Tag: significant figures and rounding of digits

$\sqrt{79.62}=8.92300397$
Since 79.62 has 4 significant digits, so we express our answer in 4 significant digits.
$\Rightarrow \sqrt{79.62}=8.923 \rightarrow$ 4 significant figures.

Volume of 1 sphere $=1.76 cm^{3}$
Volume of 25 spheres $=25\times 1.76$
                                      $=44 {cm}^{3}$
                                      $=44.0 {cm}^{3}$
                                 (Since volume is reported in 3 significant figure)

$\left ( 2.0 \right )^{10}=1024$
But since 2.0 contains 2 significant figure.
So, $1024=1.0\times 10^{3}$

Momentum$=mv=3.513\times 5.00=17.565\approx 17.57$
$\approx 17.6 kg m/s$
(Since 5.00 contains  least no. of significant figures i.e. 3)

Since, the measurement $0.500 \ cm$ gives the reading up to 3 decimel, thus $0.500 \ cm$ is the most accurate measurement among the given three measurements.

The least significant digit is the lowest digit in a number, located at the far right of a string.

If there is a decimal point, the rightmost digit is the last or least significant figure. So 0.08076 is the least significant figure

Since, all the non zero digits and zeros between the non-zero digits are considered as significant figures. Thus the number $10.02$ has $4$ significant figures.

Zeros to the left of the non-zero digits are considered as significant figures. All non-zero digits are also considered as significant figures. Thus the number  $0.0632$ has $3$ significant figures.