Significant Figures and Order of Magnitude

Here, mass of the box, $m = 2.3 kg$
Mass of one gold piece, $m _1 = 20.15 g = 0.02015 kg$
Mass of other gold piece, $m _2 = 20.17 g = 0.02017 kg$
$\therefore$ Total mass = $m + m _1 +m _2 = 2.3 kg + 0.02015 kg +0.02017 kg = 2.34032 kg$
As the result is correct only upto one place of decimal, therefore, on rounding off, we get Total mass $= 2.3 kg$

In 2.745, the digit to be rounded off (i.e., 4) is even, hence it should be left unchanged and in 2.735, the digit to be rounded off (i.e., 3) is odd, hence it should be increased by 1, i.e., changed to 4.

The number 3.845 rounded off to three significant figures becomes 3.84 since the preceding digit is even. On the other hand, the number 3.835 rounded off to three significant figures becomes 3.84 since the preceding digit is odd.

For 20.96, rounding to 3 significant figures gives 21.0. For 0.0003125, the first three significant digits are 3, 1, and 2; since the next digit is 5, we round up to 3.13 * 10^-4. Option B is the closest standard representation provided.

the order of magnitude of $0.00701$ is $-3$

Area = $ \pi (\dfrac{d}{2})^2  $

We have $ 3.9 \times 10^5- 2.5 \times 10^4$

For the number 1,000,000, we will shift the decimal to the left, stopping just before the first digit of the number. The number of moves you make to the left is the order of magnitude. Since we moved it six times, there are six orders of magnitude of 1,000,000, meaning that you can multiply 10 six times and get 1,000,000

Thickness $t = 326\ mm$

$24.25\times 10^{3} \rightarrow 2,4,2,5\rightarrow $ present by it has 4 significant digits.
If the digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
So, up to three significant figures, the number $=24.2\times 10^{3}$

a) 20.96
We have to round off for 3 significant digits. Last digit has to be rounded off.
If the digit to be dropped (6) is more than 5, then preceding digit is raised by 1.
Hence  $20.96\approx 21.0$

In general normal newspaper are in the order of 0.1 mm, but if some time in magazines the thickness may be order of 0.1 cm.

The given number:   

One light year is the distance of light at a speed $3\times 10^8 m/s$ in 1 year. 
Thus, $1 $ light year $= 3\times 10^8 m/s\times 1$ year $=3\times 10^8 m/s\times (365\times 24\times 3600 s)=9.5\times 10^{15} m$
According to rule of order of magnitude, as $9.5>5$ so 9.5 will be taken as 10.
Thus, order of magnitude of one light year $=10\times 10^{15}=10^{16} m$

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

$2.0082=2.008$, 

Given :  $R = 7\times 10^8 \ m$     and    $M = 2\times 10^{30} \ kg$
Volume of sun  $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi\times (7\times 10^8)^3 = 1.4\times 10^{27} \ m^3$
Density of sun   $\rho = \dfrac{M}{V} = \dfrac{2\times 10^{30}}{1.4\times 10^{27}} = 1.43\times 10^3 \ kg/m^3$
Thus order of magnitude of density of sun is $10^3 \ kg/m^3$.

When a number rounded to the nearest power of 10, it is called order of magnitude. Here a number which is less than 5 is taken as 1 and a number greater than 5 is taken as 10. 
As $6.4>10, $ so it would be takes as 10. 
The order of magnitude of earth's size $=10\times 10^6=10^7 m$  

Here, time taken $t=2.9$ billion years $=2.9\times 10^9$ years $ =2.9\times 10^9\times (365\times 24\times 3600) s=9.14\times 10^{16} s$
Distance $=$ velocity $\times $ time
or $d=(3\times 10^8)\times (9.14\times 10^{16})=2.74\times 10^{25} m$
As $2.74 <5$ so it will be taken as 1. 
Thus, the order of magnitude of distance $d=1\times 10^{25} m=10^{25} m$

We know that  $1\mu s = 10^{-6} \ s$
Thus,  $5\times 10^{7}\mu s = 5\times 10^{7}\mu s\times \dfrac{10^{-6} \ s}{\mu s} = 50 \ s$

Order of magnitude of $379$ is $\dfrac { 379 }{ { 10 }^{ 2 } } =3.79<{ 5 }^{ -1 }$    i.e, $379\approx { 10 }^{ 2 }$