Question

The time period of oscillation of a simple pendulum is given by T = 2p $\sqrt { \frac { \ell } { \mathrm { g } } }$ . The length of the pendulum is measured as l = 10 ± 0.1 cm and the time period as T = 0.5 ± 0.02 s. The percentage error in the value of g is –

• A. (a) 5%
• A. (b) 9%
• A. (c) 7%
• A. (d) None

Answer 1

(b)

Sol. $\frac { \Delta \mathrm { g } } { \mathrm { g } }$ × 100 = $\frac { \Delta \ell } { \ell }$ × 100 + 2 $\left( \frac { \Delta \mathrm { T } } { \mathrm { T } } \times 100 \right)$

Dl = 0.01 cm; DT = 0.023s

l = 10 cm; T = 0.5