Body 1 of mass M is dropped from a height of 1 m and body 2 of mass 3 M is dropped from a height of 9 m. Ratio of time taken by bodies 1 and 2 to reach the ground is

Options

A. `1 : 1`
B. `1 : 3`
C. `3 : 1`
D. `9 : 1 `

Answer

To solve the problem of finding the ratio of time taken by two bodies to reach the ground when dropped from different heights, we can follow these steps:

### Step 1: Identify the Given Information
- Body 1 (mass = M) is dropped from a height \( h_a = 1 \, \text{m} \).
- Body 2 (mass = 3M) is dropped from a height \( h_b = 9 \, \text{m} \).
- Both bodies are dropped (initial velocity \( u = 0 \)).

### Step 2: Use the Equation of Motion
The equation of motion we will use is:
\[ s = ut + \frac{1}{2} a t^2 \]
Since both bodies are dropped, the initial velocity \( u = 0 \). Therefore, the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]
where \( s \) is the distance fallen, \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( t \) is the time taken.

### Step 3: Calculate Time for Each Body
For Body 1:
\[ h_a = \frac{1}{2} g t_a^2 \]
Substituting \( h_a = 1 \, \text{m} \):
\[ 1 = \frac{1}{2} g t_a^2 \]
Rearranging gives:
\[ t_a^2 = \frac{2 \cdot 1}{g} \]
\[ t_a = \sqrt{\frac{2}{g}} \]

For Body 2:
\[ h_b = \frac{1}{2} g t_b^2 \]
Substituting \( h_b = 9 \, \text{m} \):
\[ 9 = \frac{1}{2} g t_b^2 \]
Rearranging gives:
\[ t_b^2 = \frac{2 \cdot 9}{g} \]
\[ t_b = \sqrt{\frac{18}{g}} \]

### Step 4: Find the Ratio of Time Taken
Now, we find the ratio \( \frac{t_a}{t_b} \):
\[ \frac{t_a}{t_b} = \frac{\sqrt{\frac{2}{g}}}{\sqrt{\frac{18}{g}}} \]
This simplifies to:
\[ \frac{t_a}{t_b} = \frac{\sqrt{2}}{\sqrt{18}} = \frac{\sqrt{2}}{3\sqrt{2}} = \frac{1}{3} \]

### Step 5: Final Ratio
Thus, the ratio of the time taken by Body 1 to Body 2 is:
\[ t_a : t_b = 1 : 3 \]

### Conclusion
The final answer is that the ratio of time taken by bodies 1 and 2 to reach the ground is \( 1 : 3 \).