2.28 The unit of length convenient on the nuclear scale is a fermi : 1 f = 10^–15 m. Nuclear sizes obey roughly the following empirical relation :

r = r₀ A^1/3

where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

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

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

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