Question

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

• A. $0.01\text{\%}$
• B. $1\text{\%}$
• C. $0.2\text{\%}$
• D. $0.1\text{\%}$

Answer 1

The correct option is C $0.2\text{\%}$

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