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Formula used:

$T=2\pi \sqrt{\dfrac{L}{g}}$

We will solve this problem by applying the direct formula for the time period of a simple pendulum.

The time period $T$ of a simple pendulum of length $L$ is given by

$T=2\pi \sqrt{\dfrac{L}{g}}$ --(1)

where $g$ is the acceleration due to gravity.

Hence, let us analyze the question.

Let the length of the pendulum be $L$.

The acceleration due to gravity on the earth is $g=9.8m.{{s}^{-2}}$.

The time period of the pendulum on earth is $T=3.5s$.

Therefore, using (1), we get,

$T=2\pi \sqrt{\dfrac{L}{g}}$ --(2)

On the moon, the length of the pendulum is still $L$.

The acceleration due to gravity on the moon is ${{g}_{moon}}=1.7m.{{s}^{-2}}$.

Let the time period of the simple pendulum on the moon be ${{T}_{moon}}$.

Therefore, using (1), we get,

${{T}_{moon}}=2\pi \sqrt{\dfrac{L}{{{g}_{moon}}}}$ --(3)

Dividing (3) by (2), we get,

$\dfrac{{{T}_{moon}}}{T}=\dfrac{2\pi \sqrt{\dfrac{L}{{{g}_{moon}}}}}{2\pi \sqrt{\dfrac{L}{g}}}=\sqrt{\dfrac{g}{{{g}_{moon}}}}$

$\therefore \dfrac{{{T}_{moon}}}{T}=\sqrt{\dfrac{9.8}{1.7}}=\sqrt{5.76}=2.4$

$\therefore {{T}_{moon}}=2.4\times T=2.4\times 3.5s=8.40s$

Hence, the time period of the pendulum on the moon is $8.40s$.

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