2.19 The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 10¹¹m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 10⁵ cm³ $\approx 27\times 10^{4}c{m}^3$  

$K=\frac{\left(\frac{Q}{r}\right)\mathrm{\Delta }x}{\mathrm{\Delta }AT}$

$\begin{array}{ll}\Rightarrow & \frac{\left({\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}\right)\left(\mathrm{L}\right)}{\left({\mathrm{L}}^2\right)\left(\theta \right)\left(\mathrm{T}\right)}\\ \Rightarrow \mathrm{}& {\mathrm{M}}^1{\mathrm{L}}^{1-}{\mathrm{T}}^{-3}{\theta }^{-1}\end{array}$

Viscosity = Pascal . Second

$P^X{A}^Y{T}^Z=\left[M^1{L}^{-1}{T}^{-1}\right]$        

=  ${\left[M^1{L}^{+1}{T}^{-1}\right]}^x\left[L^2\right]y{\left[T^1\right]}^z=M^{-1}{L}^{-1}{T}^{-1}$

= x = 1,     x + 2y = -1,      -x + z = 1

y = -1,              Z = 0

Viscosity = [P¹A^-1T⁰]

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10^-2 cm

$\Delta Q=ms\Delta T\Rightarrow s=\frac{\Delta Q}{m\Delta T}\Rightarrow \left[s\right]=\frac{M{L}^2{T}^{-2}}{MQ}=L^2{T}^{-2}{Q}^{-1}$

$\Delta Q=mL\Rightarrow \left[L\right]=\frac{M{L}^2{T}^{-2}}{M}=L^2{T}^{-2}$