Physics - Unit & Dimension Question with Solution | TestHub

Mass of the metallic cube, $m=5.580$ kg (4 significant figures)

Side length of the cube, $a=9.0$ cm $=9.0\times 10^{-2}$ m (2 significant figures)

Volume of the cube, $V=a^3=(9.0\times 10^{-2}{)}^3=729\times 10^{-6}$ m${}^3$

Density of the material, $\rho =\frac{m}{V}=\frac{5.580}{729\times 10^{-6}}=\mathrm{7.6543...}\times 10^3$ kg m${}^{-3}$

In multiplication or division, the final result should retain as many significant figures as are present in the original number with the least significant figures. Here, the side length ($9.0$ cm) has the least number of significant figures, which is 2.

Rounding off $\mathrm{7.6543...}\times 10^3$ to 2 significant figures, we look at the third digit. Since the third digit is $5$ and is followed by non-zero digits, the preceding digit is increased by $1$.

$\rho =7.7\times 10^3$ kg m${}^{-3}$

Comparing this with $X\times 10^3$ kg m${}^{-3}$, we get $X=7.7$.

Answer: $7.7$