Units And Measurements Practice Questions

$\; {\text{MSR}}=1 {\text{mm}}, {\text{CSR}}=42, \;\text{ pitch }\;=1 {\text{mm}}$ $\; {\text{LC}}=\;{\;\text{ pitch }\;}/{\;\text{ No. of }\; {\text{CSD}}}= (\;\frac{1}{100}) =0.01 {\text{mm}}$ $\;\;\text{ Diameter }\;= {\text{MSR}}+ {\text{LC}} \times {\text{CSD}}$ $\;\;\text{ Diameter }\;=1+(0.01) \times 42 {\text{mm}}$ $\;\;\text{ Diameter }\;=1.42 {\text{mm}}=\;{ {\text{x}}}/{50}$ $\; \therefore {\text{x}}=71$

\;{[\vec{\tau }]=\[{ {\text{r}}}^{→} × { {\text{F}}}^{→}]$= $[ {\text{ML}}^2 {\text{T}}^{-2}]$ }$$;{[ {\text{F}}]=[ {\text{qVB}}]}$$;\Rightarrow {\text{B}}= (;{ {\text{F}}}/{ {\text{qV}}}) = $[;{ {\text{MLT}}^{-2}}/{ {\text{ATLT}}^{-1}}]$ = [ {\text{MA}}^{-1} {\text{T}}^{-2}]$$;{[ {\text{M}}]=[ {\text{I}} \times {\text{A}}]= $[ {\text{AL}}^2]$ }$$; {\text{B}}=;\frac{\mu _0}{4 \pi } ;{ {\text{Idl}} sin ;\theta }/{ {\text{r}}^2}$$;\Rightarrow [\mu ]= $[;{ {\text{Br}}^2}/{ {\text{Idl}}}]$ = [;{ {\text{MT}}^{-2} {\text{A}}^{-1} \times {\text{L}}^2}/{ {\text{AL}}}]$$;= [ {\text{MLT}}^{-2} {\text{A}}^{-2}]$

$\;\;\text{ Angular impulse }\;=\;\text{ change \in angular momentum. }\;$ $\;\;\text{ [Angular impulse }\;]=[\;\text{ Angular momentum }\;]$ $=[ {\ mvr}]$ $\;= [ {\text{M}} {\text{L}}^2 {\text{T}}^{-1}]$

$\;n {\text{MSD}}=(n+1) {\text{VSD}}$ $\;\Rightarrow 1 {\text{VSD}}=\;\frac{n}{n+1} {\text{MSD}}$ $\;L \cdot C=1 M S D-1 {\text{VSD}}$ $\;L \cdot C=m-m (\;\frac{n}{n+1})$ $\;L \cdot C=m (\;\frac{n+1-n}{n+1})$ $\;\Rightarrow L \cdot C= (\;\frac{m}{n+1})$

$\;E=\;\frac{K Q}{R^2}$ $\;E=\;\frac{Q}{4 \pi \epsilon _0 R^2}$ $\;\epsilon _0=\;\frac{Q}{4 \pi R^2 E}$ $\;\;\text{ Now, }\; \epsilon _0 {\text{E}}^2=\;{ {\text{Q}}}/{4 \pi {\text{R}}^2 {\text{E}}} \cdot {\text{E}}^2=\;{ {\text{Q}}}/{4 \pi {\text{R}}^2} \cdot {\text{E}}$ \;{[\epsilon _0 {\text{E}}^2]\ = $[;{ {\text{QE}}}/{ {\text{R}}^2}]$ =;{[ {\text{Q}}][ {\text{E}}]}/{[ {\text{R}}^2]$ }=;{[ {\text{Q}}]}/{[ {\text{R}}^2]$ [ {\text{Q}}][ {\text{R}}]}}$$;=;{[ {\text{W}}]}/{[ {\text{R}}^3]$ }=;{ {\text{ML}}^2 {\text{T}}^{-2}}/{ {\text{L}}^3}= {\text{ML}}^{-1} {\text{T}}^{-2}$

${\text{R}}=\;{ {\text{V}}}/{ {\text{l}}}$ According to error analysis $\;\;{ {\text{dR}}}/{ {\text{R}}}=\;{ {\text{dV}}}/{ {\text{V}}}+\;{ {\text{dI}}}/{ {\text{I}}}$ $\;\;{ {\text{dR}}}/{ {\text{R}}}=\;\frac{5}{200}+\;\frac{0.2}{20}$ $\;\;{ {\text{dR}}}/{ {\text{R}}}=\;\frac{7}{200}$ $\;\% \;\text{ error }\; \;{ {\text{dR}}}/{ {\text{R}}} \times 100=\;\frac{7}{200} \times 100$ $=3.5 \%$

$\;\lambda =\;{ {\text{h}}}/{\sqrt{2 {\text{mE}}}} \;\text{ or }\; {\text{E}}= {\text{h}} v$ \;{[ {\text{ML}}^2 {\text{T}}^{-2}]\ = {\text{h}} $[ {\text{T}}^{-1}]$ }$$; {\text{h}}= [ {\text{ML}}^2 {\text{T}}^{-1}]$

