Question 118: 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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