# JEE Class main Answered

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If plancks constant h, speed of light c and universal gravitational constant G are combined to form the unit of mass ,

then we have

\ \ ................................(1)

where m is mass . , and are constants to be determined by using dimensional analysis.

SI unit of planck constant h = J s = N m s = kg m s^{-2 }m s

Hence dimension of planck's constant h = [ M L^{2} T^{-1} ]

SI unit of speed of light c = m s^{-1}

Hence dimension of speed of light = [ L T^{-1 }]

SI unit of gravitational constant = N m^{2} kg^{-2} = kg m s^{-2} m^{2} kg^{-2} = kg^{-1} m^{3} s^{-2}

Hence dimension of gravitational constant = [ M^{-1} L^{3} T^{-2} ]

Let us do dimensional analysis to eqn.(1)

By equating power of each dimesnion , we get the following equation

............................... (2)

........................(3)

..........................(4)

By solving eqn.(2) , eqn.(3) and eqn.(4) , we get

α = 1/2 , β = 1/2 and γ = -1/2

Hence we can writte eqn.(1) as

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If plancks constant h, speed of light c and universal gravitational constant G are combined to form the unit of length ,

then we have

.................................(5)

where l is length . , and are constants to be determined by using dimensional analysis.

As it is done in previous case , if we do dimensional analysis for eqn.(5) , we get

By equating power of each dimesnion , we get the following equation

α - γ = 0 ...................................(6)

2α + β + 3γ = 1 ........................(7)

α + β + 2 γ = 0 ......................(8)

By solving above equations, we get α = 1/2 , β = -3/2 and γ = 1/2

Hence eqn.(5) is written as

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