Tag: units and measurement: error analysis

Questions Related to units and measurement: error analysis

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

Order of $\frac { 1 } { 8 \times 10 ^ { 9 } }$ is:

  1. <span>$1.77\times {{10}^{-10}}$
    </span>

  2. <span>$1.25\times {{10}^{-10}}$
    </span>

  3. <span>$1.55\times {{10}^{-10}}$
    </span>

  4. <span>$1.95\times {{10}^{-10}}$
    </span>

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac{1}{8\times 10^{9}}=\left(\dfrac{1}{8}\right)\times 10^{-9}=(0.125)\times 10^{-9}$
$=1.25\times 10^{-10}$
Order of $\dfrac{1}{8\times 10^{9}}$ is $10^{-10}$
Option $B$ is correct
Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

The resultant of two equal forces acting at right angles to each other is $1414$ dyne. Find the magnitude of either force.

  1. $960$

  2. $1000$

  3. $1200$

  4. $1414$

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

$F _1=F _2=F(say)$
$\theta=90^0$
$R=1414dyne$
Resultant force,
$R=\sqrt{F _1^2+F _2^2+2F _1F _2cos\theta}$
$1414dyne=\sqrt{F^2+F^2+2F^2cos90^0}=\sqrt{2F^2}$
$1414dyne=\sqrt{2}F$
$F=\dfrac{1414}{\sqrt{2}}=999.84 \approx 1000dyne$ (by rounding off)
The correct option is B.

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

Two constant force $ \overrightarrow F _1 and \overrightarrow F _2 $ acts on a body.these forces displaces the body from point P(1, -2, 3) to Q (2, 3, 7 ) in 2s starting from rest.force $\overrightarrow F _1 $ is of magnitude 9 N and acting along vector $ ( 2 \hat i - 2 \hat j + \hat k ) $ . the positions are in meter. find work done by $ \overrightarrow F _1 $.

  1. $-12J$

  2. $+12J$

  3. $36J$

  4. $-36J$

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done is the dot product of force and displacement. Displacement vector d = Q - P = (2-1)i + (3-(-2))j + (7-3)k = 1i + 5j + 4k. Force F1 = 9 * (2i - 2j + 1k) / sqrt(2^2 + (-2)^2 + 1^2) = 9 * (2i - 2j + 1k) / 3 = 3 * (2i - 2j + 1k) = 6i - 6j + 3k. Work = (6i - 6j + 3k) dot (1i + 5j + 4k) = 6 - 30 + 12 = -12J.

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form
The Earth's radius is $6371km$. The order magnitude of the Earth's radius is
  1. $10^3$

  2. $10^2$

  3. $10^7$

  4. $10^5$

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Order of magnitude is usually written as $10 $ to the $n$th power. The represents the order of magnitude.
e.g., if we write a number $X$ in such a way that
$X = m × 10^n$ then n is order of magnitude.
here, $6371 km = 6371000 m$
$6371000 = 6.371 × 10^6$
we see, $ 6.371 ≥ \sqrt{10}$
so, $0.6371 × 10^7$
, $\dfrac{\sqrt{10}}{\sqrt{10}} ≤ 0.6371 ≤ √10$
7 is our order of magnitude.
Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

Find the order of magnitude of the mass of the star, whose radius is $384 \times 10^6$ m and average density is $4 \times 10^3$ kg $m^{-3}$ :

  1. 30

  2. 29

  3. 24

  4. 21

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Radius of star $r=384\times10^6m$

So, Volume of star is $V=4\pi r^3/3=2.371\times10^{26}m^3$
Density of star is $\rho=4\times10^3kgm^{-3}$
Mass of star is $M=V\times\rho\approx9.5\times10^{29}kg=0.95\times10^{30}kg$
So the mass is of order of $10^{30}kg$

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between.

  1. $\displaystyle \frac { 290 }{ 12.5 } $ and $\displaystyle \frac { 290 }{ 11.5 } $

  2. $\displaystyle \frac { 295 }{ 12 } $ and $\displaystyle \frac { 285 }{ 11.5 } $

  3. $\displaystyle \frac { 285 }{ 12 } $ and $\displaystyle \frac { 295 }{ 12 } $

  4. $\displaystyle \frac { 285 }{ 12.5 } $ and $\displaystyle \frac { 295 }{ 11.5 } $

  5. $\displaystyle \frac { 295 }{ 12.5 } $ and $\displaystyle \frac { 285 }{ 11.5 } $

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Here, the rounding off of numbers is tested.
Distance rounded to nearest 10 miles = 290. 
So actual distance covered may be b/w 285 and 295. 
Gasoline used rounded to nearest gallon = 12.
So the actual gas used may be b/w 11.5 and 12.5. 
Now to get range of of miles/gallon, 
least value = the least of distance/the max of gas = 285/12.5
highest value = max of distance/least of gas = 295/11.5 
Option D is the correct choice.