1.

If the length of a seconds pendulum is doubled, what will be the new period?

Answer»

T = 2\(\pi\) \(\sqrt{\frac Lg}\)

\(\therefore\) \(\frac{T_2}{T_1}\) = \(\sqrt{\frac {L_2}{2}}\) 

\(\therefore\) \(\frac{T_2}{2}\)= √2 

\(\therefore\) T2 : 2√2 s giuves the required period



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