MCQ
The acceleration due to gravity is measured on the surface of earth by using a simple pendulum. If $\alpha$ and $\beta$ are relative errors in the measurement of length and time period respectively, then percentage error in the measurement of acceleration due to gravity is
  • A
    $\left(\alpha+\frac{1}{2} \beta\right) \times 100$
  • B
    $(\alpha-2 \beta)$
  • C
    $(2 \alpha+\beta) \times 100$
  • $(\alpha+2 \beta) \times 100$

Answer

Correct option: D.
$(\alpha+2 \beta) \times 100$
d
(d)

$T=2 \pi \sqrt{\frac{L}{g}}$

$\Rightarrow T^2=4 \pi^2 \frac{L}{g}$

$\frac{\Delta g}{g} \times 100 \%=\frac{\Delta L}{L} \times 100 \%+\frac{2 \Delta T}{T} \times 100 \%$

$\frac{\Delta g}{g} \times 100 \%=(\alpha+2 \beta) \times 100$

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