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The potential energy 'U' of a particle veries with distance 'x' from a fixed origin as U=(AX)/(x^(2)+B). Where A and B are dimensional constants. Find the dimension of length in AB. |
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Answer» `therefore[B] =L^(2)` `ML""^(2)T^(-2)=([A]L)/(L^(2))` `therefore[A] =ML""^(3)T^(-2)` `[AB]=ML""^(5)T^(-2)""]` |
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