Tag: significant figures

Questions Related to significant figures

Add $3.8 \times 10^{-6} to 4.2 \times 10^{-5}$ with due regard to significant figures.

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

  1. $~4.6 \times 10^{-5}$

  2. $~4.6 \times 10^{-6}$

  3. $~5.6 \times 10^{-6}$

  4. $~6.6 \times 10^{-5}$

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Correct Option: A
Explanation:

$3.8\times { 10 }^{ -6 }+42\times { 10 }^{ -6 }=45.8\times { 10 }^{ -6 }=4.58\times { 10 }^{ -5 }$

                                                                      $=4.6\times { 10 }^{ -5 }$.

The radius of the earth is $6.37 \times 10^6 m$ and its mass is $5.975 \times 10^{24} kg$. Find the earth's average density to appropriate significant figures.

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

  1. $5 \times 10^3 kg m^{-3}$

  2. $5.52 \times 10^3 kg m^{-3}$

  3. $2 \times 10^3 kg m^{-3}$

  4. $5.52 \times 10^3 kg m^{-4}$

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Correct Option: B

The number of significant figures in $5.69 \times 10^{15}$kg is

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

  1. $1$

  2. $2$

  3. $3$

  4. $18$

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Correct Option: C
Explanation:

No. of significant figures in $\underbrace { 5.69 } \times { 10 }^{ 15 }$ Kg is $3$.

                                           Digits that cannot meaning to measurement and measurement resolution.

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 - 

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

  1. 0.020 cm

  2. 0.002 cm

  3. 0.010 cm

  4. 0.001 cm

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Correct Option: B
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 $\Rightarrow 1MSD-1VSD$
$\Rightarrow (0.050-0.048)cm$
$\Rightarrow 0.002\ cm$

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

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

  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

Show answer
Correct Option: A
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

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?

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

  1. 6.36%

  2. 4.8%

  3. 3%

  4. 1%

Show answer
Correct Option: B
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%