based on the concept of dimensional analysis derive the formula T=K
for time period of a pendulum whose experimental value of the constant K is given as 2
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The variables on which Time period(T) depends on is length(l) and acceleration due to gravity(g).
![t=Kl^ag^b\\ \\ \ [t]=[M^0L^0T^1]\\\ [l]^a=[M^0L^1T^0]^a=[M^0L^aT^0]\\\ [g]^b=[M^0L^1T^{-2}]^b=[M^0L^bT^{-2b}]\\ \\ Comparing\ dimensions\ of\ L\\a+b=0\\ \Rightarrow a=-b\\ Comparing\ dimensions\ of\ T\\-2b=1\\ \Rightarrow b = \frac{1}{-2}=- \frac{1}{2}\\ thus\ a=-b= \frac{1}{2}\\ given\ that\ K=2 \pi \\ \\So\ time\ period\ is\ given\ by\\ \\ t=Kl^{ \frac{1}{2} }g^{- \frac{1}{2}}\\ \\ \Rightarrow t=2 \pi \sqrt{ \frac{l}{g} }](https://tex.z-dn.net/?f=t%3DKl%5Eag%5Eb%5C%5C+%5C%5C+%5C+%5Bt%5D%3D%5BM%5E0L%5E0T%5E1%5D%5C%5C%5C+%5Bl%5D%5Ea%3D%5BM%5E0L%5E1T%5E0%5D%5Ea%3D%5BM%5E0L%5EaT%5E0%5D%5C%5C%5C+%5Bg%5D%5Eb%3D%5BM%5E0L%5E1T%5E%7B-2%7D%5D%5Eb%3D%5BM%5E0L%5EbT%5E%7B-2b%7D%5D%5C%5C+%5C%5C+Comparing%5C+dimensions%5C+of%5C+L%5C%5Ca%2Bb%3D0%5C%5C+%5CRightarrow+a%3D-b%5C%5C+Comparing%5C+dimensions%5C+of%5C+T%5C%5C-2b%3D1%5C%5C+%5CRightarrow+b+%3D++%5Cfrac%7B1%7D%7B-2%7D%3D-+%5Cfrac%7B1%7D%7B2%7D%5C%5C++thus%5C+a%3D-b%3D+%5Cfrac%7B1%7D%7B2%7D%5C%5C+given%5C+that%5C+K%3D2+%5Cpi+%5C%5C+%5C%5CSo%5C+time%5C+period%5C+is%5C+given%5C+by%5C%5C+%5C%5C+t%3DKl%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7Dg%5E%7B-+%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C+%5C%5C+%5CRightarrow+t%3D2+%5Cpi++%5Csqrt%7B+%5Cfrac%7Bl%7D%7Bg%7D+%7D+ "t=Kl^ag^b\\ \\ \ [t]=[M^0L^0T^1]\\\ [l]^a=[M^0L^1T^0]^a=[M^0L^aT^0]\\\ [g]^b=[M^0L^1T^{-2}]^b=[M^0L^bT^{-2b}]\\ \\ Comparing\ dimensions\ of\ L\\a+b=0\\ \Rightarrow a=-b\\ Comparing\ dimensions\ of\ T\\-2b=1\\ \Rightarrow b = \frac{1}{-2}=- \frac{1}{2}\\ thus\ a=-b= \frac{1}{2}\\ given\ that\ K=2 \pi \\ \\So\ time\ period\ is\ given\ by\\ \\ t=Kl^{ \frac{1}{2} }g^{- \frac{1}{2}}\\ \\ \Rightarrow t=2 \pi \sqrt{ \frac{l}{g} }")
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