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`.A. ` M^(1) L^(7//2) T^(-2)`B. ` M^(1) L^(11//2) T^(-2)`C. ` M^(1) L^(5//2) T^(-2)`D. ` M^(1) L^(9//2) T^(-2)` |
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Answer» Correct Answer - B Here `x^(2)` has the dimensiions of `L^(2) , B = [L^(2)]` Also `M L^(2) T^(-2) = (AL^(1//2))/(L^(2))` or `A = ML^(7//2 T^(-2))` `:. A xx B = M L^(1//2 T^(-2))` |
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