An experiment was done to find acceleration due to gravity by using the formula T=2π√l/g determine error "g" given T=2.22sec, l=121.6cm, ∆T=0.1sec and ∆l=+0.05?

Given:
T = 2.22 sec
l = 121.6 cm
∆T = 0.1 sec
∆l = +0.05 cm

Using the formula T = 2π√l/g, we can rearrange it to solve for g:

g = (4π²l) / T²

Now, let's calculate the value of g using the given values:

g = (4π² * 121.6) / (2.22)²
g = (4 * 9.8696 * 121.6) / 4.9284
g = 484.2112 / 4.9284
g = 98.27 cm/s²

Now, let's calculate the error in g using the formula:

∆g = g * √((∆l / l)² + (2 * ∆T / T)²)

∆g = 98.27 * √((0.05 / 121.6)² + (2 * 0.1 / 2.22)²)
∆g = 98.27 * √((0.0004105)² + (0.0901)²)
∆g = 98.27 * √(0.000000168 + 0.00811801)
∆g = 98.27 * √0.008118178
∆g = 98.27 * 0.0901
∆g = 8.85 cm/s²

Therefore, the acceleration due to gravity is 98.27 cm/s² with an error of ±8.85 cm/s².

Error in "g":

The formula for the period of a simple pendulum is:

```
T = 2π√l/g
```

where:

- T is the period of oscillation (in seconds)
- l is the length of the pendulum (in meters)
- g is the acceleration due to gravity (in m/s²)

To determine the error in "g", we can use the following formula:

```
Δg = |g| - [(2 - ΔT)/T - (Δl)/l]
```

where:

- Δg is the error in "g"
- g is the acceleration due to gravity
- ΔT is the error in the period of oscillation
- T is the period of oscillation
- Δl is the error in the length of the pendulum
- l is the length of the pendulum

Given values:

- T = 2.22 sec
- l = 121.6 cm = 1.216 m
- ΔT = 0.1 sec
- Δl = 0.05 cm = 0.0005 m

Calculation:

```
Δg = |g| - [(2 - 0.1)/2.22 - (0.0005)/1.216]
Δg ≈ 0.066 - g
```

Conclusion:

The error in "g" is approximately 0.066 times the value of "g".