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Question: Energy is given by E = \(\frac{a–x}{bt}\)Where E is energy, x is distance and t is time. Dimensional...

Energy is given by E = axbt\frac{a–x}{bt}Where E is energy, x is distance and t is time. Dimensional formula of a and b are –

A

L, M-1L-1T\text{L, }\text{M}^{\text{-1}}L^{\text{-1}}T

B

ML-2T, T\text{M}\text{L}^{\text{-2}}\text{T, T}

C

L, ML2 T -1\text{L, M}\text{L}^{2}\text{ T}\text{ }^{\text{-1}}

D

LT -1, ML2T -1\text{LT}\text{ }^{\text{-1}}\text{, M}\text{L}^{2}\text{T}\text{ }^{\text{-1}}

Answer

L, M-1L-1T\text{L, }\text{M}^{\text{-1}}L^{\text{-1}}T

Explanation

Solution

[a] = [x]\mathbf{\lbrack}\mathbf{a}\mathbf{\rbrack}\mathbf{\ }\mathbf{=}\mathbf{\ }\mathbf{\lbrack}\mathbf{x}\mathbf{\rbrack}

[a] = [L]\lbrack a\rbrack\ = \ \lbrack L\rbrack

And

[E] = [ax][b][t]\frac{\lbrack a–x\rbrack}{\lbrack b\rbrack\lbrack t\rbrack}

[b] = [ax][E][t]\frac{\lbrack a–x\rbrack}{\lbrack E\rbrack\lbrack t\rbrack} = [L][ML2T2][T]\frac{\lbrack L\rbrack}{\lbrack ML^{2}T^{–2}\rbrack\lbrack T\rbrack}

= [M-1L-1T+1 ]= \ \lbrack M^{\text{-1}}L^{\text{-1}}T^{+ 1}\ \rbrack