Units And Measurements Practice Questions

$40 {\text{M}}=1 {\text{cm}}$ $\Rightarrow {\text{M}}=0.025 {\text{cm}} . . . . . . . . .(1)$Also, $50 {\text{V}}=49 {\text{M}}$ $\Rightarrow$ Least count $= {\text{M}}- {\text{V}}= {\text{M}}-\;\frac{49}{50} {\text{M}}=\;\frac{M}{50}$ $\Rightarrow {\text{LC}}=\;\frac{0.025}{50} {\text{cm}}$ $=\;\frac{250}{50} \times 10^{-6} {\text{m}}$ $\Rightarrow {\text{LC}}=5 \times 10^{-6} {\text{m}}$

$\because H=i^{2} R t$ $\because \%$ error in $H=2 \times 2 \%+1 \%+3 \%$ $=8 \%$

$\;\text{ 1 M.S.D }\;=1 {\text{mm}}$ $\;\text{ 9 M.S.D }\;=10 {\text{V}} . S . D$ $\;\text{ 1 V.S.D }\;=0.9 \;\text{ M.S.D }\;=0.9 {\text{mm}}$ $\;\text{ L.C of vernier caliper }\;=1-0.9=0.1 {\text{mm}}=0.01 {\text{cm}}$ $\;\text{ zero error }\;=(10-4) \times 0.1 {\text{mm}}=-0.6 {\text{mm}}$ $\;\text{ Reading }\;=\;\text{ M.S.R }\;+\;\text{ V.S.R }\;-\;\text{ Zero error }\;$ $=3 {\text{cm}}+6 \times 0.01-[-0.06]$ $=3+0.06+0.06$ $=3.12 {\text{cm}}$Nearest given answer in the options is $3.10$

$\rho =\;\frac{m}{V}=\;\frac{m}{\pi r^2 \times l}$ $\therefore \%$ error in $\rho = (\;\frac{0.006}{0.6}+2 \times \;\frac{0.005}{0.5}+\;\frac{0.04}{4}) \times 100$ $=4 \%$

Least count $=\;\frac{0.5}{50} {\text{mm}}=0.01 {\text{mm}}$ $\therefore$ Diameter, ${\text{d}}=1.5 {\text{mm}}+7 \times 0.01$ $=1.57 {\text{mm}}$ $\therefore$ Surface area $=(2 \pi r) \times l$ $=\pi d l$ $=3.142 \times \;\frac{1.57}{10} \times 6.8 {\text{cm}}^2$ $=3.354 {\text{cm}}^2=3.4 {\text{cm}}^2$

Dimensional formula for Torque$[\tau ]=[ {\text{ML}}^2 {\text{T}}^{-2}]$NowPercentage error in torque $=\% \tau =\% {\text{M}}+2 \% {\text{L}}$ $2 \% {\text{T}}$ $\% \tau =25 \%$

${[P]=\;\frac{[\alpha ]}{[\beta ]} \Rightarrow 1 \;\frac{[\alpha ]}{[\beta ]} }$ $\Rightarrow [\alpha ]=[\beta ] . . . . . .(1)$Now, $[\;\frac{k t}{\beta t}]=1$We know, E $=\;\frac{3}{2} k T(K T G)$ $\therefore [$ Energy] $=[ {\text{KT}}]$ $\therefore [\beta x]=[k T]$ ${[\beta ]=\;\frac{M L^2 T-2}{[x]}=\;\frac{M L^2 T^{-2}}{L} }$ $\therefore [\alpha ]=[\beta ]=M L T{ }^{-2}$

$\%$ error in $z=3 \times 4+\;\frac{1}{2} \times 12$ $=12+6=18 \%$

$\;\text{ [pascal-second] }\;=\;\frac{M L T^{-2}}{L^2} \times T$ $=M L^{-1} T^{-1}$

Velocity gradient $=\;\frac{d v}{d x}$ $\Rightarrow$ Dimensions are \;\frac{[L T^{-1}]\}{[L]}=[T^{-1}]$Decay constant$λ$has dimensions of$[T^{-1}]$because of the relation$;\frac{d N}{d t}=-λ N$$\Rightarrow $ Velocity gradient and decay constant have same dimensions.

