A simple pendulum designed on the moon as a seconds pendulum is taken to a planet where the acceleration due to gravity on the surface is twice that on the Earth . If `g_("earth") : g_("moon") =6 :1` find the period of oscillation of the pendulum on the planet mentionedabove.

Explanation

(i) Express 1 m.s.d in the different portions of the scale as a fraction of an inch and convert into cm . Count the total number of divisions and convert into cm to arrive at the length of object in cm. Alternately find the length in inches and convert into cm
(ii) Count the number of M.S.D. 's from left block to 2 say x
`1 M.S.D. =(1)/(16) inch`
Count the number of M.S.D.'s from 2 to right block say y
here ` 1 M.S.D. =(1)/(8)`inch
Length of the object `=(x xx (1)/(16) +yxx(1)/(8) ) `inch `=((x+2y)/(16))` inch
Convert this value to cm
(iii) 5.87 cm