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Check the accuracy of the relation `T=sqrt((L)/(g))` for a simple pendulum using dimensional analysis. |
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Answer» The dimensions of LHS =the dimension of `T=|M^(0)L^(0)T^(1)|` The dimensions of RHS `=(("dimensions of length")/("dimensions of acceleration"))^(1//2)` (`therefore 2pi` is a dimensionless constant) `[(L)/(LT^(2))]^(1//2)=[T^(2)]^(1//2)=[T]=[M^(0)L^(0)T^(1)]` Since the dimensions are same on both the sides, the relation is correct. |
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