Tag: rounding off digits

Questions Related to rounding off digits

Which of the following is the smallest unit?

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

  1. millimetre

  2. angstrom

  3. fermi

  4. metre

Show answer
Correct Option: C
Explanation:

Metre is the SI unit of length.
$1 \ millimetre = 10^{-3} \ metre$
$1 \ angstrom = 10^{-10} \ metre$
$1 \ fermi = 10^{-15} \ metre$
So, fermi is the smallest unit.

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

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

  1. $1.77\times {{10}^{-10}}$

  2. $1.25\times {{10}^{-10}}$

  3. $1.55\times {{10}^{-10}}$

  4. $1.95\times {{10}^{-10}}$

Show answer
Correct Option: B
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

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

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

  1. $960$

  2. $1000$

  3. $1200$

  4. $1414$

Show answer
Correct Option: B
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.

A particle is moving with a velocity $35\ m/s$ along positive x-axis.It's acceleration is towards negative X-axis with the magnitude $4m/s^ {2}$ then the distance covered by the particle in the $9th$ second:-

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

  1. $1m$

  2. $\dfrac {9}{8}m$

  3. $\dfrac {5}{4}m$

  4. $153m$

Show answer
Correct Option: B

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 $.

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

  1. $-12J$

  2. $+12J$

  3. $36J$

  4. $-36J$

Show answer
Correct Option: A

The vector sum of three forces having magnitudes $ | \overrightarrow F _1 | = 100 N $,  $ | \overrightarrow F _2 | = 80 N $ &  $ | \overrightarrow F _3 | = 60 N $ acting on a particle is zero. the angle between $ \overrightarrow F _1$  & $ \overrightarrow F _2 $ is nearly:-

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

  1. $ 53^0 $

  2. $ 143^0 $

  3. $ 37^0 $

  4. $ 127^0 $

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Correct Option: C
The Earth's radius is $6371km$. The order magnitude of the Earth's radius is
physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $10^3$

  2. $10^2$

  3. $10^7$

  4. $10^5$

Show answer
Correct Option: C
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.

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 $m^{-3}$ :

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

  1. 30

  2. 29

  3. 24

  4. 21

Show answer
Correct Option: A
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 $10^{30}kg$

Round off following number upto four significant figures.

$45.689$

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

  1. $45.689$

  2. $45.69$

  3. $45.68$

  4. $45.7$

Show answer
Correct Option: B
Explanation:
The given number has $5$ significant figures.
As  $9>5$, thus the digit $8$ gets changed to  $9$ after rounding off upto four significant figures.
$\therefore$ After rounding off, $45.689$ becomes $45.69$

Round off 2.0082 to four significant figures.

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

  1. 2.008

  2. 2.0082

  3. 2.009

  4. 2

Show answer
Correct Option: A
Explanation:

$2.0082=2.008$, 

Here the digit $2$ is less than $5$ so $2.008$ is the closest value in four significant figures.