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Question: If the velocity of light (3), gravitational constant (G) and Planck's constant (h) are chosen as fun...

If the velocity of light (3), gravitational constant (G) and Planck's constant (h) are chosen as fundamental units, then the dimensions of mass in new system is

A

c1/2G1/2h1/2c^{1/2}G^{1/2}h^{1/2}

B

c1/2G1/2h−1/2c^{1/2}G^{1/2}h^{- 1/2}

C

c1/2G−1/2h1/2c^{1/2}G^{- 1/2}h^{1/2}

D

c−1/2G1/2h1/2c^{- 1/2}G^{1/2}h^{1/2}

Answer

c1/2G−1/2h1/2c^{1/2}G^{- 1/2}h^{1/2}

Explanation

Solution

Let m∝cxGyhzm \propto c^{x}G^{y}h^{z} or m=KcxGyhzm = Kc^{x}G^{y}h^{z}

By substituting the dimension of each quantity in both sides

[M1L0T0]=K[LT−1]x[M−1L3T−2]y[ML2T−1]z=[M−y+zLx+3y+2zT−x−2y−z]\lbrack M^{1}L^{0}T^{0}\rbrack = K\lbrack LT^{- 1}\rbrack^{x}\lbrack M^{- 1}L^{3}T^{- 2}\rbrack^{y}\lbrack ML^{2}T^{- 1}\rbrack^{z} = \lbrack M^{- y + z}L^{x + 3y + 2z}T^{- x - 2y - z}\rbrack

By equating the power of M, L and T in both sides : −y+z=1- y + z = 1, x+3y+2z=0x + 3y + 2z = 0, −x−2y−z=0- x - 2y - z = 0

By solving above three equations x=1/2x = 1/2, y=−1/2y = - 1/2 and z=1/2z = 1/2.

∴\therefore m∝c1/2G−1/2h1/2m \propto c^{1/2}G^{- 1/2}h^{1/2}