Question

The time period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$. The measured value of the length of pendulum is $10\text{cm}$ known to $1\text{mm}$ accuracy. The time for $200$ oscillations of the pendulum is found to be $100$ second using a clock of $1\text{s}$ resolution. The percentage accuarcy in the determination of $g$ using this pendulum is $x$. The value of $x$ to the nearest integer is.

• A. $5$%
• B. $4$%
• C. $3$%
• D. $2$%

Answer 1

The correct option is C $3$%

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

Now, $\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$

$\Rightarrow \frac{\Delta g}{g}=\frac{1\times {10}^{-3}}{10\times {10}^{-2}}+\frac{2\times 1}{100}$

$\Rightarrow \frac{\Delta g}{g}=0.01+0.02=0.03$

$\Rightarrow \frac{\Delta g}{g}\times 100=0.03\times 100=3$%