Question

$1Joule$in terms of mass, length, and time is:

• A. $M{L}^{-2}{T}^{-2}$
• B. $M{L}^2{T}^{-2}$
• C. $M{L}^2{T}^{-1}$
• D. $M{L}^2{T}^{-3}$

Answer 1

The correct option is B

$M{L}^2{T}^{-2}$

1. $Joule\left(J\right)$ is the SI unit of work done. Therefore, the dimensional formula of $Joule$is same as that of the work done.
2. Work done $\left(W\right)=$Force $\left(F\right)\times$Displacement $\left(s\right)$.
3. And, Force $\left(F\right)=$mass $\left(m\right)\times$acceleration $\left(a\right)$$=M^1\times [M^{0}L{T}^{-2}]$
4. The dimensions of force$\left(F\right)=M^1{L}^1{T}^{-2}$
5. Also, the dimensions of displacement$\left(s\right)=M^0{L}^1{T}^0$
6. On substituting the dimensions of force and displacement in the formula of work done, we get:-

$$\begin{gathered} W=F\times s \\ W=[M^1{L}^1{T}^{-2}]\times \left[L^1\right] \\ \therefore W=[M^1{L}^2{T}^{-2}] \end{gathered}$$

Hence, the correct answer is option (B) i.e. $M{L}^2{T}^{-2}$