JEE Physics 2023 Question with Solution

Given: Force ($F$), velocity ($V$), and time ($T$) are taken as fundamental quantities.

Find: The dimensional formula of density $\rho$ in terms of $F$, $V$, and $T$.

Let

$$[\rho] = F^x V^y T^z$$

Now use the usual dimensions:

• $[F] = [MLT^{-2}]$
• $[V] = [LT^{-1}]$
• $[T] = [T]$
• $[\rho] = [ML^{-3}]$

So,

$$[ML^{-3}] = [MLT^{-2}]^x [LT^{-1}]^y [T]^z$$

Expanding,

$$[ML^{-3}] = M^x L^{x+y} T^{-2x-y+z}$$

Equating powers of $M$, $L$, and $T$:

• $x = 1$
• $x + y = -3$
• $-2x - y + z = 0$

From $x = 1$,

$$y = -4$$

Then,

$$-2(1) - (-4) + z = 0$$ $$-2 + 4 + z = 0$$ $$z = -2$$

Hence,

$$[\rho] = F^1 V^{-4} T^{-2}$$

Therefore, the dimensional formula of density is $FV^{-4}T^{-2}$.

The solution concludes this value, but it inconsistently labels the correct option as D. Among the listed options, $FV^{-4}T^{-2}$ is option C, so the correct option is C.