Q.2.4 Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:

(a) Atoms are very small objects

(b) A jet plane moves with great speed

(c) The mass of Jupiter is very large

(d) The air inside this room contains a large number of molecules

(e) A proton is much more massive than an electron

(f) The speed of sound is much smaller than the speed of light

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

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

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