Compare the time periods of two pendulums of length 1m and 9m

As we know that,

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

Time period is directly proportional to the square root of the length of the pendulum.

In the case when the length is 1 m,

T₁ = 2 \pi \sqrt{1\over g}

and

In the case when the length is 9m,

T₂ = 2 \pi \sqrt{9\over g}

So, a comparison of T₁ and T₂ gives

T₁ : T₂ = 2 \pi \sqrt{l\over g} : 2 \pi \sqrt{9\over g}

⇒ T₁ : T₂ = \sqrt{1\over 9}

⇒ T₁ : T₂ = 1\over 3

Hence, T₁ : T₂ = 1 : 3