Question

The period of oscillation of a simple pendulum is given by $T = 2 \pi \sqrt { \frac { l } { g } }$ where l is about 100 cm and is known to have 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is

• A. (a) 0.1%
• A. (b) 1%
• A. (c) 0.2%
• A. (d) 0.8%

Answer 1

(c)

Sol. $\mathrm { T } = 2 \pi \sqrt { 1 / \mathrm { g } }$ $\Rightarrow T ^ { 2 } = 4 \pi ^ { 2 } l / g$ $\Rightarrow g = \frac { 4 \pi ^ { 2 } l } { T ^ { 2 } }$

Here % error in l = $\frac { 1 \mathrm {~mm} } { 100 \mathrm {~cm} } \times 100 = \frac { 0.1 } { 100 } \times 100 = 0.1 \%$ and % error in T = $\frac { 0.1 } { 2 \times 100 } \times 100 = 0.05 \%$

∴ % error in g = % error in l + 2(% error in T)

$= 0.1 + 2 \times 0.05$ = 0.2 %

( $T = 2 \pi \sqrt { l / g }$ $\Rightarrow T ^ { 2 } = 4 \pi ^ { 2 } l / g$ $\Rightarrow g = \frac { 4 \pi ^ { 2 } l } { T ^ { 2 } }$