If the length of seconds pendulum on the surface of the earth is 1 m. The length of the seconds pendulum on the surface of the planet is \ g_ planet =(7)/(2)g_ earth \ periodic time of the seconds pendulum is same on

Question

Quantrex Academy · Accepted Answer

Correct answer: C. (7)/(2)\ m

The time period of a simple pendulum is given by

T = 2\pi \sqrt{\dfrac{l}{g}}

.

Since the time period of a seconds pendulum is constant (

T = 2\ s

), we have

\dfrac{l}{g} = \text{constant}

.

\dfrac{l_{earth}}{g_{earth}} = \dfrac{l_{planet}}{g_{planet}}

Substituting the given values

l_{earth} = 1\ m

and

g_{planet} = \dfrac{7}{2}g_{earth}

:

\dfrac{1}{g_{earth}} = \dfrac{l_{planet}}{\dfrac{7}{2}g_{earth}}

l_{planet} = \dfrac{7}{2}\ m

Answer:

\dfrac{7}{2}\ m

If the length of seconds pendulum on the surface of the earth is 1...

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