Units & Dimensions

JEE Main level, single correct, medium

71 of 149

Question 71 of 149

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

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Answer: (B) FV−4T−2F{V}^{-4}{T}^{-2}

The dimensional formula of force is
F=MLT−2F=ML{T}^{-2}
The dimensional formula of velocity is
V=LT−1V=L{T}^{-1}
The formula for density is
Density =ML3=ML−3=\dfrac{M}{{L}^{3}}=M{L}^{-3}
Comparing the density dimensions with the assumed fundamental quantities in the question,
ML−3=[F]x[V]y[T]z⇒ML−3=[MLT−2]x[LT−1]y[T]z⇒ML−3=MxL(x+y)T−2x−y+zM{L}^{-3}={\left[F\right]}^{x}{\left[V\right]}^{y}{\left[T\right]}^{z} \Rightarrow M{L}^{-3}={\left[ML{T}^{-2}\right]}^{x}{\left[L{T}^{-1}\right]}^{y}[T{]}^{z} \Rightarrow M{L}^{-3}={M}^{x}{L}^{(x+y)}{T}^{-2x-y+z}
Comparing the exponents,
x=1...(i)x+y=−3...(ii)−2x−y+z=0...(iii)x=1...(i) x+y=-3...(ii) -2x-y+z=0...(iii)
Solving equations (i), (ii) & (iii), we get
y=−4,z=−2y=-4,z=-2.
The dimensions of density =FV−4T−2=F{V}^{-4}{T}^{-2}.

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