Find the acceleration due to gravity on the surface of the moon by taking a seconds pendulum from the earth's surface. Arrange the following steps in sequential order to solvbe the above problem.
(A) Find the time period of this simple pendulum on the surface of the moon using a stop clock `[ T_(M)]`
(B) The acceleration due to gravity on the moon (gm) is 1/6th the accleration due to gravity on the earth (ge).
(C) Use the formula , ` T = 2 pi sqrt(l/g)`
(D) first , take pendulum to the surface of the moon without altering the length of the pendulum (l), .
(E) Substibe the value of `T_(M)` in (a), then ` (T_(E))/(T_(M)) sqrt ((g_(M))/(g_(E))`, and obtain the value of `(g_(m))`

Options

A. ABCDE
B. DACEB
C. CBADE
D. CABED

Answer

B

Explanation

Take the pendulum to the surface of the moon and keep the length of the pendulum constant (D).find the time period fo this simple pendulum using stop watch on the surface of the moon `(T_(m)) (A)` ,Use the formula ` T = 2pi sqrt(l/g)` (C ) , substitute the values of ` T_(m) " in " T = 2pi sqrt(l/g)` and obtain ` (T_(E))/(T_(m))= sqrt((g_(M)/(g_(E)))`
From this obtain the value of (gm) (E). the acceleration due to gravity on the moon (`g_(m)`) is 1/6 th the acceleration due to gravity on the earth (`g_(E)`) (B)