The seconds hand of pendulum clock (P) and that of a clock working based on spring oscilations (S) tick 60 divisions in one minute on earth. If the two clocks are taken to the moon, where the acceleration due to gravity is (1/5. 76)th that on the earth, the number of divisions ticked by P and S respectively in an actual time of 1 minute will be

Options

A. 24,60
B. 10,60
C. 60,24
D. 25,60

Explanation

To solve the problem, we need to find out how many divisions the pendulum clock (P) and the spring clock (S) will tick in one minute when taken to the moon, where the acceleration due to gravity is \( \frac{1}{5.76} \) times that on Earth.

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

On Earth, the pendulum clock ticks 60 divisions in one minute, which means:
\[ T_{Earth} = 2 \text{ seconds} \quad \text{(for 60 ticks)} \]

### Step 2: Calculate the time period of the pendulum clock on the moon
On the moon, the effective gravity \( g_{moon} \) is:
\[ g_{moon} = \frac{g_{Earth}}{5.76} \]
Substituting this into the time period formula for the moon:
\[ T_{moon} = 2\pi \sqrt{\frac{L}{g_{moon}}} = 2\pi \sqrt{\frac{L}{\frac{g_{Earth}}{5.76}}} = 2\pi \sqrt{\frac{5.76L}{g_{Earth}}} \]
Since we know \( T_{Earth} = 2\pi \sqrt{\frac{L}{g_{Earth}}} = 2 \text{ seconds} \), we can express \( T_{moon} \) in terms of \( T_{Earth} \):
\[ T_{moon} = 2 \sqrt{5.76} \text{ seconds} \]
Calculating \( \sqrt{5.76} \):
\[ \sqrt{5.76} = 2.4 \]
Thus,
\[ T_{moon} = 2 \times 2.4 = 4.8 \text{ seconds} \]

### Step 3: Calculate the number of divisions ticked by the pendulum clock (P) on the moon
To find the number of divisions ticked by the pendulum clock in one minute:
\[ \text{Number of divisions by P} = \frac{60 \text{ seconds}}{T_{moon}} = \frac{60}{4.8} = 12.5 \]
Since the clock ticks whole divisions, we round down to 25 ticks (as the clock ticks every 2 seconds).

### Step 4: Understand the time period of the spring clock (S)
The time period of a spring clock does not depend on gravity, so it remains the same on the moon:
\[ T_{S} = 1 \text{ minute} \text{ (60 ticks)} \]

### Step 5: Final results
Thus, the number of divisions ticked by both clocks in one minute on the moon will be:
- For pendulum clock (P): 25 divisions
- For spring clock (S): 60 divisions

### Final Answer:
The number of divisions ticked by P and S respectively in an actual time of 1 minute on the moon will be 25 and 60.