Units And Measurements Practice Questions

(i) Spring constant $\; {\text{F}}= {\text{Kx}}$ $\; {\text{K}}=\;{ {\text{F}}}/{ {\text{x}}}=\;{ {\text{kgm}} {\ sec}}/{ {\text{m}}}= {\text{kg}} \cdot {\ sec}^{-2}= [ {\text{ML}}^0 {\text{T}}^{-2}]$ (ii) Angular speed : (ω)$\omega \;=\;{2 \pi }/{ {\text{T}}}$ $\omega \;= [ {\text{M}}^0 {\text{L}}^0 {\text{T}}^{-1}]$(iii) Angular momentum$\; {\text{L}}= {\text{MVR}} \Rightarrow {\text{Kg}} \cdot {\text{m}} {\ sec}{ }^{-1} {\text{m}}= {\text{kgm}}^2 {\ sec}^{-1}$ $\;= [ {\text{ML}}^2 {\text{T}}^{-1}]$ $\;$(iv) Moment of Inertia${\text{I}}\;= {\text{MR}}^2= {\text{Kg}} \cdot {\text{m}}^2$ $\;= [ {\text{ML}}^2 {\text{T}}^0]$

By dimensional analysis,$\; {\text{V}}=\lambda ^{ {\text{a}}} {\text{g}}^{ {\text{b}}} \rho ^{ {\text{c}}}$ \;{[ {\text{L}}^1 {\text{T}}^{-1}]\ = $[ {\text{L}}^1]$ ^{ {\text{a}}} $[ {\text{L}}^1 {\text{T}}^{-2}]$ ^{ {\text{b}}} $[ {\text{M}}^1 {\text{L}}^{-3}]$ ^{ {\text{c}}}}$$;{[ {\text{L}}^{ {\text{l}}} {\text{T}}^{-1}]$ = $[ {\text{M}}^{ {\text{c}}} {\text{L}}^{ {\text{a}}+ {\text{b}}-3 {\text{c}}} {\text{T}}^{-2 {\text{b}}}]$ }$On comparing respective powers,$; {\text{c}}=0,-2 {\text{b}}=-1, {\text{a}}+ {\text{b}}-3 {\text{c}}=1,$$; {\text{b}}=;\frac{1}{2}, {\text{a}}=;\frac{1}{2}$

\vec{L}=\vec{r} \times \vec{p} \Rightarrow [ {\text{L}}]\;= \[ {\text{M}}^0 {\text{L}}^1 {\text{T}}^0]$ [ {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-1}]$$=;{[ {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-1}]$ }$$\vec{τ}=\vec{r} × \vec{F} \Rightarrow [τ];= $[ {\text{L}}^1]$ [ {\text{MLT}}^{-2}]$$;= [ {\text{ML}}^2 {\text{T}}^{-2}]$$;\text{ Stress }; ≡ ;\text{ Pressure };=;\frac{F}{A} \Rightarrow [;\text{ Stress };]= [ {\text{ML}}^{-1} {\text{T}}^{-2}]$Pressure Gradient$=;\frac{d P}{d x} \Rightarrow [$Pressure Gradient$]= [ {\text{ML}}^{-2} {\text{T}}^{-2}]$

A.V., Par sec and light year are the unit of distance Light year $\to$ distance travelled by light in one year $1 {\text{ly}}=9.5 \times 10^{15} {\text{m}}$ parcec $=3.262$ light year A.V. $=1.58 \times 10^{-5}$ light year A. V. $< 1 {\text{y}} <$ Parsec. Statement I correct and statement II incorrect.

