Measurement MCQs for NEET — Physics Questions with Answers

$mv=kg\left(\frac{m}{\sec }\right)$

Quantities of similar dimensions can be added or subtracted so unit of a will be same as that of velocity.

$1MeV={10}^{6}eV$

Energy (E) = F × d $\Rightarrow$ $F=\frac{E}{d}$

so Erg/metre can be the unit of force.

$\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{Q}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.}}=\frac{\mathrm{Watt}}{\mathrm{Ampere}}=\mathrm{volt}$

Kilowatt-hour is a unit of energy.

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{IdlSin\theta }}{{\mathrm{r}}^2} \\ {\mathrm{\mu }}_0=\mathrm{B}\times \frac{4{\mathrm{\pi r}}^2}{\mathrm{IdlSin\theta }} \\ \mathrm{SI} \mathrm{unit} = \mathrm{Telsa}-\mathrm{metre}/\mathrm{Amp}. \left(\mathrm{If} \mathrm{SI} \mathrm{unit} \mathrm{of} \mathrm{B}=\mathrm{Tesla}\right) \\ \& \mathrm{SI} \mathrm{unit} = \mathrm{webre}/\mathrm{metre}-\mathrm{Ampere} \left(\mathrm{If} \mathrm{SI} \mathrm{unit} \mathrm{of} \mathrm{B}=\frac{\mathrm{webre}}{{\mathrm{m}}^2}\right) \\ \& \mathrm{SI} \mathrm{unit} = \mathrm{Henry}/\mathrm{metre} \left[\mathrm{If} \mathrm{SI} \mathrm{unit} \mathrm{of} \mathrm{B}=\frac{\mathrm{Henry}-\mathrm{Amp}}{{\mathrm{metre}}^2}\right] \\ \left[\mathrm{Henry} = {\mathrm{kgm}}^2{\mathrm{S}}^{-2}{\mathrm{A}}^{-2}\right] \end{gathered}$$

Weber is a unit of magnetic flux. 

Flux = Magnetic Field X Area

        = Tesla X (meter)^2

This is called Weber which is in SI units.

Magnetic induction and Magnetic intensity.

Energy = force × distance, so if both are increased by 4 times then energy will increase by 16 times.