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Answer 1 · 2016-05-14T15:10:42

#2sqrt2s#

For small angle of deviation #theta# from the mean position the time period #T# of a pendulum is given as

#T=2pisqrt(L/g)# .......(1)
where #L# is the length of the pendulum and #g# is the local gravity.

We know that seconds pendulum is a pendulum which has its period of two seconds. Its each swing takes one second, #T=2s#
Also local acceleration due to gravity #g=G (M_e)/R_e^2#.
#G# is constant #=6.67408 xx 10^-11 m^3 kg^-1 s^-2#. #M_e# is mass and #R_e# is radius of earth respectively.

Now the acceleration due gravity of the planet is

#g_p=G (M_p)/R_p^2#

Inserting given quantities we get

#g_p=G (2M_e)/(2R_e)^2#
#=>g_p=G (M_e)/(2(R_e)^2)#,
in terms of #g#
#g_p=g/2#

Inserting in (1)

#T_p=2pisqrt(L/g_p)# ........(2)

in terms of #g#

#T_p=2pisqrt(2L/g)#

Using (1)

#T_p=sqrt2T#
#=>T_p=2sqrt2s#