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Based on the concept of dimensional analysis derive the formula T=K

 \sqrt{ \frac{l}{g}

for time period of a pendulum whose experimental value of the constant K is given as 2 \pi



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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} }
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