$\because [ {\text{V}}]=[ {\text{b}}]$ $\therefore$ Dimension of $b= [ {\text{L}}^3]$ $\&[ {\text{P}}]= [\;{ {\text{a}}}/{ {\text{V}}^2}]$ [ {\text{a}}]= \[ {\text{PV}}^2]$ = $[ {\text{ML}}^{-1} {\text{T}}^{-2}]$ [ {\text{L}}^6]$Dimension of${\text{a}}= [ {\text{ML}}^5 {\text{T}}^{-2}]$$\therefore {\text{ab}}^{-1}=;{[ {\text{ML}}^5 {\text{T}}^{-2}]$ }/{[ {\text{L}}^3]$ }= [ {\text{ML}}^2 {\text{T}}^{-2}]$

To match the physical quantities with their SI units, we derive the units for each term:
A. Torque is defined as force multiplied by perpendicular distance. Its unit is $({\text{kg}} \cdot {\text{m/s}}^2) \times {\text{m}} = {\text{kg}} {\text{m}}^2 {\text{s}}^{-2}$, which corresponds to IV.

B. Energy density is energy per unit volume. Its unit is $\frac{\text{Energy}}{\text{Volume}} = \frac{{\text{kg}} {\text{m}}^2 {\text{s}}^{-2}}{{\text{m}}^3} = {\text{kg}} {\text{m}}^{-1} {\text{s}}^{-2}$, which corresponds to I.

C. Pressure gradient is pressure per unit length. Its unit is $\frac{\text{Pressure}}{\text{Length}} = \frac{{\text{kg}} {\text{m}}^{-1} {\text{s}}^{-2}}{{\text{m}}} = {\text{kg}} {\text{m}}^{-2} {\text{s}}^{-2}$, which corresponds to III.

D. Impulse is the change in momentum, or force multiplied by time. Its unit is $({\text{kg}} \cdot {\text{m/s}}^2) \times {\text{s}} = {\text{kg}} {\text{m}} {\text{s}}^{-1}$, which corresponds to II.

The correct matching sequence is A-IV, B-I, C-III, D-II.

$\;\;{ {\text{dP}}}/{ {\text{P}}} \times 100= (2 \;{ {\text{da}}}/{ {\text{a}}}+3 \;{ {\text{db}}}/{ {\text{b}}}+\;{ {\text{dc}}}/{ {\text{c}}}+\;\frac{1}{2} \;{ {\text{d}}( {\text{d}})}/{ {\text{d}}}) \times 100$ $\;=2 \times 1+3 \times 2+3+\;\frac{1}{2} \times 4$ $\;=2+6+3+2$ $\;=13 \%$

\;\;\text{ Pressure gradient }\;=\;{ {\text{dp}}}/{ {\text{dx}}}=\;{[ {\text{ML}}^{-1} {\text{T}}^{-2}]\ }/{[ {\text{L}}]}$$;= [M^1 {\text{L}}^{-2} {\text{T}}^{-2}]$$;;\text{ Energy density };=;\frac{;\text{ energy };}{;\text{ volume };}=;{[ {\text{ML}}^2 {\text{T}}^{-2}]$ }/{[ {\text{L}}^3]$ }$$;= [ {\text{M}}^1 {\text{L}}^{-1} {\text{T}}^{-2}]$$;;\text{ Electric field };=;\frac{;\text{ Force };}{;\text{ charg e };}=;{[ {\text{MLT}}^{-2}]$ }/{[ {\text{A}} \cdot {\text{T}}]}$$;= [ {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-3} {\text{A}}^{-1}]$$;;\text{ Latent heat };=;\frac{;\text{ heat };}{;\text{ mass };}=;{[ {\text{ML}}^2 {\text{T}}^{-2}]$ }/{[ {\text{M}}]}$$;= [ {\text{M}}^0 {\text{L}}^2 {\text{T}}^{-2}]$$;$

In parallel combination, $\;\frac{1}{R_{e q}}=\;\frac{1}{R_1}+\;\frac{1}{R_2}$ $\Rightarrow \;\frac{1}{R_{e q}}=\;\frac{1}{10}+\;\frac{1}{15}=\;\frac{5}{30}=\;\frac{1}{6}$Now, for error calculation,$\;\;\frac{d R_{e q}}{R_{e q}^2}=\;\frac{d R_1}{R_1^2}+\;\frac{d R_2}{R_2^2}$ $\;\Rightarrow \;\frac{d R_{e q}}{36}=\;\frac{0.5}{100}+\;\frac{0.5}{225}$ $\;d R_{e q}=36 \times 0.5 \times (\;\frac{13}{900}) =18 \times \;\frac{13}{900}=\;\frac{26}{100}=0.26$Now, $\;\frac{d R_{e q}}{R_{e q}} \times 100=\;\frac{0.26}{6} \times 100=\;\frac{26}{6}=4.33$

