1.

The length of a seconds pendulum on the surface of earth is 100 cm. what will be the length of a seconds pendulum on the surface of moon?

Answer»

For a simple pendulum, time period (T) is given by; T = \(2\pi\sqrt{\frac{l}{g}}\)

At Earth, T = \(2\pi\sqrt{\frac{l}{g}}\)    (i)

At Moon, T’=\(v2\pi\sqrt{\frac{l'}{g'}}\)   (ii)

Here, T = T’ and g’ = \(\frac{1}{6}g\)

So, from equation (i) & (ii)

\(\frac {l}{g}=\frac{l'}{g'}\)

or,\(l'=\frac{g'}{g}\times l\)

= \(\frac{1}{6}g\times\frac{1}{g}\times l\) 

= \(\frac{1}{6}\times l\) 

= \(\frac{1}{6}\)× 100 cm

= 16.7 cm


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