(A) Planck's constant${\text{h}} v\;= {\text{E}}$ ${\text{h}}\;=\;{ {\text{E}}}/{v}=\;{ {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-2}}/{ {\text{T}}^{-1}}= {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-1}$(B) ${\text{E}}= {\text{qV}}$ $V=\;\frac{E}{q}=\;\frac{M^1 L^2 T^{-2}}{A^1 T^1}=M^1 L^2 T^{-3} A^{-1}(I V)$(C) $\varphi ($ work function $)=$ energy$= {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-2}$(D) Momentum (p) = F.t$\;= {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-2} {\text{T}}^1$ $\;= {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-1}$

$\;\;\text{ Least count }\;=\;\frac{\;\text{ Pitch }\;}{\;\text{ No. of circular divisions }\;}$ $\;=\;{0.5 {\text{mm}}}/{100}$ $\;\;\text{ Least count }\;=5 \times 10^{-3} {\text{mm}}$ $\;\;\text{ Positive Error }\;= {\text{MSR}}+ {\text{CSR}}( {\text{LC}})$ $\;=0 {\text{mm}}+6 (5 \times 10^{-3} {\text{mm}})$ Reading of Diameter $=$ MSR $+$ CSR $( {\text{LC}})-$ Positive zero error$\;=4 \times 0.5 {\text{mm}}+ (46 (5 \times 10^{-3}) ) -6 (5 \times 10^{-3}) {\text{mm}}$ $\;=2 {\text{mm}}+40 \times 5 \times 10^{-3} {\text{mm}}=2.2 {\text{mm}} \;\text{ (Ans.) }\;$

$\;{[\;\text{ Torque }\;]=\;\text{ F.L }\;}$ $\; {\text{MLT}}^{-2} \cdot {\text{L}}= {\text{ML}}^2 {\text{T}}^{-2}$ $\;{[\;\text{ Stress }\;]=\;{ {\text{F}}}/{ {\text{A}}}}$ $\;\;{M L T^{-2}}/{ {\text{L}}^2}= {\text{ML}}^{-1} {\text{T}}^{-2}$ $\;{[\;\text{ Pressure gradient }\;]=\;{\Delta {\text{P}}}/{\Delta {\text{L}}}=\;{ {\text{F}}}/{ {\text{A}}^2 . {\text{L}}}}$ $\;=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}^3}$ $\;= {\text{ML}}^{-2} {\text{T}}^{-2}$ $\; {\text{F}}= {\text{nA}} \;{ {\text{dv}}}/{ {\text{dy}}}$ $\;η=\text{ML}^{-1} {\text{T}}^{-1}$

Say dimensional formale of mass is ${\text{H}}^x C^y G^z$ $\; {\text{M}}^1= ( {\text{ML}}^2 {\text{T}}^{-1}) ^{ {\text{x}}} ( {\text{LT}}^{-1}) ( {\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}) ^z$ $\; {\text{M}}^1 {\text{L}}^0 {\text{T}}^0= {\text{M}}^{ {\text{x}}- {\text{z}}} {\text{L}}^{2 {\text{x}}+ {\text{y}}+3 z} {\text{T}}^{- {\text{x}}- {\text{y}}-2 z}$on comparing both side$\;x-z=1$ $\;2 x+y+3 z=0$ $\;-x-y-2 z=0$On solving above equations we get$x=\;\frac{1}{2} \;\; {\text{y}}=\;\frac{1}{2} \;\; {\text{z}}=\;\frac{-1}{2}$

$10 {\text{VSD}}=9 {\text{MSD}}$ $1 {\text{VST}}=.9 {\text{MSD}}$L.C. $=.1 {\text{mm}}=.01 {\text{cm}}$ $+$ ve zero error $=.4 {\text{mm}}$ $=0.04 {\text{cm}}$Negative zero error $=4.1 {\text{cm}}+6 \times .01$ $=4.12 {\text{cm}}$ $=412 \times 10^{-2} {\text{cm}}$

Torque $\to {\text{Nm}}$Stress $\to N / {\text{m}}^2$Latent heat $\to {\text{J}} / {\text{kg}}$Power $\to {\text{Nm}} / {\ s}$A-III, B-IV, C-II, D-I

