Tag: units and measurements

Questions Related to units and measurements

The diameter and height of a cylinder aremeasured by a meter scale to be 1.6$\pm 0.1$$\mathrm { cm }$ and 5.0$\pm 0.1 \mathrm { cm }$ , respectively. What will be the value of its volume inappropriate significant figures? 

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

  1. 4300$\pm 80 \mathrm { cm } ^ { 3 }$

  2. 4260$\pm 80 \mathrm { cm } ^ { 3 }$

  3. 4264$\pm 81 \mathrm { cm } ^ { 3 }$

  4. 10.0$\pm .325 \mathrm { cm } ^ { 3 }$

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Correct Option: D

In the final answer of the expression  $\dfrac { ( 29.2 - 20.2 ) \left( 1.79 \times 10 ^ { 5 } \right) } { 1.37 }.$  The number of significant figures 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. $4$

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Correct Option: C

The number of significant figures reduced in

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

  1. Addition

  2. Substraction

  3. Multiplication

  4. division

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Correct Option: C

The measurement $2.3456789\ km$ is rounded to $4$ significant figures. The value of measurement will be written as

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

  1. $2.3457$

  2. $2.346$

  3. $2.345$

  4. $2.3456$

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Correct Option: B

Find the value of following on the basis of significant figure rule:
The height of a man is $5.87532$ ft. But measurement is correct upto three significant figures. The correct height is?

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

  1. $5.86$ ft

  2. $5.87$ ft

  3. $5.88$ ft

  4. $5.80$ ft

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Correct Option: C
Explanation:

Significant figures upto three significant figures for the given question is 5.875 . Therefore, 5 is equal to 5 and we can round off the digit preceding 5 is increased by 1.


Therefore , now the correct answer vis 5.88 ft.
Therefore answer  c is correct.

Find the value of following on the basis of significant figure rule:
$1.00\times 2.88$ is equal to:

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

  1. $2.88$

  2. $2.880$

  3. $2.9$

  4. None of these

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Correct Option: A

Two resistances $r _1=(5.0\pm 0.2)\Omega$ and $r _2=(10.0\pm 0.1)\Omega$ are connected in parallel. Find the value of equivalent resistance with limits of percentage error.

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

  1. $r _p=4\Omega\pm $7%

  2. $r _p=3.3\Omega\pm $14%

  3. $r _p=3.3\Omega\pm $3.5%

  4. $r _p=3.3\Omega\pm $7%

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Correct Option: D
Explanation:

Here, $r _1=5 \Omega , r _2=10 \Omega, \Delta r _1=0.2 \Omega $ and $\Delta r _2=0.1 \Omega$


The equivalent resistance, $r _p=\dfrac{r _1r _2}{r _1+r _2}=\dfrac{5\times 10}{5+10}=3.3 \Omega$

$\dfrac{\Delta r _p}{r _p} = \dfrac{\Delta r _1}{r _1}+\dfrac{\Delta r _2}{r _2}+\dfrac{\Delta (r _1+r _2)}{r _1+r _2}=\dfrac{0.2}{5}+\dfrac{0.1}{10}+\dfrac{0.2+0.1}{5+10}=0.07$

Thus, The value of equivalent resistance with limits of  % error $=r _p\pm\Delta r _p=3.3 \Omega \pm 7 \%$

The length of on rod $l _1=3.323 cm$ and the other is $l _2=3.321 cm$. Both rods were measured with one measuring instrument with least count 0.001 cm. Then $(l _1-l _2)$ is :

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

  1. $(0.002\pm 0.001)cm$

  2. $(0.002\pm 0.000)cm$

  3. $(0.002\pm 0.002)cm$

  4. none of these

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Correct Option: C
Explanation:

The error in two readings always adds up.
$l _1-l _2=(3.323\pm 0.001)-(3.321\pm 0.001)=(0.002\pm 0.002)cm$

Area of a square is $(100\pm 2)m^2$. Its side is:

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

  1. $(10\pm 1)m$

  2. $(10\pm 0.1)m$

  3. $(10\pm \sqrt 2)m$

  4. $10\pm \sqrt 2$ %

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Correct Option: A
Explanation:

$Area=(Length)^2$
$Length =(Area)^{1/2}$
             $=(100\pm 2)^{1/2}$
             $=(100)^{1/2}\pm \dfrac {1}{2}\times 2$
             $=(10\pm 1)m$

Relative density of a metal may be found with the help of spring balance. In air the spring balance reads $(5.00\pm 0.05)$ N and in water it reads $(4.00\pm 0.05)$N. Relative density would be:

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

  1. $(5.00\pm 0.05)N$

  2. $(5.00\pm$ 11%)

  3. $(5.00\pm 0.10)$

  4. $(5.00\pm$ 6%)

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Correct Option: D
Explanation:

Relative density $=\dfrac {\text {Weight of body in air}}{\text {Loss of weight in water}}$$=\dfrac {5.00}{5.00-4.00}=\dfrac {5.00}{1.00}$

$\dfrac {\Delta \rho}{\rho}\times 100=\left (\dfrac {0.05}{5.00}+\dfrac {0.05}{1.00}\right )\times 100$
                    $=(0.01+0.05)\times 100$
                    $=0.06\times 100$
                    $ =6$% 
$\therefore$ Relative density $=5.00\pm$ 6%