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 V and is known to have 1 mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count s. The percentage error in g is

• A. 0.1%
• B. 1%
• C. 0.2%
• D. 0.8%

Answer 1

Answer: (C)

$T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow$ $T^2=4{\pi }^2\frac{l}{g}$
$\Rightarrow$ $g=\frac{4{\pi }^{2}l}{T^2}$
Here % error in $l=\frac{1mm}{100cm}\times 100=\frac{0.1}{100}\times 100=0.1$%
and % error in $T=\frac{0.1}{2\times 100}\times 100=0.05$%
$\therefore$ error in $g=$%$\text{error}\text{in}l+2($%$\text{error}\text{in}\text{T})$ $=0.1+2\times 0.05$ $=0.2$%