THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period of the pendulum with uncertainty in it

  2.4583 ± 0.0207 seconds

Explanation:

The time period of a pendulum is approximately given by the formula ...

  T = 2π√(L/g)

The maximum period will be achieved when length is longest and gravity is smallest:

  Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds

The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:

  Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds

If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.

  T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2

  T ≈ 2.4583 ± 0.0207 . . . seconds

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We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.

i think the answer would be b