The decimal equivalent of $\frac{1}{20}$ up to three significant figures is

The area of a rectangular field (in $m^{2}$) of length $55.3 \ m$ and breadth $25 \ m$ after rounding off the value for correct significant digits is

$A$ person measures the mass of $3$ different particles as $435.42 \ g$,$226.3 \ g$,and $0.125 \ g$. According to the rules for arithmetic operations with significant figures,the sum of the masses of the $3$ particles will be: (in $g$)

Statement-$1$: All reliable digits plus the first uncertain digit together are called significant figures.
Statement-$2$: Trailing zero$(s)$ in a number with a decimal point are never significant.

The number of significant figures in the simplification of $\frac{0.501}{0.05}(0.312-0.03)$ is

The number of significant figures in $4.870 \ m$ is