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Q.

The period of oscillation of a simple pendulum is $T = 2 π \sqrt{\frac{L}{g}}$ . Measured value of ‘L’ is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stop watch of 0.01s resolution. The percentage error in the determination of ‘g’ will be:

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a

1.33%

b

1.30%

c

1.13%

d

1.03%

answer is C.

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Detailed Solution

$g = \frac{4 π^{2} L}{T^{2}} \Rightarrow \frac{Δ g}{g} \times 100 = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) \times 100 = (\frac{10^{- 3}}{1} + \frac{2 (0.01)}{1.95}) \times 100 = 1.13 %$

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The period of oscillation of a simple pendulum is T=2πLg. Measured value of ‘L’ is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stop watch of 0.01s resolution. The percentage error in the determination of ‘g’ will be: