Respuesta :
Answer:
[tex]g'=3.71\ m/s^2[/tex]
Explanation:
Given that,
Time period of a pendulum on the earth's surface, T₁ = 1.2 s
Time period of the same pendulum on Mercury, T₂ = 1.95 s
The time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
On earth :
[tex]T_1=2\pi \sqrt{\dfrac{l}{g}}[/tex]
[tex]1.2=2\pi \sqrt{\dfrac{l}{9.8}}[/tex].............(1)
Let g' is the acceleration due to gravity on Mercury. So,
[tex]1.95=2\pi \sqrt{\dfrac{l}{g'}}[/tex]............(2)
From equation (1) and (2) :
[tex]\dfrac{1.2}{1.95}=\sqrt{\dfrac{g'}{9.8}}[/tex]
[tex]g'=(\dfrac{1.2}{1.95})^2\times 9.8[/tex]
[tex]g'=3.71\ m/s^2[/tex]
So, the acceleration due to gravity on the mercury is [tex]3.71\ m/s^2[/tex]. Hence, this is the required solution.
