Solve

Guides

0

Question

Open in App

Solution

Verified by Toppr

$$T = \dfrac{30 sec}{20} $$ $$\Delta T = \dfrac{1}{20} sec$$

$$L = 55 cm$$ $$\Delta L = 1 \, mm = 0.1 \, cm$$

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

percentage error in g is

$$\dfrac{\Delta g}{g} \times 100 \% = \left(\dfrac{\Delta L}{L} + \dfrac{2 \Delta T}{T} \right) 100\%$$

$$= \left(\dfrac{0.1}{55} + \dfrac{2 \left(\dfrac{1}{20} \right)}{\dfrac{30}{20}} \right) 100 \% = 6.8 \%$$

Was this answer helpful?

38