A mass falls from a height `h` and its time of fall `t` is recorded in terms of time period `T` of a simple pendulum. On the surface of earth it is found that `t=2T`.The entire sete up is taken on the surface of another planet whose mass is half of that of earth and raidus the same experiment is repeated and corresponding times noted as `t'` and `T'`
Then we can say

Options

A. `t'=2T'`
B. `t'=sqrt12T'`
C. `t'gt2T'`
D. `t'lt2T'`

Answer

To solve the problem, we need to analyze the situation step by step, focusing on the relationships between the time of fall, the time period of the pendulum, and the gravitational acceleration on both Earth and the other planet.

### Step 1: Understand the time of fall on Earth
The time of fall \( t \) of a mass dropped from height \( h \) is related to the acceleration due to gravity \( g \) on Earth. The equation of motion for free fall is:
\[ h = \frac{1}{2} g t^2 \]
Given that \( t = 2T \) (where \( T \) is the time period of a simple pendulum), we can express \( T \) in terms of \( g \):
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
where \( l \) is the length of the pendulum.

### Step 2: Relate \( h \) and \( T \) on Earth
Substituting \( t = 2T \) into the equation for height:
\[ h = \frac{1}{2} g (2T)^2 = 2gT^2 \]

### Step 3: Analyze the situation on the other planet
On the other planet, the mass is half that of Earth, and the radius remains the same. The gravitational acceleration \( g' \) on this planet can be calculated as:
\[ g' = \frac{G \cdot \frac{M_E}{2}}{R_E^2} = \frac{1}{2} g \]

### Step 4: Calculate the time of fall on the other planet
Using the same free fall equation:
\[ h = \frac{1}{2} g' (t')^2 \]
Substituting \( g' = \frac{1}{2} g \):
\[ h = \frac{1}{2} \left(\frac{1}{2} g\right) (t')^2 = \frac{1}{4} g (t')^2 \]

### Step 5: Set the two expressions for height \( h \) equal
From the first planet (Earth):
\[ h = 2gT^2 \]
From the second planet:
\[ h = \frac{1}{4} g (t')^2 \]
Setting these equal gives:
\[ 2gT^2 = \frac{1}{4} g (t')^2 \]

### Step 6: Solve for \( t' \)
Cancel \( g \) from both sides:
\[ 2T^2 = \frac{1}{4} (t')^2 \]
Multiplying both sides by 4:
\[ 8T^2 = (t')^2 \]
Taking the square root:
\[ t' = 2\sqrt{2}T \]

### Step 7: Find the relationship between \( T' \) and \( T \)
The time period \( T' \) on the other planet can be calculated as:
\[ T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{\frac{1}{2}g}} = 2\pi \sqrt{\frac{2l}{g}} = \sqrt{2}T \]

### Step 8: Final relationship
Now, we can relate \( t' \) to \( T' \):
\[ t' = 2\sqrt{2}T = 2\sqrt{2} \cdot \frac{T'}{\sqrt{2}} = 2T' \]

### Conclusion
Thus, the final conclusion is:
\[ t' = 2T' \]