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JEE Main 2016 · Physics · Units and Measurements · medium

In a system of units, force F, velocity V and time T are taken as the fundamental quantities. The dimensions of mass density in this system are

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In a system of units, force FF, velocity VV and time TT are taken as the fundamental quantities. The dimensions of mass density in this system are
  1. A. [F V−4 T−2][F\,V^{-4}\,T^{-2}]
  2. B. [F V−3 T−1][F\,V^{-3}\,T^{-1}]
  3. C. [F V−4 T2][F\,V^{-4}\,T^{2}]
  4. D. [F−1 V4 T2][F^{-1}\,V^{4}\,T^{2}]
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Answer: (A) [F V−4 T−2][F\,V^{-4}\,T^{-2}]

Express length and mass using FF, VV, TT. Length L=VTL=VT (velocity times time). Force is MLT−2MLT^{-2}, so M=FLT−2=F(VT)T−2=FTVM=\dfrac{F}{LT^{-2}}=\dfrac{F}{(VT)T^{-2}}=\dfrac{FT}{V}. Density is mass per unit volume: ρ=ML3=FTV−1V3T3=F V−4 T−2\rho=\dfrac{M}{L^3}=\dfrac{FTV^{-1}}{V^3T^3}=F\,V^{-4}\,T^{-2}. Check in the usual system: FV4T2=MLT−2L4T−4T2=ML−3\dfrac{F}{V^4T^2}=\dfrac{MLT^{-2}}{L^4T^{-4}T^{2}}=ML^{-3}, correct.
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