Tag: units and measurement: error analysis

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Questions Related to units and measurement: error analysis

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

A vernier callipers has its main scale of 10 cm equally divided into 200 equal parts. Its vernier scale of 25 divisions coincides with 12 mm on the main scale. The least count of the instrument is - 

  1. 0.020 cm

  2. 0.002 cm

  3. 0.010 cm

  4. 0.001 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$10\ cm$ of Main Scale $=200$ div of Main scale 

$\Rightarrow IMSD=0.05\ cm$
given $25$ vernier scale division coincide with $12\ mm$ on main scale
$\Rightarrow 25VSD=12\ mm$ on main scale
$=\dfrac{12}{0.5}=24\ MSD$
$\Rightarrow 25VSD=24MSD$
$\Rightarrow 1VSD=\dfrac{24}{25}MSD=0.048\ cm$
Least Count $\Rightarrow 1MSD-1VSD$
$\Rightarrow (0.050-0.048)cm$
$\Rightarrow 0.002\ cm$

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

Calculate focal length of a spherical mirror from the following observations.

Object distance, u=(50.1±0.5)u=(50.1±0.5) cm
Image distance, v=(20.1±0.2) cm

  1. $\displaystyle \left ( 14.3\pm 0.4\right )$ cm

  2. $\displaystyle \left ( 14.3\pm 0.2\right )$ cm

  3. $\displaystyle \left ( 12.3\pm 0.4\right )$ cm

  4. $\displaystyle \left ( 12.3\pm 0.2\right )$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let focal length  of the mirror be $f$  cm.

$\dfrac{1}{v} + \dfrac{1}{u} =  \dfrac{1}{f}$

$\dfrac{1}{20.1} + \dfrac{1}{50.1} =  \dfrac{1}{f}$            $\implies  f =  14.3$ cm

As, $f = \dfrac{uv}{u+v}$
$\therefore$ Taking log and differentiating, we get $    \Delta f = f \bigg( \dfrac{\Delta u}{u}  + \dfrac{\Delta v}{v}  + \dfrac{\Delta u + \Delta v}{u+ v} \bigg)$
  $    \Delta f = 14.3 \bigg( \dfrac{0.5}{50.1}  + \dfrac{ 0.2}{20.1}  + \dfrac{0.5 +0.2}{50.1+20.1} \bigg)  = 0.4$  cm

Thus focal length of the mirror $ = (14.3 \pm 0.4)$  cm

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

In an experiment the following observations were recorded 
$L=2.890$ metre
$\mathrm{M}=3.0$ kg
$\ell=0.87$ cm
$\mathrm{D}=0.041$ cm
$\mathrm{g}=981\mathrm{c}\mathrm{m}/\sec^{2}$
and the formula used for calculatlon of Young's modulus ($\mathrm{Y}$) is $\displaystyle \mathrm{Y}=\frac{\mathrm{M}\mathrm{g}\mathrm{L}}{\pi \mathrm{r}^{2}l}$
What is the maximum possible error expressed in percentage?

  1. 6.36%

  2. 4.8%

  3. 3%

  4. 1%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Y=\dfrac{4MgL}{\pi D^2 l}$

Hence, $log(Y)=log4+log(M)+log(g)+log(L)-2log(D)-log(l)$
$\dfrac{\Delta Y}{Y}=\dfrac{\Delta M}{M}+\dfrac{\Delta g}{g}+\dfrac{\Delta L}{L}+\dfrac{2\Delta D}{D}+\dfrac{\Delta l}{l}$
For maximum possible error,
$\Delta M=0.05kg$
$\Delta g=0.5cm/sec^2$
$\Delta L=0.0005metre$
$\Delta D=0.0005cm$
$\Delta l=0.005cm$
Putting the values in $\dfrac{\Delta Y}{Y}\times 100$
Hence maximum possible error is 4.8%