In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is The values of R and r are measured to be and , respectively. [2016]

Question

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 0.01 s. Which of the following statement(s) is (are) true? [2016]

Smart Achievers · Accepted Answer

(1, 2, 4)

$T_{mean} = \frac{0.52 + 0.56 + 0.57 + 0.54 + 0.59}{5} = 0.556 \approx 0.56 s$

$Error in reading = T_{mean} - T_{1}; T_{mean} - T_{2}; T_{mean} - T_{3}; T_{mean} - T_{4}; T_{mean} - T_{5}$

$Mean error = \frac{0.04 + 0 + 0.01 + 0.02 + 0.03}{5} = 0.016 \approx 0.02 s$

$∴$ $% error in the measurement of T$

$\frac{Δ T}{T} \times 100 = \frac{0.02}{0.56} \times 100 = 3.57 %$

$% error in the measurement of g$

$\frac{Δ g}{g} \times 100 = 2 \frac{Δ T}{T} \times 100 + (\frac{Δ R + Δ r}{R - r}) \times 100$

$= 2 (3.57) + (\frac{1 + 1}{60 - 10}) \times 100 \approx 11 %$

$% error in the measurement of r$

$\frac{Δ r}{r} \times 100 = \frac{1}{10} \times 100 = 10 %$

Solution

1 The error in the measurement of $r$ is 10%

2 The error in the measurement of $T$ is 3.75%

3 The error in the measurement of $T$ is 2%

4 The error in the determined value of $g$ is 11%

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Solution

1 The error in the measurement of r is 10%

2 The error in the measurement of T is 3.75%

3 The error in the measurement of T is 2%

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

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1 The error in the measurement of $r$ is 10%

2 The error in the measurement of $T$ is 3.75%

3 The error in the measurement of $T$ is 2%

4 The error in the determined value of $g$ is 11%