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 . 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 0.01s. Which of the following statement(s) is (are) true?

% error in measurement of ‘r’ = 1 10 × 100 = 10 % T m e a n = 0.52 + 0.56 + 0.57 + 0.54 + 0.59 5 = 0.556 ≈ 0.56 S Δ T = 0.04 + 0 + 0.01 + 0.02 + 0.03 5 = 0.016 ≈ 0.02 S ∴ % Error in the measurement of ‘T’ = 0.02 0.56 × 100 = 3.57 % = 0.02 0.56 × 100 = 3.57 % % error in the value of g = 2 Δ T T × 100 + ( Δ R + Δ r R − r ) × 100 = 2 ( 3.57 ) + ( 1 + 1 60 − 10 ) × 100 ≈ 11 %

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In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is $T = 2 π \sqrt{\frac{7 (R - r)}{5 g}}$ . The values of R and r are measured to be $(60 \pm 1)$ mm and $(10 \pm 1)$ 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 0.01s. Which of the following statement(s) is (are) true?

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The error in the measurement of r is 10%

The error in the measurement of T is 2%

The error in the determined value of g is nearly 11%

The error in the measurement of T is 3.57 %

answer is A, B, D.

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Detailed Solution

% error in measurement of ‘r’ $= \frac{1}{10} \times 100 = 10 %$
$T_{m e a n} = \frac{0.52 + 0.56 + 0.57 + 0.54 + 0.59}{5} = 0.556 \approx 0.56 S$
$Δ_{T} = \frac{0.04 + 0 + 0.01 + 0.02 + 0.03}{5} = 0.016 \approx 0.02 S$
$∴ %$ Error in the measurement of ‘T’
$= \frac{0.02}{0.56} \times 100 = 3.57 %$
$= \frac{0.02}{0.56} \times 100 = 3.57 %$
% error in the value of g
$= 2 \frac{Δ T}{T} \times 100 + (\frac{Δ R + Δ r}{R - r}) \times 100$
$= 2 (3.57) + (\frac{1 + 1}{60 - 10}) \times 100 \approx 11 %$

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