A pendulum beats, faster than a standard pendulum. In order to bring it to the standard beat, the length of th pendulum is to be

Options

A. reduced
B. increased
C. reduced and the mass of the bob increased
D. reduced and also the mass of the bob reduced

Explanation

To solve the problem of adjusting the length of a pendulum that beats faster than a standard pendulum, we can follow these steps:

### Step 1: Understand the relationship between the time period and length of a pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( T \) is the time period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Set up the relationship for the standard and faster pendulum
Let:
- \( T_s \) = time period of the standard pendulum (10 seconds),
- \( T_f \) = time period of the faster pendulum (5 seconds),
- \( L_s \) = length of the standard pendulum (10 cm),
- \( L_f \) = length of the faster pendulum (unknown).

From the formula, we know that:
\[ \frac{T_f}{T_s} = \sqrt{\frac{L_f}{L_s}} \]

### Step 3: Substitute known values into the equation
Substituting the known values into the equation:
\[ \frac{5}{10} = \sqrt{\frac{L_f}{10}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{5}{10}\right)^2 = \frac{L_f}{10} \]
\[ \frac{25}{100} = \frac{L_f}{10} \]

### Step 5: Solve for \( L_f \)
Now, multiply both sides by 10:
\[ L_f = 10 \times \frac{25}{100} \]
\[ L_f = 2.5 \text{ cm} \]

### Step 6: Conclusion
To bring the faster pendulum to the standard beat, the length of the pendulum must be reduced to 2.5 cm.