Time period (T) of pedulum is directly proprortional to the square root of length of string by which bob is attached to a fixed point and iknversely proportional to the square root of gravitational constant g. Time period of a bob is 3 seconds when the gravitational constannt g is `4m//"sec"^(2)` and length of string is 9 metre, what is the time period of a bob having astring of length 64 metre and gravitational constant `16m//"sec"^(2)`?

Options

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

Answer

To solve the problem, we will use the relationship given in the question. The time period \( T \) of a pendulum is directly proportional to the square root of the length of the string \( L \) and inversely proportional to the square root of the gravitational constant \( g \). This can be expressed mathematically as:

\[ T \propto \frac{\sqrt{L}}{\sqrt{g}} \]

We can introduce a constant of proportionality \( k \):

\[ T = k \cdot \frac{\sqrt{L}}{\sqrt{g}} \]

### Step 1: Find the constant \( k \)

We know from the problem that when \( T = 3 \) seconds, \( L = 9 \) meters, and \( g = 4 \, \text{m/s}^2 \).

Substituting these values into the equation:

\[ 3 = k \cdot \frac{\sqrt{9}}{\sqrt{4}} \]

Calculating the square roots:

\[ \sqrt{9} = 3 \quad \text{and} \quad \sqrt{4} = 2 \]

So we have:

\[ 3 = k \cdot \frac{3}{2} \]

Now, solving for \( k \):

\[ k = 3 \cdot \frac{2}{3} = 2 \]

### Step 2: Use \( k \) to find the new time period

Now we need to find the time period \( T \) when the length \( L = 64 \) meters and \( g = 16 \, \text{m/s}^2 \).

Substituting these values into the equation:

\[ T = 2 \cdot \frac{\sqrt{64}}{\sqrt{16}} \]

Calculating the square roots:

\[ \sqrt{64} = 8 \quad \text{and} \quad \sqrt{16} = 4 \]

So we have:

\[ T = 2 \cdot \frac{8}{4} \]

Simplifying:

\[ T = 2 \cdot 2 = 4 \]

### Final Answer

The time period of the bob with a string length of 64 meters and gravitational constant of \( 16 \, \text{m/s}^2 \) is \( 4 \) seconds.

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