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Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The mass of a beaker is found to be ($10.1 \pm 0.1)gm $ when empty and $(17.3 \pm 0.1)gm$ when partially filled with a liquid. What is the best value for the mass of the liquid together with the possible limits of accuracy?

  1. $(7.2  \pm 0.2)gm$

  2. $(7.2  \pm  0.1)gm$

  3. $(7.1  \pm  0.2)gm$

  4. $(7.2  \pm  0.3)gm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here mass of the liquid , $m=m _2-m _1=17.3-10.1=7.2 gm$

So the relative error, $\dfrac{\Delta m}{m}=\pm\left[\dfrac{\Delta m _1}{m _1}+\dfrac{\Delta m _2}{m _2}\right]$
or $\Delta m=\pm\left[\dfrac{\Delta m _1}{m _1}+\dfrac{\Delta m _2}{m _2}\right]\times m=\pm[0.1/10.1+0.1/17.3]\times 7.2=\pm 0.1 gm$
Thus mass of liquid with possible accuracy $=(7.2\pm 0.1) gm$

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Measure of two quantities along with the precision of respective measuring instrument is:
$A = 2.5 ms^{-1}\pm 0.5 ms^{-1}$
$B = 0.10s \pm 0.01 s$
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. $(0.25\pm 0.135)m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given :  $\Delta A = 0.5 \ m/s$        $\Delta B = 0.01 \ s$

The absolute value of AB  $ = 2.5\times 0.10 = 0.25 \ m$

Relative error in AB,    $\dfrac{\Delta AB}{AB} = \dfrac{\Delta A}{A} +\dfrac{\Delta B}{B}$

$\therefore$   $\dfrac{\Delta AB}{0.25} = \dfrac{0.5}{2.5} +\dfrac{0.01}{0.10}$

$\implies \ \Delta AB = 0.075 = 0.08 \ m$
Thus value of AB is  $(0.25\pm0.08) \ m$