The period of oscillation of a simple pendulum is T =2π√(L/g) . Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?

The correct option is 3%.

Step-by-Step Explanation:

We are given the formula for the period of oscillation of a simple pendulum:
$T=2\pi \sqrt{\frac{L}{g}}$

Squaring both sides of the equation gives:
$T^2=4{\pi }^2\frac{L}{g}$

Rearranging the equation to solve for the acceleration due to gravity, $g$:
$g=4{\pi }^2\frac{L}{T^2}$

The relative error in the determination of $g$ is given by:
$\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$

Let $t$ be the time for $n$ oscillations. Then the time period is $T=\frac{t}{n}$ and the error in the time period is $\Delta T=\frac{\Delta t}{n}$.
Thus, the relative error in the time period is:
$\frac{\Delta T}{T}=\frac{\Delta t/n}{t/n}=\frac{\Delta t}{t}$

Substituting this back into the relative error equation for $g$:
$\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta t}{t}$

Now, let's identify the given values and their respective absolute errors from the problem statement:
- Measured length, $L=20.0\text{cm}$
- Accuracy (absolute error) in length, $\Delta L=1\text{mm}=0.1\text{cm}$
- Measured time for 100 oscillations, $t=90\text{s}$
- Resolution of the watch (absolute error in time), $\Delta t=1\text{s}$

Substitute these values into the relative error equation:
$\frac{\Delta g}{g}=\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)$

Calculate the terms:
$\frac{\Delta g}{g}=0.005+\frac{2}{90}\approx 0.005+0.0222=0.0272$

To find the percentage accuracy (percentage error) in $g$, we multiply the relative error by 100:
$\frac{\Delta g}{g}\times 100\%=0.0272\times 100\%\approx 2.72\%\approx 3\%$

Therefore, the accuracy in the determination of $g$ is approximately 3%.