A person in lift which ascents up with acceleration `10 ms^(-2)` drops a stone from a height 10 m. The time of decent is `[g=10ms^(-2)]`

Options

A. 1 s
B. 2 s
C. 1.5 s
D. 3 s

Answer

To solve the problem, we need to find the time of descent of a stone dropped from a height of 10 meters in a lift that is ascending with an acceleration of \(10 \, \text{m/s}^2\).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height from which the stone is dropped, \(h = 10 \, \text{m}\)
- Acceleration of the lift, \(a = 10 \, \text{m/s}^2\)
- Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)

2. **Determine the Effective Acceleration:**
- Since the lift is moving upwards with an acceleration of \(10 \, \text{m/s}^2\), the effective acceleration acting on the stone relative to the lift can be calculated as:
\[ a_{\text{relative}} = g + a = 10 \, \text{m/s}^2 + 10 \, \text{m/s}^2 = 20 \, \text{m/s}^2 \]

3. **Use the Equation of Motion:**
- We can use the second equation of motion to find the time of descent. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \(s\) is the displacement (which is \(10 \, \text{m}\)), \(u\) is the initial velocity (which is \(0 \, \text{m/s}\) since the stone is dropped), and \(a\) is the effective acceleration (\(20 \, \text{m/s}^2\)).
- Plugging in the values, we get:
\[ 10 = 0 \cdot t + \frac{1}{2} \cdot 20 \cdot t^2 \]
- Simplifying, we have:
\[ 10 = 10t^2 \]
- Dividing both sides by \(10\):
\[ 1 = t^2 \]

4. **Solve for Time \(t\):**
- Taking the square root of both sides gives:
\[ t = 1 \, \text{s} \]

### Final Answer:
The time of descent of the stone is \(1 \, \text{s}\).