Answer:
Its period if its length is increased by a factor of four is 5 s.
Explanation:
The period of a simple pendulum is given by;
[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g} \ be \ constant \\\\\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\[/tex]
Given;
initial period, T₁ = 2.5
initial length, = L₁
new length, L₂ = 4L₁
the new period, T₂ = ?
[tex]\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\ T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\ T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s[/tex]
Therefore, its period if its length is increased by a factor of four is 5 s.
