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Q.

What is the accuracy of g determined by a simple pendulum of length (100 ± 0.1)cm whose period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution?

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a

0.2 %

b

0.5%

c

0.1%

d

2%

answer is A.

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Detailed Solution

$\begin{array}{l} T = \frac{t}{n} and Δ T = \frac{Δ t}{n} \\ ∴ \frac{Δ T}{T} \times 100 = \frac{Δ t}{t} \times 100 \\ = \frac{0.1}{2 \times 100} \times 100 = 0.05 % \\ \frac{Δ l}{l} \times 100 = \frac{0.1 cm}{100 cm} \times 100 \\ = 0.1 % \end{array}$

Now, $T = 2 π \sqrt{\frac{l}{g}}$

or $g = \frac{4 π^{2} l}{T^{2}} \propto \frac{l}{T^{2}}$

So % error in g is

$\frac{Δ g}{g} \times 100 = (\frac{Δ l}{l} + 2 \frac{Δ T}{T}) \times 100$

so

$\frac{Δ g}{g} \times 100 = (0.1 + 2 \times 0.05) = 0.2 %$

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