Question

The length of a simple pendulum is about 100 cm known to an accuracy of 1 mm. Its period of oscillation
2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the
accuracy in the determined value of g?
A) 0.2%
C) 0.5%
B) 0.1%
D) 2%​

Answer 1

Answer:

$\boxed{\mathfrak{0.2 \ \%}}$

Given:

Length of pendulum (l) = 100 cm

Accuracy in length of pendulum (∆l) = 1 mm = 0.1 cm

Time period of 1 oscillation = 2 s

Time period for 100 oscillation (T) = 2 × 100 s

Resolution of clock (∆T) = 0.1 s

To Find:

Accuracy in the value of g i.e. Percentage error in g

Explanation:

From question we can write:

$\sf \dfrac{\Delta l}{l} = \dfrac{0.1}{100} \\ \\ \sf \dfrac{\Delta T}{T} = \dfrac{0.1}{2 \times 100}$

Time period of Simple Pendulum:

$\boxed{ \bold{T = 2\pi \sqrt{ \frac{l}{g} }}}$

$\sf \implies g = \dfrac{4 {\pi}^{2}l }{{T}^{2} } \\ \\ \sf \implies \dfrac{\Delta g}{g} = \dfrac{\Delta l}{l} + 2\dfrac{\Delta T}{T} \\ \\ \sf \implies \dfrac{\Delta g}{g} \times 100 = \dfrac{\Delta l}{l} \times 100 + 2\dfrac{\Delta T}{T} \times 100 \\ \\ \sf \implies \dfrac{\Delta g}{g} \times 100 = \dfrac{ 0.1}{ \cancel{100}} \times \cancel{100} + \cancel{2} \times \dfrac{ 0.1}{ \cancel{2} \times \cancel{100}} \times \cancel{100} \\ \\ \sf \implies \dfrac{\Delta g}{g} \times 100 = 0.1 \%+ 0.1 \% \\ \\ \sf \implies \dfrac{\Delta g}{g} \times 100 = 0.2 \%$

$\therefore$

Accuracy in the value of g = 0.2 %