The time period T of oscillation of a simple pendulum of length l is given by `T=2pi. sqrt((l)/(g))`.
Find the percentage error in T corresponding to
(i) on increase of `2%` in the value of l.
(ii) decrease of `2%` in the value of l.`

Answer

To solve the problem, we need to find the percentage error in the time period \( T \) of a simple pendulum when the length \( l \) is increased or decreased by \( 2\% \).

### Given:
The formula for the time period \( T \) of a simple pendulum is:

\[ T = 2\pi \sqrt{\frac{l}{g}} \]

Where:
- \( T \) = Time period
- \( l \) = Length of the pendulum
- \( g \) = Acceleration due to gravity (constant)

### Step 1: Differentiate the formula
To find the percentage error in \( T \) with respect to \( l \), we can differentiate \( T \) with respect to \( l \).

Using the formula:

\[ T = 2\pi \sqrt{\frac{l}{g}} = K \sqrt{l} \quad \text{(where \( K = \frac{2\pi}{\sqrt{g}} \) is a constant)} \]

Differentiating \( T \):

\[ \frac{dT}{dl} = K \cdot \frac{1}{2\sqrt{l}} \]

### Step 2: Relate changes in \( T \) and \( l \)
The change in \( T \) can be expressed as:

\[ \Delta T = \frac{dT}{dl} \cdot \Delta l \]

Substituting the derivative:

\[ \Delta T = K \cdot \frac{1}{2\sqrt{l}} \cdot \Delta l \]

### Step 3: Find the percentage error
The percentage error in \( T \) is given by:

\[ \frac{\Delta T}{T} \times 100 \]

Substituting the expressions for \( \Delta T \) and \( T \):

\[ \frac{\Delta T}{T} = \frac{K \cdot \frac{1}{2\sqrt{l}} \cdot \Delta l}{K \sqrt{l}} = \frac{\Delta l}{2l} \]

Thus, the percentage error in \( T \) is:

\[ \text{Percentage Error} = \frac{\Delta l}{2l} \times 100 \]

### Step 4: Calculate for the first part (increase of \( 2\% \) in \( l \))
If \( l \) is increased by \( 2\% \):

\[ \Delta l = 0.02l \]

Substituting into the percentage error formula:

\[ \text{Percentage Error} = \frac{0.02l}{2l} \times 100 = \frac{0.02}{2} \times 100 = 1\% \]

### Step 5: Calculate for the second part (decrease of \( 2\% \) in \( l \))
If \( l \) is decreased by \( 2\% \):

\[ \Delta l = -0.02l \]

Substituting into the percentage error formula:

\[ \text{Percentage Error} = \frac{-0.02l}{2l} \times 100 = \frac{-0.02}{2} \times 100 = -1\% \]

### Final Answer:
(i) The percentage error in \( T \) for an increase of \( 2\% \) in \( l \) is **1%**.

(ii) The percentage error in \( T \) for a decrease of \( 2\% \) in \( l \) is **-1%** (but in terms of absolute error, we consider it as **1%**).