JEE MainPhysicsOscillations
A student performs an experiment to determine the acceleration due to gravity using a simple pendulum. They plot a graph of the square of the time period ( T^2 ) on the y-axis against the length ( L ) on the x-axis, which yields a straight line passing through the origin. If the slope of the graph is 4 s ^2/ m , the value of acceleration due to gravity ( g ) at that location is: (Take ^2 = 10 )
Options
- A10 m/s ^2
- B5 m/s ^2
- C2.5 m/s ^2
- D20 m/s ^2
Correct answer
A. 10 m/s ^2
Step-by-step solution
The time period T of a simple pendulum is given by: T = 2 L g Squaring both sides yields: T^2 = 4 ^2 g L Comparing this with the equation of a straight line passing through the origin, y = mx , where y = T^2 and x = L , the slope m is: m = 4 ^2 g Given that the slope is 4 s ^2/ m : 4 ^2 g = 4 Solving for g : g = ^2 Substituting ^2 = 10 : g = 10 m/s ^2 Answer: 10 m/s ^2