QJEEM23Ph7acf: JEE Main 2023 Physics - Class 11 JEEM23Ph7acf Solution: If force (F), velocity (V) and time (T) are considered as…

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:

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