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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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