Units and Measurements

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 $\approx 27*10^{4}c{m}^3$  

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

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

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

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

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

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

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

Sol. $\Rightarrow \mathrm{}{\stackrel{?}{\mathrm{V}}}_1=2\stackrel{?}{{\mathrm{U}}_{\mathrm{c}\mathrm{m}}}-\stackrel{?}{{\mathrm{U}}_1}\Rightarrow \mathrm{}2\left[\frac{\mathrm{m}/2{\mathrm{V}}_{\mathrm{o}}}{\frac{\mathrm{m}}{2}+\frac{\mathrm{m}}{3}}\right]-{\mathrm{V}}_{\mathrm{o}}$

$\Rightarrow \mathrm{}\frac{6}{5}{\mathrm{V}}_{\mathrm{o}}-{\mathrm{V}}_{\mathrm{o}}\Rightarrow \frac{{\mathrm{V}}_0}{5}$

$\Rightarrow \mathrm{}{\lambda }_{\mathrm{o}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{m}}{2}{\mathrm{V}}_{\mathrm{o}}\right)}\mathrm{}{\lambda }_{\mathrm{f}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{M}}{2}\frac{{\mathrm{V}}_{\mathrm{o}}}{5}\right)}$

$\Rightarrow \mathrm{\Delta }\lambda =\frac{8\mathrm{h}\mathrm{c}}{{\mathrm{m}\mathrm{V}}_{\mathrm{o}}}$

S. I. unit Pascal – second   $=\frac{N}{m^2}-sec$  

= $\frac{ML{T}^{-2}T}{L^2}$  

= $M{L}^{-1}{T}^{-1}$  

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) We know that one light year = 9.46 $*{10}^{11}m$ = distance that light travels in 1 year with a speed 3 $*{10}^8$ m/s

1 par sec =3.08 $*{10}^{16}m$ = distance at which average radius of earth's orbit subtends an angle of par second.

Here second and year represent time.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) we know that pressure = force/area

Pressure= $\frac{force*distance}{area*distance}=\frac{work}{volume}=\frac{energy}{volume}$

So we can say that pressure is also equal to energy per unit volume.