Measurement MCQs for NEET — Physics Questions with Answers

$ F = ma $ $ a = { f \over m} = { N \over Kg } = N Kg ^ {-1} $

$$\begin{gathered} \frac{\mathrm{Heat}\left(\mathrm{Q}\right)}{\mathrm{Time}\left(\mathrm{t}\right)} = \frac{\mathrm{K} \times \mathrm{A}\left(\mathrm{Area}\right) \times ∆\mathrm{T}\left(\mathrm{Temperature} \mathrm{difference}\right)}{\mathrm{l}\left(\mathrm{Length}\right)} \\ Where K is the thermal conductivity \\ \mathrm{So}, \mathrm{expression} for \mathrm{K} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{other} va\mathrm{riables}, \\ \mathrm{K} = \frac{\mathrm{Q} \times \mathrm{l}}{\mathrm{t} \times \mathrm{A} \times ∆\mathrm{T}} \\ \mathrm{Putting} \mathrm{dimension} \mathrm{of} \mathrm{individual} \mathrm{quantities} \\ \mathrm{K} = \frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right] \times \left[\mathrm{L}\right]}{\left[\mathrm{T}\right] \times \left[{\mathrm{L}}^2\right] \times \left[\mathrm{K}\right]} = \left[{\mathrm{MLT}}^{-3}{\mathrm{K}}^{-1}\right] \end{gathered}$$

$$\begin{gathered} Q=mL \\ L=\frac{Q}{m}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[M\right]}=\left[M^0{L}^2{T}^{-2}\right] \end{gathered}$$

$\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{{\mathrm{MLT}}^{-2}}{{\mathrm{L}}^2}={\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}$

Light year is a distance which light travels in one year.

Because magnitude is absolute.

Impulse = change in momentum = F × t

So the unit of momentum will be equal to Newton-sec.

Option a : Stress = $\frac{Force}{Area} = \frac{Newton \left(N\right)}{{\left(Metre\right)}^2} = N/m^2$

                $Strain = Fractional change in length = \frac{∆l}{l}= Dimensionless$

So , $\frac{Stress}{Strain }= M{L}^{-1}{T}^{-2}$

Option b : Surface tension = $\frac{Force}{Length}=\frac{Newton \left(N\right)}{metre \left(m\right)} = N/m$

Option c : Energy = Force X Displacement = Newton X Metre = N X m

Option d :$Pressure =\frac{Force}{Area} = \frac{Newton \left(N\right)}{{\left(Metre\right)}^2} = N/m^2$

It is by standard definition.