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Physics · UNITS AND MEASUREMENT
हिन्दीAs per the revised SI base unit definitions, Planck's constant \( h \) has dimensions of \( [\text{M L}^2 \text{T}^{-1}] \). In a new system of units, the unit of length is halved (\( L' = 0.5\text{ L} \)) and the unit of time is doubled (\( T' = 2\text{ T} \)). If the numerical value of Planck's constant in the new system is twice its value in the SI system, then the new unit of mass \( M' \) in terms of the SI unit of mass \( M \) is:
Correct answer · Option 1
\( M' = 4M \)
Explanation
Using \( N_1 U_1 = N_2 U_2 \), we get \( N_1 [M L^2 T^{-1}] = (2N_1) [M' (0.5L)^2 (2T)^{-1}] \). Simplifying gives \( 1 = 2 \times M' \times 0.25 \times 0.5 \implies 1 = 0.25 M' \implies M' = 4M \).
