Units And Measurements Practice Questions

To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope

Least count of screw gauge $=\;\frac{0.5}{50} {\text{mm}}=0.01 {\text{mm}}$Main scale reading $=3 {\text{mm}}$Vernier scale reading $=35$ $\therefore$ Reading $=$ [Main scale reading $+$ circular scale reading $\times$ L.C] $-$ (zero error)$=[3+35 \times 0.01]-(-0.03)=3.38 {\text{mm}}$

Momentum, $p=m \times v$ $\;=3.513 \times 5.00=17.565 {\text{kg}} \frac{\text{m}}{\ s}$ $\;=17.6 {\text{kg}} \frac{\text{m}}{\ s}$Here number of significant digits in $m$ is 4 and in $v$ is 3 , so, $p$ must have minimum (which is 3) significant digit.

We know that,Lorentz force $|\vec{F}|=|\vec{q v} \times \vec{B}|$ \therefore [B]=\;\frac{[F]}{[q][v]}=\;\frac{[M L T^{-2}]\ }{[C] × $[L T^{-1}]$ }= [M T^{-1} C^{-1}]$

From stokes law, Viscous force ${\text{F}}=6 \pi η r v$ $\;\Rightarrow η=\;\frac{F}{6 \pi r v}$ \; \therefore [η]=\;\frac{[M L T^{-2}]\ }{[L] $[L T^{-1}]$ }$$;\Rightarrow [η]= [M L^{-1} T^{-1}]$

The velocity of light in vacuum is $c=\;{1}/{{\sqrt{\mu _0 \epsilon _0}}} ;$ \therefore [\;\frac{1}{\mu _0 \epsilon _0}] = \[ {\text{c}}^2]$ = [ {\text{L}}^2 {\text{T}}^{-2}]$$\therefore $Dimension of$;\frac{1}{\mu _0 \epsilon _0}= [ {\text{L}}^2 {\text{T}}^{-2}]$

Momentum $= {\ mv}= [M L T^{-1}]$ Planck's constant, {\text{h}}=\;\frac{E}{v}=\;\frac{[M L^2 T^{-2}]\ }{[T^{-1}]$ }= [M L^2 T^{-1}]$ So Momentum and Planck's constant do not have same dimensions.

Work $( {\text{W}})=\vec{F} \cdot \vec{s}= {\text{Fs}} \cos \theta$ =\[ {\text{MLT}}^{-2}]$[L]=[ {\text{ML}}^2 {\text{T}}^{-2}]$Torque$\vec{(\tau )}=\vec{r} \times \vec{F} \Rightarrow \tau =r \vec{F} \sin \theta $$=[L]$[ {\text{MLT}}^{-2}]$=[ {\text{ML}}^2 {\text{T}}^{-2}]$ So dimension of torque and work are same.