The period of oscillation of a simple pendulum is T = 2π √(L/g). 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 :

Solution

Physics sir · 4 steps

Write g in terms of L and T, then add the fractional errors, with T's error counted twice.

1. Make g the subject. Squaring T = 2π√(L/g) gives g = 4π² L/T². The constant 4π² carries no error, so only L and T contribute.

   (Δ g)/g = (Δ L)/L + 2 (Δ T)/T

2. Read the errors. The least count gives the error. So Δ L = 1 mm = 0.001 m and Δ T = 0.01 s. The factor 2 comes from the power of T in g = 4π² L/T².
3. Put in numbers. (Δ L)/L = 0.001/1.0 = 0.001 and 2 (Δ T)/T = 2 × 0.01/1.95 = 0.01026. Adding them gives (Δ g)/g = 0.01126.
4. So. The percentage error in g is 0.01126 × 100 ≈ 1.13%, which is option (c).

The idea to keep. For a product of powers, multiply each quantity's fractional error by the magnitude of its power and add. Errors always add, even when the quantity sits in the denominator.

Answer from the NTA final answer key. Solution by PrepWiser.