If $E, m, J$ and $G$ denote energy, mass, angular momentum and gravitational constant respectively. Then the dimensions of $\frac{{E J}^{2}}{m^{5} G^{2}}$ are same as that of

angle

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length

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mass

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time

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Solution

The correct option is B angle
$E = M^{1} L^{2} T^{- 2}$
$M = M^{1}$
$J = M^{1} L^{2} T^{- 1}$
$G = (M^{1} L^{1} T^{- 2}) \times \frac{L^{2}}{M^{2}} = M^{- 1} L^{3} T^{- 2}$

$\frac{E J^{2}}{m^{5} G^{2}} = \frac{(M^{1} L^{2} T^{- 2}) \times {(M^{1} L^{2} T^{- 1})}^{2}}{M^{5} \times {(M^{- 1} L^{3} T^{- 2})}^{2}} = \frac{M^{3} \times L^{6} T^{- 4}}{M^{3} \times L^{6} T^{- 4}} = M^{0} L^{0} T^{0}$

Angle $= M^{0} L^{0} T^{0}$
Length $= M^{0} L^{1} T^{0}$
Mass $= M^{1} L^{0} T^{0}$
Time $= M^{0} L^{0} T^{1}$
Hence, option A is correct.

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The correct option is B angleE=M1L2T−2M=M1J=M1L2T−1G=(M1L1T−2)×L2M2=M−1L3T−2EJ2m5G2=(M1L2T−2)×(M1L2T−1)2M5×(M−1L3T−2)2=M3×L6T−4M3×L6T−4=M0L0T0Angle =M0L0T0Length =M0L1T0Mass =M1L0T0Time =M0L0T1Hence, option A is correct.

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