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JEE Main
Exam
JEE Main
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Physics
Chapter
Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

In an experiment to determine the acceleration due to gravity ( g )...

In an experiment to determine the acceleration due to gravity (gg) using a simple pendulum, the length of the pendulum is measured as 25.025.0 cm using a scale of least count 0.50.5 cm. The time for 5050 oscillations is measured as 100100 s using a stopwatch of least count 11 s. The maximum percentage error in the determination of gg is:

Options

  1. A44
  2. B33
  3. C55
  4. D102102

Correct answer

A. 44

Step-by-step solution

The acceleration due to gravity gg using a simple pendulum is given by:
g=4π2LT2g = \dfrac{4\pi^2 L}{T^2}

If tt is the total time for nn oscillations, then T=tnT = \dfrac{t}{n}. Substituting this, we get:
g=4π2n2Lt2g = \dfrac{4\pi^2 n^2 L}{t^2}

The maximum percentage error in gg is:
Δgg×100=ΔLL×100+2(Δtt×100)\dfrac{\Delta g}{g} \times 100 = \dfrac{\Delta L}{L} \times 100 + 2 \left(\dfrac{\Delta t}{t} \times 100\right)

Given values:
L=25.0L = 25.0 cm, ΔL=0.5\Delta L = 0.5 cm
t=100t = 100 s, Δt=1\Delta t = 1 s

Calculating individual percentage errors:
ΔLL×100=0.525.0×100=2%\dfrac{\Delta L}{L} \times 100 = \dfrac{0.5}{25.0} \times 100 = 2\%

Δtt×100=1100×100=1%\dfrac{\Delta t}{t} \times 100 = \dfrac{1}{100} \times 100 = 1\%

Therefore, the maximum percentage error in gg is:
Δgg×100=2%+2(1%)=4%\dfrac{\Delta g}{g} \times 100 = 2\% + 2(1\%) = 4\%

Answer: 44
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