Time period (T) of pendulum is directly proportional to the square root of length of string by which bob is attached to a fixed point and inversely proportional to the square root of gravitational constant 'g'. Time period of a bob is 3 seconds when the gravitational constant g is 4 `m/sec^2` and length of string is 9 metre, what is the time period of a bob having a string length of 64 metre and gravitational constant g is 16 `m/sec^2`?

Options

A. 4 seconds
B. 12 seconds
C. 16 seconds
D. 10 seconds

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship
The time period \( T \) of a pendulum is given by the formula:
\[ T \propto \frac{\sqrt{L}}{\sqrt{g}} \]
This means that \( T \) is directly proportional to the square root of the length \( L \) and inversely proportional to the square root of the gravitational constant \( g \).

### Step 2: Establish the proportionality constant
From the information provided, we know that when \( g = 4 \, \text{m/s}^2 \) and \( L = 9 \, \text{m} \), the time period \( T = 3 \, \text{s} \). We can express this relationship as:
\[ T = k \cdot \frac{\sqrt{L}}{\sqrt{g}} \]
where \( k \) is the proportionality constant.

### Step 3: Substitute known values to find \( k \)
Substituting the known values into the equation:
\[ 3 = k \cdot \frac{\sqrt{9}}{\sqrt{4}} \]
Calculating the square roots:
\[ 3 = k \cdot \frac{3}{2} \]
To isolate \( k \), multiply both sides by \( \frac{2}{3} \):
\[ k = 3 \cdot \frac{2}{3} = 2 \]

### Step 4: Write the complete formula for \( T \)
Now that we have \( k \), we can write the complete formula for the time period:
\[ T = 2 \cdot \frac{\sqrt{L}}{\sqrt{g}} \]

### Step 5: Substitute the new values
Now we need to find the time period when \( L = 64 \, \text{m} \) and \( g = 16 \, \text{m/s}^2 \):
\[ T = 2 \cdot \frac{\sqrt{64}}{\sqrt{16}} \]

### Step 6: Calculate the square roots
Calculating the square roots:
\[ T = 2 \cdot \frac{8}{4} \]

### Step 7: Simplify the expression
Now simplify the expression:
\[ T = 2 \cdot 2 = 4 \, \text{s} \]

### Conclusion
The time period of the bob with a string length of 64 meters and gravitational constant \( g = 16 \, \text{m/s}^2 \) is:
\[ \boxed{4 \, \text{s}} \]

---