Tag: standard form

Questions Related to standard form

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 $10^{30}kg$