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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Physics
Chapter
Mathematics in Physics

JEE Advanced 2016 · Paper 2 · 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 g, the formula used for the time period of a periodic motion is T=2π7(R−r)5g.T = 2\pi\sqrt{\dfrac{7\left(R - r\right)}{5g}} . The values of R and r are measured to be \left(60 \pm 1\right)mmand\left(10 \pm 1\right)mm,$ respectively. In five successive measurements, the time period is found to be 0.52s, 0.56s, 0.57s, 0.54s and 0.59s. The least count of the watch used for the measurement of time period is 0.01s. Which of the following statements (s) is (are) true?

Options

  1. AThe error in the measurement of r is 10%.
  2. BThe error in the measurement of T is 3.57 %.
  3. CThe error in the measurement of T is 2%.
  4. DThe error in the determined value of g is 11%.

Correct answer

A. The error in the measurement of r is 10%.

Step-by-step solution

T Absolute error
1 0.52 −0.04- 0.04
2 0.56 00
3 0.57 +0.01+ 0.01
4 0.54 −0.02- 0.02
5 0.59 +0.03+ 0.03
Tavg=0.556{T}_{avg} = 0.556 =0.56= 0.56



Avg. absolute error 0.04+0+0.01+0.02+0.035=0.02\dfrac{0 . 04 + 0 + 0 . 01 + 0 . 02 + 0 . 03}{5} = 0 . 02
Percentage error in time period=∆TT×100= \dfrac{∆ T}{T} \times 100% = \dfrac{0 . 02}{0 . 56} \times 100% \approx 3.57%
Percentage error in radius=∆rr×100= \dfrac{∆ r}{r} \times 100% = \dfrac{1}{10} \times 100 = 10%
Also, g=28π(R−r)5T2g = \dfrac{28\pi\left(R - r\right)}{5{T}^{2}}

Percentage error in gg,=∆gg=∆R+∆rR−r+2∆TT\dfrac{∆ g}{g} = \dfrac{∆ R + ∆ r}{R - r} + \dfrac{2 ∆ T}{T}
∆gg×100\dfrac{∆ g}{g} \times 100% = \dfrac{1 + 1}{50} \times 100% + 2\left(3.57\right)%≈11\approx 11%

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