Test: Significant Figures (NCERT)

The power of 10 is irrelevant to the determination of significant figures.

The final result should be 3 significant figures.

Here
Length of the cube, L = 1.2 x 10^-2 m
Volume of the cube, V = (1.2 x 10^-2m)³ = 1.728 x 10^-6 m³
As the result can have only two significant figures, therefore, on rounding off, we get, V = 1.7 x 10^-6 m³

Radius of the sphere, r = 1.41cm
(3 significant figures)
Volume of the sphere,

= 11.736 cm³
Rounded off upto 3 significant figures = 11.7 cm³.

Here, mass of the box, m = 2.3 kg
Mass of one gold piece, m₁ = 20.15 g = 0.02015 kg
Mass of other gold piece, m₂ = 20.17 g = 0.02017 kg
∴ Total mass = m + m₁ + m₂
= 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

Difference in masses
= m₂ - m₁ = 20.17 g - 20.15 g
= 0.02 g (correct upto two places of decimal)

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.

According to the rules of significant figures 6.320 has four significant figures.
6.032 has four significant figures. 0.0006032 has four significant figures.

As per rule of significant figures, 4.8000 x 10⁴ has 5 significant figures and 48000.50 has 7 significant figures.

Here, s = (13.8 ± 0.2) m, t = (4.0 ± 0.3)s
∴ veliocity, v = s/t = 13.8/4.0 = 3.45 m s⁻¹
(Rounded off to first place of decimal)


= 0.0865
or Δv = v × 0.0865 = 3.45 × 0.0865
= 0.3087.
∴ velocity =(3.5 ± 0.3) m s⁻¹.