Units And Measurements Practice Questions

According to principle of homogeneity dimension of LHS should be equal to dimensions of RHS so option (3) is correct.$\; {\text{T}}^2=\;{4 \pi ^2 {\text{r}}^3}/{ {\text{GM}}}$ \;{[ {\text{T}}^2]\ =;{[ {\text{L}}^3]$ }/{[ {\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}]$ [ {\text{M}}]}}$(Dimension of${\text{G}}$is$[ {\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}]$)$[ {\text{T}}^2] =;{[ {\text{L}}^3]$ }/{[ {\text{L}}^3 {\text{T}}^{-2}]$ }= [ {\text{T}}^2]$

$\;\;\text{ Least count of vernier calipers }\;=\;{1}/{20 {\text{N}}} {\text{cm}}$ $\; \because \;\text{ Least count }\;=1 {\text{MSD}}-1 {\text{VSD}}$let ${\text{x}}$ no. of divisions of main scale coincides with ${\text{N}}$ division of vernier scale, then$\;1 {\text{VSD}}=\;{ {\text{x}} \times 1 {\text{mm}}}/{ {\text{N}}}$ $\; \therefore \;{1}/{20 {\text{N}}} {\text{cm}}=1 {\text{mm}}-\;{ {\text{x}} \times 1 {\text{mm}}}/{ {\text{N}}}$ $\;\;{1}/{2 {\text{N}}} {\text{mm}}=1 {\text{mm}}-\;{ {\text{x}}}/{ {\text{N}}} {\text{mm}}$ $\; {\text{x}}= (1-\;{1}/{2 {\text{N}}}) {\text{N}}$ $\; {\text{x}}=\;{2 {\text{N}}-1}/{2}$

$\;a \times 10^b$ $\;\;\text{ if }\; a \le 5 \;\text{ order is }\; b$ $\;a > 5 \;\text{ order is }\; b+1$

Comparing the given equation with standard\;\;\text{ equation of standing }\; \;{2 \pi {\text{n}}}/{\lambda }=\omega & \;\frac{2 \pi }{\lambda }= {\text{k}} $\;{ [\;{ {\text{n}}}/{\lambda }] =[\omega ]= {\text{T}}^{-1}}$ $\;{[ {\text{nt}}]=[\lambda ]= {\text{L}}}$ $\;{[ {\text{n}}]=[\lambda \omega ]= {\text{LT}}^{-1}}$ $\;{[ {\text{x}}]=[\lambda ]= {\text{L}}}$Hence option (3)

$\rho =\; {\text{R}} \;\frac{\rho }{ℓ}$ $\;\;\frac{\Delta \rho }{\rho }=\;{\Delta {\text{R}}}/{ {\text{R}}}+2 \;{\Delta {\text{r}}}/{ {\text{r}}}+\;\frac{\Delta ℓ}{ℓ}$ $\;=\;\frac{10}{100}+2 \times \;\frac{0.05}{0.35}+\;\frac{0.2}{15}$ $\;=\;\frac{1}{10}+\;\frac{2}{7}+\;\frac{1}{75}$ $\;\;\frac{\Delta \rho }{\rho }=39.9 \%$

$\;50 {\text{V}}+ {\text{S}}=49 {\text{S}}+ {\text{S}}$ $\; {\text{S}}=50( {\text{S}}- {\text{V}})$ $\;.5=50( {\text{S}}- {\text{V}})$ $\; {\text{S}}- {\text{V}}=\;\frac{0.5}{50}=\;\frac{1}{100}=0.01 {\text{mm}}$

$\; {\text{R}}=\;{\rho {\text{L}}}/{\pi \;{ {\text{d}}^2}/{4}}$ $\;\;{\Delta {\text{R}}}/{ {\text{R}}}=\;{\Delta {\text{L}}}/{ {\text{L}}}+\;{2 \Delta {\text{d}}}/{ {\text{d}}}$ $\;\;{\Delta {\text{L}}}/{ {\text{L}}}=0.1 \% \;\text{ and }\; \;{\Delta {\text{d}}}/{ {\text{d}}}=0.1 \%$ $\;\;{\Delta {\text{R}}}/{ {\text{R}}}=0.3 \%$

$\;20 {\text{VSD}}=19 {\text{MSD}}$ $\;1 {\text{VSD}}=\;\frac{19}{20} {\text{MSD}}$L.C. $=1 {\text{MSD}}-1 {\text{VSD}}$ $0.1 {\text{mm}}=1 {\text{MSD}}-\;\frac{19}{20} {\text{MSD}}$ $\;0.1=\;\frac{1}{20} {\text{MSD}}$ $\;1 {\text{MSD}}=2 {\text{mm}}$

${\text{T}}\;=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ ${\text{g}}\;=\;{4 \pi ^2 ℓ}/{ {\text{T}}^2}$ $\;{\Delta {\text{g}}}/{ {\text{g}}}\;=\;\frac{\Delta ℓ}{ℓ}+\;{2 \Delta {\text{T}}}/{ {\text{T}}}$ $\;=\;\frac{0.2}{20}+2 (\;\frac{1}{40})$ $=\;\;\frac{0.3}{20}$Percentage change $=\;\frac{0.3}{20} \times 100=6 \%$

