The period of oscillation of a simple pendulum is given by T = 2π√(l/g) where l is about 100 cm and is known to 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

The correct option is 0.2%.

Step-by-step Explanation:

The relationship between the time period of a simple pendulum $T$, its length $l$, and the acceleration due to gravity $g$ is given by:

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

Squaring both sides of the equation to express $g$ explicitly:

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

Rearranging the equation for $g$:

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

The maximum fractional error in $g$ is given by the sum of the fractional errors in the measured quantities:

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

Multiplying by 100 to get the percentage errors:

$\frac{\Delta g}{g}\times 100=(\frac{\Delta l}{l}\times 100)+2(\frac{\Delta T}{T}\times 100)$

Let's calculate each term individually:
1. Error in length ($l$):
Length $l$ = 100 cm = 1000 mm
Accuracy (absolute error) $\Delta l$ = 1 mm
Percentage error in length:

$\frac{\Delta l}{l}\times 100=\frac{1\text{ mm}}{1000\text{ mm}}\times 100=0.1\%$

2. Error in time period ($T$):
The time for 100 oscillations is measured using a stopwatch. Let $t$ be the total time for $n=100$ oscillations.
Since $T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n}$, the fractional error in $T$ is equal to the fractional error in the total time $t$:

$\frac{\Delta T}{T}=\frac{\Delta t}{t}$

Given details:
Least count of stopwatch (absolute error in total time) $\Delta t$ = 0.1 s
Time period of one oscillation $T$ ≈ 2 s
Total time for 100 oscillations $t=n\times T=100\times 2\text{ s}=200\text{ s}$
Percentage error in the time period:

$\frac{\Delta T}{T}\times 100=\frac{0.1}{200}\times 100=0.05\%$

3. Total percentage error in $g$:
Now we substitute the values back into the percentage error formula:

$\frac{\Delta g}{g}\times 100=0.1\%+2\times 0.05\%$

$\frac{\Delta g}{g}\times 100=0.1\%+0.1\%=0.2\%$

Therefore, the percentage error in the acceleration due to gravity $g$ is 0.2%.