(i) Mass of box = 2.3 kg Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg. Total mass = 2.3 + 0.04032 = 2.34032 kg In correct significant figure mass = 2.3 kg (as least decimal) (ii) Difference in mass of gold pieces = 0.02 g In correct significant figure (2 significant fig. minimum decimaRead more
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
In correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box. (i) What is the total mass of the box ? (ii) The difference in masses of the pieces to correct significant figures.
(i) Mass of box = 2.3 kg Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg. Total mass = 2.3 + 0.04032 = 2.34032 kg In correct significant figure mass = 2.3 kg (as least decimal) (ii) Difference in mass of gold pieces = 0.02 g In correct significant figure (2 significant fig. minimum decimaRead more
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
See lessIn correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.
The sides of a rectangle are (10..5 ± 0.2) cm and (5.2 ± 0.1) cm. Calculate its perimeter with error limits.
P = 2 (l + b) ± 2 (∆l + ∆b) = 2(10.5 + 5.2) ± 2(0.2 + 0.1) = (31.4 ± 0.6) cm.
P = 2 (l + b) ± 2 (∆l + ∆b)
See less= 2(10.5 + 5.2) ± 2(0.2 + 0.1)
= (31.4 ± 0.6) cm.
Determine the number of light years in one metre.
1 Iy = 9.46 × 10¹⁵ m 1 m = 1/9.46x10¹⁵ = 1.057 x 10⁻16 Iy
1 Iy = 9.46 × 10¹⁵ m
See less1 m = 1/9.46×10¹⁵ = 1.057 x 10⁻16 Iy
Magnitude of force experienced by an object moving with speed v is given by F = kv². Find dimensions of k.
[k] = [F]/[v²] = M¹L¹T⁻²/ M⁰L²T⁻² = [M¹ L⁻¹]
[k] = [F]/[v²] = M¹L¹T⁻²/ M⁰L²T⁻² = [M¹ L⁻¹]
See lessIn Vander Wall’s gas equation (P+a/v²) (v-b) = RT. Determine the dimensions of a and b.
Since dimensionally similar quantities can only be added [P] = [a/v²] ⇒[a]=[PV²]=[M¹L⁵T²] [b] = [v] = [L³]
Since dimensionally similar quantities can only be added
See less[P] = [a/v²] ⇒[a]=[PV²]=[M¹L⁵T²]
[b] = [v] = [L³]
If the unit of force is 100N, unit of length is 10m and unit of time is 100s. What is the unit of Mass in this system of units ?
[F] = [MLT ⁻ ²] [M] = [F] / [L][T⁻ ²] = [100N] / [10m][100s] ⁻ ² =10⁵kg.
[F] = [MLT ⁻ ²]
See less[M] = [F] / [L][T⁻ ²] = [100N] / [10m][100s] ⁻ ² =10⁵kg.
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm. Calculate the minimum inaccuracy in the measurement of distance.
Minimum inaccuracy = Vernier constant = 1 MSD – 1 VS.D = 1 MSD - 49/50 MSD = 1/50 (0.5mm) = 0.01
Minimum inaccuracy = Vernier constant
See less= 1 MSD – 1 VS.D
= 1 MSD – 49/50 MSD
= 1/50 (0.5mm) = 0.01