Units & Dimensions

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62 of 149

Question 62 of 149

Consider two physical quantities AA and BB related to each other as E=B−x2AtE=\dfrac{B-{x}^{2}}{At} where E,xE,x and tt have dimensions of energy, length and time respectively. The dimension of ABAB is

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Answer: (B) L2M−1T1{L}^{2}{M}^{-1}{T}^{1}

From the dimensional checking of the equation, we can say the dimensions of [B]=[x2]=[L2]\left[B\right]=\left[{x}^{2}\right]=\left[{L}^{2}\right]
And dimensions of [A]=[x2][tE]=[L2][T][ML2T−2]=1[MT−1]\left[A\right]=\dfrac{\left[{x}^{2}\right]}{\left[tE\right]}=\dfrac{\left[{L}^{2}\right]}{\left[T\right]\left[M{L}^{2}{T}^{-2}\right]}=\dfrac{1}{\left[M{T}^{-1}\right]}
Hence, [A]=[M−1T]\left[A\right]=\left[{M}^{-1}T\right]
Required value, [AB]=[L2M−1T1]\left[AB\right]=\left[{L}^{2}{M}^{-1}{T}^{1}\right]

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