Question

iii) The length of second's pendulum on the
surface of earth is nearly 1 m. Its length
on the surface of moon should be (Given:
acceleration due to gravity (g) on moon
is 1/6 th of that on the earth's surface]​

Answer 1

Answer:

Explanation:

A second's pendululm means a simple pendulum having time period T=2s.

For a simple pendulum, T=2πlg−−√

where, l=length of the pendulum and

g= acceleration due to gravity on surface of the earth.

Te=2πlege−−−√ .....(i)

On the surface of the moon, Tm=2πlmgm−−−√ .....(ii)

∴TeTm=2π2πlege−−−√×gmlm−−−√

Te=Tm to maintain the second's pendululm time period.

∴1=lege×gmle−−−−−−−−√ .....(iii)

but the acceleration due to gravity at moon is 1/6 of the acceleration due to gravity at earth,

i.e., gm=ge6

Squaring Eq. (iii) and putting this value,

1=lelm×ge/6ge=lelm×16

⇒lelm=1⇒6lm=le

lm=16le=16×1=16m