A stone drop from height `h` reaches to earth surface in 1 sec. if the same stone taken to moon and drop freely then it will reaches from the surface of the moon in the time (the 'g' of moon is 1/6 times of earth)-

Options

A. `sqrt(6)` second
B. 9 second
C. `sqrt(3)` second
D. 6 second

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step-by-Step Solution:

**Step 1: Understand the problem**
- A stone dropped from a height \( h \) on Earth takes 1 second to reach the surface. We need to find out how long it will take to reach the surface of the Moon when dropped from the same height.

**Step 2: Use the second equation of motion**
- The second equation of motion states:
\[ h = ut + \frac{1}{2} a t^2 \]
where \( h \) is the height, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

**Step 3: Apply the equation for Earth**
- On Earth, the stone is dropped from rest, so the initial velocity \( u = 0 \). The acceleration \( a \) is the acceleration due to gravity \( g \).
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \implies h = \frac{1}{2} g t^2 \]
- Given that \( t = 1 \) second, we can substitute this value:
\[ h = \frac{1}{2} g (1)^2 = \frac{1}{2} g \]

**Step 4: Apply the equation for the Moon**
- For the Moon, the acceleration due to gravity is \( g' = \frac{1}{6} g \). Using the same equation:
\[ h = \frac{1}{2} g' t'^2 \]
- Substituting \( g' \):
\[ h = \frac{1}{2} \left(\frac{1}{6} g\right) t'^2 = \frac{1}{12} g t'^2 \]

**Step 5: Equate the two expressions for height \( h \)**
- Since the height \( h \) is the same in both cases, we can set the two equations equal to each other:
\[ \frac{1}{2} g = \frac{1}{12} g t'^2 \]

**Step 6: Solve for \( t'^2 \)**
- Cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{2} = \frac{1}{12} t'^2 \]
- Multiply both sides by 12:
\[ 6 = t'^2 \]

**Step 7: Find \( t' \)**
- Taking the square root of both sides:
\[ t' = \sqrt{6} \text{ seconds} \]

### Final Answer:
The time taken for the stone to reach the surface of the Moon is \( \sqrt{6} \) seconds.

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