Cylindrical wire ${\text{m}}=(0.4 ± 0.01) {\text{g}}$ $ℓ=(8 ± 0.04) {\text{cm}}$ ${\text{r}}=(6 ± 0.03) {\text{mm}}$Density $\rho =\;{ {\text{m}}}/{\pi r^2 ℓ} \Rightarrow \rho r^2 ℓ m^{-1}=\;\frac{1}{\pi }=$ const.Differentiating after taking $\log$ on both side$\;\;{ {\text{d}} \rho }/{\rho }+\;{2 {\text{dr}}}/{ {\text{r}}}+\;{ {\text{d}} ℓ}/{ℓ}-\;{ {\text{dm}}}/{ {\text{m}}}=0$ $\;\;\frac{\Delta \rho }{\rho }=\;{\Delta {\text{m}}}/{ {\text{m}}}-\;\frac{\Delta ℓ}{ℓ}-\;{2 \Delta {\text{r}}}/{ {\text{r}}}$ $\; (\;\frac{\Delta \rho }{\rho }) _{\max }= [\;\frac{0.01}{0.4}+\;\frac{0.04}{8}+2 (\;\frac{0.03}{6}) ]$ $\; (\;\frac{\Delta \rho }{\rho }) _{\max }=0.04$ $\;\;\text{ Percentage error }\;=0.04 \times 100=4 \%$Percentage error $=0.04 \times 100=4 \%$

$\;( {\text{x}}- {\text{At}})^{2+} ( {\text{y}}-\;{ {\text{t}}}/{ {\text{B}}}) ^2= {\text{a}}^2$ $\;{[ {\text{At}}]= {\text{A}} \times \;{1}/{ {\text{T}}}= {\text{L}}}$ $\; \therefore \;\; [ {\text{A}}]= {\text{T}}^1 {\text{L}}^1$ $\; \;\; \;{ {\text{t}}}/{ {\text{B}}} \;\text{ is \in meters }\;$ $\; \therefore \;\; \;{1}/{ {\text{T}}[ {\text{B}}]}= {\text{L}}$ $\; \therefore \;\; [ {\text{B}}]= {\text{T}}^{-1} {\text{L}}^{-1}$ $\; \therefore \;\; \;\text{ Correct Ans. (2) }\;$

Kinetic energy, ${\text{KE}}\;=\;\frac{1}{2} {\ mv}^2$ ${\text{KE}}\;=\;\frac{1}{2} \times 5 \times 20^2$ ${\text{KE}}=1000 {\text{J}}$ $\;\;{\Delta {\text{K}}}/{ {\text{K}}}=\;{\Delta {\text{m}}}/{ {\text{m}}}+\;{2 \Delta {\text{v}}}/{ {\text{v}}}$ $\;\;{\Delta {\text{K}}}/{1000}=\;\frac{0.5}{5}+2 \times \;\frac{0.4}{20}$ $\;\Delta {\text{K}}=1000(0.1+0.04)$ $\;\Delta {\text{K}}=1000 \times 0.14$ $\;\Delta {\text{K}}=140 {\text{J}}$ $\; {\text{KE}}=(1000 ± 140) {\text{J}}$

The zero error in verniel scale is $=6 \times 0.1 {\text{mm}}=0.6 {\text{mm}}$ (+ve zero error)Note: +ve zero error will have to be subtractedFrom the reading of the object.Now, the diameter measured with the help of Vernier scale isGiven by $\to$ M.S.D + V.S.D $\times$ L.S$\;\Rightarrow 3.2 {\text{cm}}+0.1 {\text{mm}} \times 4$ $\;=3.24 {\text{cm}}$The actual diameter is $\Rightarrow 3.24 {\text{mm}}$ - (zero error) $=3.24-0.6=3.18 {\text{cm}}$

Dimension of $\;\frac{1}{\mu _0 \epsilon _0}$ $\; {\text{C}}=\;{1}/{{\sqrt{\mu _0 \epsilon _0}}} \Rightarrow \;\frac{1}{\mu _0 \epsilon _0}= {\text{c}}^2$ \;{ [\;\frac{1}{\mu _0 \epsilon _0}] = \[ {\text{c}}^2]$ }$$;= [;{ {\text{L}}^2}/{ {\text{T}}^2}]$

\; {\text{Y}}=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;{ {\text{F}} / {\text{A}}}/{\Delta ℓ / ℓ}=\;{[ {\text{MLT}}^{-2}]\ }/{[ {\text{L}}^2]$ }= [ {\text{ML}}^{-1} {\text{T}}^{-2}]$$; {\text{F}}=6 \pi η {\text{rv}} \Rightarrow η=;{ {\text{F}}}/{6 \pi {\text{rv}}}$$;{[η]=;{[ {\text{MLT}}^{-2}]$ }/{[ {\text{L}}] $[ {\text{LT}}^{-1}]$ }= $[ {\text{ML}}^{-1} {\text{T}}^{-1}]$ }$$; {\text{E}}= {\text{h}} v \Rightarrow {\text{h}}=;{ {\text{E}}}/{v}=;{[ {\text{ML}}^2 {\text{T}}^{-2}]$ }/{[ {\text{T}}^{-1}]$ }= [ {\text{ML}}^2 {\text{T}}^{-1}]$Work function has same dimension as that of energy, so$[φ]= [ {\text{ML}}^2 {\text{T}}^{-2}]$