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JEE Main 202118 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Morning Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2021 · 18 March · Morning Shift · Physics · Mathematics in Physics

The time period of a simple pendulum is given by T = 2π (l)/(g) ....

The time period of a simple pendulum is given by T=2πlg.T = 2\pi\sqrt{\dfrac{l}{g}} . The measured value of the length of the pendulum is 10cm10cm known to a 1mm1mm accuracy. The time for 200200 oscillations of the pendulum is found to be 100100 second using a clock of 1s1s resolution. The percentage accuracy in the determination of gg using this pendulum is x.x . The value of xx to the nearest integer is:-

Options

  1. A22 %
  2. B33 %
  3. C55 %
  4. D44 %

Correct answer

B. 33 %

Step-by-step solution

g=4π2lT2g = \dfrac{4{\pi}^{2}l}{{T}^{2}}
\Deltagg=\Deltall+2\DeltaTT=0.110+2(12000.5)\dfrac{\Deltag}{g} = \dfrac{\Deltal}{l} + 2\dfrac{\DeltaT}{T} = \dfrac{0 . 1}{10} + 2\left(\dfrac{\dfrac{1}{200}}{0 . 5}\right)
\Deltagg=1100+150\dfrac{\Deltag}{g} = \dfrac{1}{100} + \dfrac{1}{50}
\Deltagg×100=3\dfrac{\Deltag}{g} \times 100 = 3 %
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