Question

Check whether the relation $S=ut+\frac{1}{2}a{t}^2$ is dimensionally correct or not. Where symbols have their

usual meaning?

• A. Yes, dimensionally correct
• B. No, dimensionally incorrect

Answer 1

The correct option is A

Yes, dimensionally correct

We have $S=ut+\frac{1}{2}a{t}^2$

Checking the dimensions on both sides, $L.H.S.=\left[M^0{L}^1{T}^0\right]$

R.H.S. = $\left[L{T}^{-1}\right]\left[T\right]+\left[L{T}^{-2}\right]\left[T^2\right]=\left[M^{\circ }{L}^1{T}^{\circ }\right]+\left[M^{\circ }{L}^1{T}^{\circ }\right]=\left[M^{\circ }{L}^1{T}^{\circ }\right]$

Comparing the L.H.S. and R.H.S.

Hence the formula is dimensionally correct.