Question bank/JEE Main/Physics

JEE MainPhysicsModerateUnit and dimension

If the velocity of light denoted by acceleration due to
gravity are taken as fundamental
units, then the dimensions of length will be

Show answer and solution
Let LαcagbpcL\alpha c^{a}g^{b}p^{c}
L=kcagbpcL = kc^{a}g^{b}p^{c}
Putting dimensions on both sides:
[M0LT0]=[LT−1]a[LT−2]b[ML−1T−2]c\left\lbrack M^{0}LT^{0} \right\rbrack = \left\lbrack LT^{- 1} \right\rbrack^{a}\left\lbrack LT^{- 2} \right\rbrack^{b}\left\lbrack ML^{- 1}T^{- 2} \right\rbrack^{c}
[M0LT0]=[McLa+b−cT−a−2b−2c]\left\lbrack M^{0}LT^{0} \right\rbrack = \left\lbrack M^{c}L^{a + b - c}T^{- a - 2b - 2c} \right\rbrack
Comparing dimensions of both sides
c=0,     a+b−c=1c = 0,\ \ \ \ \ a + b - c = 1
−a−2b−2c=0- a - 2b - 2c = 0
solving we get a=2,b=−1a = 2,b = - 1
hence[L]=[c2/g]\lbrack L\rbrack = \left\lbrack c^{2}/g \right\rbrack

Source: Grass-G / P-05 Physics paper 10 October.docx

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