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Q.

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T=2π7R−r5g. The values of R and r are measured to be 60±1  mm and 10±1  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 0.01 s. Which of the following statements (s) is (are) true

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a

The error in the measurement of r is 10%

b

The error in the measurement of T is 3.57%

c

The error in the measurement of T is 2%

d

The error in the determined value of g is 11%

answer is A.

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Detailed Solution

A  Δrr×100=110×100=10%​B  T¯=0.52+0.56+0.57+0.54+0.595=0.556≈0.56s△T¯=0.04+0+0.01+0.02+0.035=0.02s​So, ΔT¯T¯×100=0.020.56×100=3.57%D  g=R−rT2​Δgg×100=ΔR+ΔrR−r+2ΔTT×100=1+160−10+20.0357×100=11.14%
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