Find Dimensions of ABC Using Principle of Homogeneity

 

❓ Question

The expression given below shows the variation of velocity $(v)$ with time $(t)$:

$$v=At^2+\frac{Bt}{C+t}$$

Find the dimensions of $ABC$.

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✍️ Solution

Using the Principle of Homogeneity of Dimensions,

$$[v]=[At^2]=\left[\frac{Bt}{C+t}\right]$$

We know,

$$[v]=[LT^{-1}]$$

Finding dimension of $A$

$$[LT^{-1}]=[A][T^2]$$

Therefore,

$$[A]=[LT^{-3}]$$

Finding dimension of $C$

Since $C$ is added to $t$,

$$[C]=[t]$$

Therefore,

$$[C]=[T]$$

Finding dimension of $B$

From,

$$[v]=\left[\frac{Bt}{C+t}\right]$$

Since,

$$[C+t]=[T]$$

Therefore,

$$[LT^{-1}] = \frac{[B][T]}{[T]}$$
$$[B]=[LT^{-1}]$$

Now,

$$[ABC]=[A][B][C]$$
$$=[LT^{-3}][LT^{-1}][T]$$
$$\boxed{[ABC]=[L^2T^{-3}]}$$

✅ Final Answer

$$\boxed{[L^2T^{-3}]}$$