Chapter MH, Problem 14PP

Solution

Interpretation: Whether the given equation is dimensionally correct or not needs to be determined.

Concept Introduction: Dimensional analysis: Whether any equation is dimensionally correct or not can be easily identified by studying the relation between the physical quantities using their dimensions and units. This is known as dimensional analysis. It helps in performing calculations as it keeps units the same. For any equation to be dimensionally correct, it doesn't have to be true.

No.

The given equation is as follows:

  $t=\frac{v}{s}$

Here, it is given that time is in seconds, v is in m/s and s is in m.

Substituting the units and checking the RHS and LHS of the equation.

  $\begin{array}{l}t=\frac{v}{s}\\ \text{s}=\frac{\text{m/s}}{\text{m}}\\ \mathrm{s}={\text{s}}^{-1}\end{array}$

Here, the equation is not dimensionally correct. The unit for v is m/s and that of s and t is m and s respectively. To obtain the same unit s must be divided by v. Thus, the correct dimensionally correct equation will be:

  $t=\frac{s}{v}$

Thus, the given equation is not dimensionally correct.

For any equation to be dimensionally correct, the units must be the same on both sides of the equation.