$\; {\text{T}}=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{k}}}}$ $\; {\text{T}}^2 \propto \;{ {\text{m}}}/{ {\text{k}}}$ $\;\;{2 \Delta {\text{T}}}/{ {\text{T}}} \%=\;{\Delta {\text{m}}}/{ {\text{m}}} \%-\;{\Delta {\text{k}}}/{ {\text{k}}} \%$ $\;\;{\Delta {\text{k}}}/{ {\text{k}}} \%=\;{\Delta {\text{m}}}/{ {\text{m}}} \%-\;{2 \Delta {\text{T}}}/{ {\text{T}}} \%$ $\;\;{\Delta {\text{k}}}/{ {\text{k}}} \%=(-1) \%-2(2) \%=|-5 \%|=5 \%$

$\;\;\text{ Given }\; 9 {\text{MSD}}=10 {\text{VSD}}$ $\;\;\text{ mass }\;=8.635 {\text{g}}$ $\; {\text{LC}}=1 {\text{MSD}}-1 {\text{VSD}}$ $\; {\text{LC}}=1 {\text{MSD}}-\;\frac{9}{10} {\text{MSD}}$ $\; {\text{LC}}=\;\frac{1}{10} {\text{MSD}}$ $\; {\text{LC}}=0.01 {\text{cm}}$ $\;\;\text{ Reading of diameter }\;= {\text{MSR}}+ {\text{LC}} \times {\text{VSR}}$ $\;=2 {\text{cm}}+(0.01) \times (2)$ $\;=2.02 {\text{cm}}$ $\;\;\text{ Volume of sphere }\;=\;\frac{4}{3} \pi (\;{ {\text{d}}}/{2}) ^3=\;\frac{4}{3} \pi (\;\frac{2.02}{2}) ^3$ $\;=4.32 {\text{cm}}^3$ $\;\;\text{ Density }\;=\;\frac{\;\text{ mass }\;}{\;\text{ volume }\;}=\;\frac{8.635}{4.32}=1.998 ∼ 2.00 {\text{g}}$ $\;$

$\;{[ {\text{B}}]= {\text{L}}^2}$ $\; {\text{A}}=\;{ {\text{x}}^2}/{ {\text{tE}}}=\;{ {\text{L}}^2}/{ {\text{TML}}^2 {\text{T}}^{-2}}=\;{1}/{ {\text{MT}}^{-1}}$ $\;{[ {\text{A}}]= {\text{M}}^{-1} {\text{T}}}$ \;{[ {\text{AB}}]= \[ {\text{L}}^2 {\text{M}}^{-1} {\text{T}}^1]$ }$

$\;10 {\text{MSD}}=11 {\text{VSD}}$ $\;1 {\text{VSD}}=\;\frac{10}{11} {\text{MSD}}$ $\; {\text{LC}}=1 {\text{MSD}}-1 {\text{VSD}}$ $\;=1 {\text{MSD}}-\;\frac{10}{11} {\text{MSD}}$ $\;=\;{1 {\text{MSD}}}/{11}$ $\;=\;\frac{5}{11} \;\text{ units }\;$

Assertion & Reason both are correct Theory

$\;F=η A \;\frac{d v}{d y}$ \;{[M L T^{-2}]\ =η $[L^2]$ $[T^{-1}]$ }$$;η= [M L^{-1} T^{-1}]$$;S . T=;\frac{F}{ℓ}=;\frac{[M L T^{-2}]$ }{[L]}$$= [M L^0 T^{-2}]$$;L=m v r= [M L^2 T^{-1}]$$;K . E=;\frac{1}{2} I \omega ^2= [M L^2 T^{-2}]$

Latent heat is specific heat$\Rightarrow \;{ {\text{ML}}^2 {\text{T}}^{-2}}/{ {\text{M}}}= {\text{M}}^0 {\text{L}}^2 {\text{T}}^{-2}$

$\;{[ {\text{h}}]= {\text{ML}}^2 {\text{T}}^{-1}}$ $\;{[ {\text{L}}]= {\text{ML}}^2 {\text{T}}^{-1}}$ $\;{[ {\text{P}}]= {\text{MLT}}^{-1}}$ $\;{[\tau ]= {\text{ML}}^2 {\text{T}}^{-2}}$(Here ${\text{h}}$ is Planck's constant, ${\text{L}}$ is angular momentum, ${\text{P}}$ is linear momentum and $\tau$ is moment of force)

$\; {\text{m}}= {\text{kc}}^{ {\text{P}}} {\text{G}}^{-1 / 2} {\text{h}}^{1 / 2}$ \; {\text{M}}^1 {\text{L}}^0 {\text{T}}^0= \[ {\text{LT}}^{-1}]$ ^{ {\text{P}}} $[ {\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}]$ ^{-1 / 2}$$[ {\text{ML}}^2 {\text{T}}^{-1}] ^{1 / 2}$By comparing${\text{P}}=1 / 2$

$\; {\text{F}}= {\text{ax}}^{2+} {\text{bt}}^{1 / 2}$ \;{[ {\text{a}}]=\;{[ {\text{F}}]}/{[ {\text{x}}^2]\ }= $[ {\text{M}}^1 {\text{L}}^{-1} {\text{T}}^{-2}]$ }$$;{[ {\text{b}}]=;{[ {\text{F}}]}/{[ {\text{t}}^{1 / 2}]$ }= $[ {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-5 / 2}]$ }$$;{[;{ {\text{b}}^2}/{ {\text{a}}}]$ =;{[ {\text{M}}^2 {\text{L}}^2 {\text{T}}^{-5}]$ }/{[ {\text{M}}^1 {\text{L}}^{-1} {\text{T}}^{-2}]$ }= $[ {\text{M}}^1 {\text{L}}^3 {\text{T}}^{-3}]$ }$