$[η]=[M L^{-1} T^{-1}]$Now if $[η]=[P]^{a}[A]^{b}[T]^{c}$ \Rightarrow \[M L^{-1} T^{-1}]$=$[M L^{1} T^{-1}]^{a}$[L^{2}]^{b}[T]^{c}$$\Rightarrow a=1, a+2 b=-1,-a+c=-1$$\Rightarrow a=1, b=-1, c=0$$\Rightarrow [η]=[P][A]^{-1}[T]^{0}$$=[P A^{-1} T^{0}]$

${\text{MSR}}=2.5 {\text{mm}}$ ${\text{CSR}} \;\; =45 \times \;\frac{0.5}{50} {\text{mm}}$ $=0.45 {\text{mm}}$ $\;\text{ Diameter reading }\; \;\; = {\text{MSR}}+ {\text{CSR}}-\;\text{ zero error }\;$ $\;\; =2.5+0.45-(-0.03)$ $\;\; =2.98 {\text{mm}}$

$1 {\text{MSD}}=\;\frac{1}{20} {\text{cm}}$ $1 {\text{VSD}}=\;\frac{24}{25} {\text{MSD}}=\;\frac{24}{25} \times \;\frac{1}{20} {\text{cm}}$ $\therefore$ Least count $=\;\frac{1}{20} (1-\;\frac{24}{25}) {\text{cm}}$ $=\;\frac{1}{20} \times \;\frac{1}{25}=\;\frac{1}{500} {\text{cm}}$ $=0.002 {\text{cm}}$

Since zero error is negative, we will add $0.05 {\text{cm}}$.$\Rightarrow \;\text{ Corrected reading }\;=1.7 {\text{cm}}+5 \times 0.1 {\text{mm}}+0.05 {\text{cm}}$ $=180 \times 10^{-2} {\text{cm}}$

$u=\;\frac{\alpha }{\beta } \ sin (\;\frac{\alpha x}{k t})$ ${[\alpha ]=[\;\frac{k t}{x}]=\;\frac{[\;\text{ Energy }\;]}{[\;\text{ Dis }\; \tan c e]} }$ ${[\beta ]=\;\frac{[\alpha ]}{[u]} }$ $=\;\frac{[\;\text{ Energy }\;] /[\;\text{ Distan }\; c e]}{[\;\text{ Energy }\;] / \;\text{ Volume }\;]}$ $=[L^2]$

$[S]=\;\frac{[C]}{[m] \times [\Delta T]}$and, $[L]=\;\frac{[Q]}{[m]}$ $\Rightarrow$ They have different dimensions.

As we know that, The dimensional formula of following terms are (a) [ Torque ] = [ Radius ] × [ Force ] =\[{\text{M}}^{0} {\text{LT}}^{0}]$$[{\text{M}} {\text{L}} {\text{T}}^{-2}]$=[{\text{ML}}^{2} {\text{T}}^{-2}]$(b) [Impulse] = [Force] [Time]$=$[{\text{MLT}}^{-2}]$[{\text{T}}]=[{\text{MLT}}^{-1}]$(c) [ Tension ] = [ Force ]$= [{\text{M}} {\text{L}} {\text{T}}^{-2}]$(d) [Surface tension] =${[{ \text{Force} }]}/{[{ \text{Length} }]}= {[{\text{ML}} {\text{L}}^{-2}]$}/{[{\text{L}}]}=[{\text{MT}}^{-2}]$ Hence, option (a) is the correct.

$\frac{\Delta{y}}{{y}}=\frac{2 \Delta{m}}{{m}}+\frac{4 \Delta{r}}{{r}}+\frac{{x} \Delta{g}}{{g}}+\frac{3}{2} \frac{\Deltal}{l}$ $18=2(1)+4(0.5)+x p+\frac{3}{2}(4)$ $8={xp}$ By checking from options. $x=\frac{16}{3}, p=\±\frac{3}{2}$