The period of oscillation of a simple pendulum is T = 2π. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be:

  1. 1.12%
  2. 1.03%
  3. 1.13%
  4. 1.30%

Answer (Detailed Solution Below)

Option 1 : 1.12%
Free
JEE Main 04 April 2024 Shift 1
90 Qs. 300 Marks 180 Mins

Detailed Solution

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CONCEPT:

To calculate the percentage error we use the formulation as

       ---- (1)

CALCULATION:

Given: T = 2π       ----(2)

L = 1.0 m

T = 1.98 s

Let us simplify the equation (2) we have;

T = 2π

T2 = 4π2

⇒ 

For error we have;

Now, On putting all the given values we have;

⇒ 

%

Hence, option 1) is the correct answer.

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