$I_{\;\text{mean }\;}=\;\frac{1.22+1.23+1.19+1.20}{4}=1.21$ $\Delta I_{\;\text{mean }\;}=\;\frac{0.01+0.02+0.02+0.01}{4}=0.015$So $\% I=\;{\Delta I_{\;\text{mean }\;}}/{I_{\;\text{mean }\;}} \times 100=\;\frac{0.015}{1.21} \times 100=\;\frac{150}{121} \%$ $x=150$

$T=2 \pi \sqrt{\;\frac{l}{g}}$ $\;\frac{d g}{g} \times 100=\;\frac{2 d T}{T} \times 100+\;\frac{d l}{l} \times 100$ $=2 \times \;\frac{1}{50} \times 100+\;\frac{1}{100} \times 100=5 \%$

{[\;\frac{B^2}{\mu _0}]\=}$[Energy density]$=;\frac{M L^2 T^{-2}}{L^3}=M L^{-1} T^{-2}$

[\;\frac{\rho r^3}{T^{3 / 2}}] =\;\frac{[M L^{-3}]\ $[L^3]$ }{[M L^0 T^{-2}]$ ^{3 / 2}} \ne [T]$As the equation for first statement is wrong dimensionally.$\Rightarrow A$is false and$R$ is true.

{[L]=\;\frac{[v^2]\ }{[a]}}$$;\text{ so }; ;\frac{ [v_2] ^2}{ [a_2] }=;\frac{[\frac{n}{m^2} v_1]$ ^2}{ [;\frac{a_1}{m_n}] }$$;\frac{ [v_2] ^2}{ [a_2] }=;\frac{n^3}{m^3} ;\frac{ [v_1] ^2}{ [a_1] }$$;\text{ or }; [L_2] =;\frac{n^3}{m^3} [L_1]$Similarly,$[T]=;\frac{[v]}{[a]}$So,$[T_2] =;\frac{n^2}{m} [T_1]$

$1 {\text{MSD}}=\;\frac{1}{20} {\text{cm}}$ $\because 10 {\text{VSD}}=9 {\text{MSD}}$ $1 {\text{VSD}}=\;\frac{9}{10} \times \;\frac{1}{20} {\text{cm}}=\;\frac{9}{200} \times 10 {\text{mm}}=0.45 {\text{mm}}$Now, $1 {\text{MSD}}=\;\frac{1}{20} \times 10 {\text{mm}}=0.50 {\text{mm}}$ ${\text{LC}}=(0.50-0.45) {\text{mm}}=0.05 {\text{mm}}$ $=5 \times 10^{-2} {\text{mm}}$

$\because U=\;\frac{1}{2} M i^2$ \Rightarrow [M]=\;\frac{[U]}{[i^2]\}=;\frac{M L^2 T^{-2}}{A^2}$$=[M L^2 T^{-2} A^{-2}]$

5 parts of main scale division $=1 {\text{cm}}$ $\therefore 1$ part of main scale division $=\;\frac{1}{5} {\text{cm}}$ $\therefore 1$ M.S.D. $=\;\frac{1}{5} {\text{cm}}$ $(n+1)$ vernier scale division $=n$ main scale division.$\therefore 1 \;\text{ V.S.D. }\;=\;\frac{n}{n+1} \;\text{ M.S.D. }\;$ $=\;\frac{n}{n+1} \times 1 \;\text{ M.S.D. }\;$ $=\;\frac{n}{n+1} \times \;\frac{1}{5} {\text{cm}}$We know,$\;\text{ L.C. }\;=1 \;\text{ M.S.D. }\;-1 \;\text{ V,S.D. }\;$ $=\;\frac{1}{5} {\text{cm}}-\;\frac{n}{5(n+1)} {\text{cm}}$ $=\;\frac{n+1-n}{5(n+1)} {\text{cm}}$ $=\;\frac{1}{5(n+1)} {\text{cm}}$