1.0 \; m \pm 0.2 \; m, B = 2.0 \; m \pm 0.2 \; m. We should report correct value for √ AB as:

a) 1.4 \; m \pm 0.4 \; m

b) 1.41 \; m \pm 0.15\; m

c) 1.4 \; m \pm 0.3 \; m

d) 1.4 \; m \pm 0.2\; m

Answers (1)

The answer is the option (d) 1.4\pm 0.2\; m

There are two significant figures in 1.0 & 2.0

\sqrt{AB}=\sqrt{1.0\times 2.0}=\sqrt{2}

=1.414\; m

After rounding up 1.0 &2.0, we get two significant features, i.e.

X=\sqrt{AB}

=1.4

Now, \frac{\Delta X}{X}=\frac{1}{2}[\frac{\Delta A}{A}+\frac{\Delta B}{B}]

\frac{\Delta X}{1.4}=0.1\left [ \frac{2.0+1.0}{1.0\times 2.0} \right ]

\Delta X=0.1(\frac{3}{2})((1.4)

=0.21\ m after rounding up we get 0.2

Thus, option (d) is the right answer.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question