Question

If force is proportional to the square of velocity, then the dimensions of the proportionality constant are

• A. ${\text{[ML}}^{-1}\text{T]}$
• B. $\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^0\right]$
• C. $[{\mathrm{MLT}}^0]$
• D. ${\text{[ M}}^0{\mathrm{LT}}^{-1}]$

Answer 1

The correct option is B

$\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^0\right]$

Step 1: Given information

As given force is proportional to the square of the velocity

$\therefore F\propto V^2$

Now, $\therefore F=k{V}^2$ , where $k$ is the proportional constant

Step 2: Find the dimension of Force

$\mathrm{Force}=\mathrm{Mass}\times \mathrm{Acceleration}$ . . . . (1)

$\text{∴Acceleration = Velocity ×}{\left[\text{Time}\right]}^{\text{–1}}$

$\begin{array}{l}=\left[{\mathrm{LT}}^{-1}\right]\times {\left[\mathrm{T}\right]}^{-1}\\ =\left[{\mathrm{LT}}^{-2}\right]...\left(2\right)\end{array}$

Where, M = Mass, L = Length, and T = Time

Substituting equation (2) in equation (1) we get,
$\begin{array}{c}\mathrm{Force}=\mathrm{m}\times \mathrm{a}\\ =\left[\mathrm{M}\right]\times \left[{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]\\ =\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]\end{array}$
Therefore, force is dimensionally represented as $\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]$

Step 3: Find the dimension of Velocity

$\begin{array}{c}\text{∴Velocity}=\frac{\text{Distance}}{\text{Time}}\\ =\frac{\left[\mathrm{L}\right]}{\left[\mathrm{T}\right]}\\ =\left[{\mathrm{LT}}^{-1}\right]\end{array}$

$\therefore \mathrm{V}=\left[{\mathrm{LT}}^{-1}\right]$

Step 4: Put the dimension of Velocity and Force in $F=k{V}^2$

Therefore,$\left[{\mathrm{MLT}}^{-2}\right]=\mathrm{k}\left[{\mathrm{LT}}^{-1}\right]$

$\begin{array}{c}k=\frac{\left[ML{T}^{-2}\right]}{{\left[L{T}^{-1}\right]}^2}\\ =\frac{\left[ML{T}^{-2}\right]}{\left[L^2{T}^{-2}\right]}\\ =\left[M{L}^1{L}^{-2}{T}^{-2}{T}^2\right]\\ =\left[M{L}^{-1}{T}^0\right]\end{array}$

$\therefore k=\left[M{L}^{-1}{T}^0\right]$

Hence, the correct option is (B).