The time period of oscillation of a simple pendulum is given by T = 2 π l g . The length of the pendulum is measured as l = 10 ± 0.01 c m and the time period as T = 0.5 ± 0.02 s . The percentage error in the value of g is

Δ g g × 100 = Δ l l × 100 + 2 Δ T T × 100 Δ l = 0.01 c m Δ T = 0.02 s l = 10 c m T = 0.5 s Δ l l × 100 = 0.01 10 × 100 = 0.1 Δ T T × 100 = 0.02 0.5 × 100 = 4 Δg g × 100 = 0.1 + 2 × 4 ≈ 8 %

Physics - Error Question with Solution | TestHub

PhysicsErrorMiscellaneousEasy2 minPYQ_2020

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The time period of oscillation of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ . The length of the pendulum is measured as $l = 10 \pm 0.01 c m$ and the time period as $T = 0.5 \pm 0.02 s$ . The percentage error in the value of $g$ is

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Solution:

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

$Δ l = 0.01 c m$

$Δ T = 0.02 s$

$l = 10 c m$

$T = 0.5 s$

$\frac{Δ l}{l} \times 100 = \frac{0.01}{10} \times 100 = 0.1$

$\frac{Δ T}{T} \times 100 = \frac{0.02}{0.5} \times 100 = 4$

$\frac{Δg}{g} \times 100 = 0.1 + 2 \times 4 \approx 8 %$

Stream:NTA_ABHYASSubject:PhysicsTopic:ErrorSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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