The period of oscillation T of a pendulum of length l at a place of acceleleration due to gravity g is given by `T=2pisqrt(l/g`. If the calculated length is 0.992 times the actual length and if the value assumed for g is 1.002times its actal value, the relative error in the computed value of T, is

Options

A. `0.005`
B. `-0.005`
C. `0.003`
D. `-0.003`

Explanation

To find the relative error in the computed value of the period of oscillation \( T \) of a pendulum, we start with the formula for the period:

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

### Step 1: Identify the given values
- The calculated length \( l' = 0.992l \) (where \( l \) is the actual length).
- The assumed value of \( g' = 1.002g \) (where \( g \) is the actual acceleration due to gravity).

### Step 2: Calculate the errors in length and gravity
1. The error in length \( \Delta l \):
\[ \Delta l = l' - l = 0.992l - l = -0.008l \]

2. The error in gravity \( \Delta g \):
\[ \Delta g = g' - g = 1.002g - g = 0.002g \]

### Step 3: Use the formula for relative error
The relative error in \( T \) can be found using the formula:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( \frac{\Delta l}{l} - \frac{\Delta g}{g} \right) \]

### Step 4: Substitute the values
1. Calculate \( \frac{\Delta l}{l} \):
\[ \frac{\Delta l}{l} = \frac{-0.008l}{l} = -0.008 \]

2. Calculate \( \frac{\Delta g}{g} \):
\[ \frac{\Delta g}{g} = \frac{0.002g}{g} = 0.002 \]

### Step 5: Substitute into the relative error formula
Now substitute these values into the relative error formula:
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( -0.008 - 0.002 \right) \]
\[ \frac{\Delta T}{T} = \frac{1}{2} \left( -0.010 \right) = -0.005 \]

### Conclusion
The relative error in the computed value of \( T \) is:
\[ \frac{\Delta T}{T} = -0.005 \]