If energy of a body moving along x-axis at time t andposition x isgivenby E=A−x2Bt,then the dimensions of A × B will be

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If energy of a body moving along x-axis at time $t and position x is given by E = \frac{A - x^{2}}{Bt}, then$ the dimensions of A × B will be

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$[{ML}^{- 3} T^{2}]$

$M^{- 1} L^{3} T^{- 2}]$

$[M^{- 1} L^{2} T^{1}]$

$[{MLT}^{- 2}]$

answer is A.

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Detailed Solution

$A] = [x^{2}] = [L^{2}] Also [B] = \frac{[x^{2}]}{(E) (t)} = \frac{(L^{2})}{({ML}^{2} T^{- 2}) (T)} = [M^{- 1} T^{1}] Hence [AB] = [M^{- 1} L^{2} T^{1}]$

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