InterviewSolution
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InterviewSolution
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The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`. |
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Answer» From the principle of dimensional homogenity , `[x^(2)] = [ B]` :. `[B] = [L^(2)]` As well as `[U] = ([A] [x^(1//2)])/( [x^(2)] + [B]) rArr [ ML^(2)T^(-2)] = ([A] [L^(1//2)])/([L^(2)])` :. `[A] = [ ML^(7//2)T^(-2)]` Now `[AB] = [ ML^(7//2)T^(-2)] xx [L^(2)] = [ML^(11//2)T^(-2)]` |
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