Chapter 1, Problem 32P

To determine

The dimensions and S.I units for A and B.

Answer to Problem 32P

Solution:

Dimensions of A and B are $\left[{\text{LT}}^{-4}\right]$ and $\left[{\text{LT}}^{-2}\right]$ respectively. The S.I units are $\text{m}\cdot {\text{s}}^{-4}$ and $\text{m}\cdot {\text{s}}^{-2}$ respectively.

Explanation of Solution

Given:

The speed $v$ is given by $v=A{t}^3-Bt$ , where $t$ refers to time.

The dimension of $v$ is $\left[{\text{LT}}^{-1}\right]$ . The dimensions of both $A{t}^3$ and $Bt$ should be same since only physical quantities of same dimensions can be added. The dimension of both terms should be equal to $\left[{\text{LT}}^{-1}\right]$ which is the dimension of $v$ .

Hence,dimension of:

  $\begin{array}{l}A{t}^3=\left[{\text{LT}}^{-1}\right]\\ A\left[{\text{T}}^3\right]=\left[{\text{LT}}^{-1}\right]\\ A=\left[{\text{LT}}^{-4}\right]\end{array}$

Similarly,

  $\begin{array}{l}Bt=\left[{\text{LT}}^{-1}\right]\\ B\left[\text{T}\right]=\left[{\text{LT}}^{-1}\right]\\ B=\left[{\text{LT}}^{-2}\right]\end{array}$

Dimensions of A and B are $\left[{\text{LT}}^{-4}\right]$ and $\left[{\text{LT}}^{-2}\right]$ respectively.

The S.I unit for $\left[\text{L}\right]$ and $\left[\text{T}\right]$ are mands respectively. Hence, S.I unit for

  $A=\left[{\text{LT}}^{-4}\right]=\text{m}\cdot {\text{s}}^{-4}$

  $B=\left[{\text{LT}}^{-2}\right]=\text{m}\cdot {\text{s}}^{-2}$

Conclusion: Dimensions of A and B are $\left[{\text{LT}}^{-4}\right]$ and $\left[{\text{LT}}^{-2}\right]$ respectively. The S.I units are $\text{m}\cdot {\text{s}}^{-4}$ and $\text{m}\cdot {\text{s}}^{-2}$ respectively.