Question 103: Units , Dimensions and Error Analysis

Question

If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :

Accepted Answer

We know that force (F) has dimensions of $$\mathrm{MLT}^{-2}$$, velocity (V) has dimensions of $$\mathrm{LT}^{-1}$$, and time (T) has dimensions of $$\mathrm{T}$$. To express density ($$\rho$$), which has dimensions of $$\mathrm{ML}^{-3}$$, in terms of F, V, and T, we need to find the exponents of F, V, and T.
Let the dimensions of density be expressed as:
$$\mathrm{[M]}^a \mathrm{[L]}^b \mathrm{[T]}^c$$
Substituting the dimensions of F, V, and T:
$$\mathrm{[F]}^a \mathrm{[V]}^b \mathrm{[T]}^c = (\mathrm{MLT}^{-2})^a (\mathrm{LT}^{-1})^b (\mathrm{T})^c$$
Since density has dimensions of $$\mathrm{ML}^{-3}$$, we can set up the following equations:
$$a = 1$$ (for the mass term M)
$$a + b = -3$$ (for the length term L)
$$-2a - b + c = 0$$ (for the time term T)
Solving these equations, we get:
$$a = 1$$
$$b = -4$$
$$c = -2$$
Thus, the dimensional formula of density in terms of F, V, and T is:
$$\mathrm{FV}^{-4} \mathrm{T}^{-2}$$

Answer

$$\mathrm{FV}^{-2} \mathrm{~T}^{2}$$

Answer

$$\mathrm{FV}^{4} \mathrm{~T}^{-6}$$

Answer

$$\mathrm{F}^{2} \mathrm{~V}^{-2} \mathrm{~T}^{6}$$

Answer

$$\mathrm{FV}^{-4} \mathrm{~T}^{-2}$$

Units , Dimensions and Error Analysis

If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :

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