A 10 kg stone is suspended with a rope of breaking strength 30 kg-wt. The minimum time in which the stone can be raised through a height 10 m starting from rest is (Take, `g = 10 N kg^(-1)`).

Options

A. 0.5 s
B. 1.0 s
C. `sqrt((2)/(3))s`
D. 2.0 s

Explanation

To solve the problem, we need to find the minimum time in which a 10 kg stone can be raised through a height of 10 m, given that the rope has a breaking strength of 30 kg-wt and the acceleration due to gravity (g) is 10 N/kg.

### Step-by-Step Solution:

1. **Determine the maximum tension in the rope:**
The breaking strength of the rope is given as 30 kg-wt. We can convert this to Newtons:
\[ \text{Maximum tension (T)} = 30 \text{ kg} \times 10 \text{ N/kg} = 300 \text{ N} \]

2. **Calculate the weight of the stone:**
The weight (W) of the stone can be calculated using the formula:
\[ W = mg = 10 \text{ kg} \times 10 \text{ N/kg} = 100 \text{ N} \]

3. **Set up the equation of motion:**
Using Newton's second law, the net force acting on the stone when it is being lifted is given by:
\[ T - W = ma \]
Substituting the values we have:
\[ 300 \text{ N} - 100 \text{ N} = 10 \text{ kg} \times a \]
Simplifying this gives:
\[ 200 \text{ N} = 10 \text{ kg} \times a \]
Therefore, the acceleration (a) is:
\[ a = \frac{200 \text{ N}}{10 \text{ kg}} = 20 \text{ m/s}^2 \]

4. **Use the equation of motion to find the time:**
We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, the initial velocity (u) is 0 (starting from rest), the distance (s) is 10 m, and the acceleration (a) is 20 m/s². Plugging in these values:
\[ 10 = 0 + \frac{1}{2} \times 20 \times t^2 \]
Simplifying this:
\[ 10 = 10 t^2 \]
Dividing both sides by 10:
\[ 1 = t^2 \]
Taking the square root of both sides gives:
\[ t = 1 \text{ second} \]

### Final Answer:
The minimum time in which the stone can be raised through a height of 10 m is **1 second**.