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Time period of simple pendulum of length l and a place where acceleration due to gravity is g is T. what is the period of a simple pendulum of the same length at a place where the acceleration due to gravity is 1.029 is, (a) T (b) 1.02 T (c) 0.99 T (d) 1.01 T

Time period of simple pendulum of length l and a place where acceleration due to gravity is g is T. what is the period of a simple pendulum of the same length at a place where the acceleration due to gravity is 1.029 is,

(a) T (b) 1.02 T
(c) 0.99 T (d) 1.01 T

Grade:Upto college level

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
3 years ago
Dear student

The Time period = 2pi*(sqrt (l/g))

Now we can calculate the needful

Regards
Vikas TU
14149 Points
3 years ago
When acceleration due to gravity (g) is constant, the time period (T) of oscillation of a simple pendulum is directly proportional to the square root of its effective length (L).

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