A body of mass 10 kg is to be raised by a massless string from rest to rest, through a height 9.8 m. The greatest tension which the string can safely bear is 20 kg-wt. The least time of ascent is

Options

A. 1 s
B. 3 s
C. 4 s
D. None of the above

Explanation

To solve the problem step by step, we will follow the principles of Newton's laws of motion and the equations of motion.

### Step 1: Identify the forces acting on the body
The body has a mass \( m = 10 \, \text{kg} \). The weight of the body, which acts downwards, is given by:
\[ W = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

The tension \( T \) in the string acts upwards. The maximum tension that the string can bear is given as \( 20 \, \text{kg-wt} \):
\[ T_{\text{max}} = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

### Step 2: Set up the equation of motion
Using Newton's second law, we can write the equation of motion for the body when it is being raised:
\[ T - W = ma \]
Substituting the values we have:
\[ T - 98 \, \text{N} = 10 \, \text{kg} \cdot a \]

### Step 3: Find the maximum permissible acceleration
Since the maximum tension \( T \) is \( 196 \, \text{N} \):
\[ 196 \, \text{N} - 98 \, \text{N} = 10 \, \text{kg} \cdot a \]
\[ 98 \, \text{N} = 10 \, \text{kg} \cdot a \]
\[ a = \frac{98 \, \text{N}}{10 \, \text{kg}} = 9.8 \, \text{m/s}^2 \]

### Step 4: Use the equation of motion to find the time of ascent
The total height \( h \) to be raised is \( 9.8 \, \text{m} \). The body accelerates for the first half of the ascent and decelerates for the second half. For the first half of the ascent, the distance \( s \) is:
\[ s = \frac{h}{2} = \frac{9.8 \, \text{m}}{2} = 4.9 \, \text{m} \]

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ 4.9 \, \text{m} = 0 + \frac{1}{2} (9.8 \, \text{m/s}^2) t_1^2 \]
\[ 4.9 \, \text{m} = 4.9 \, t_1^2 \]
\[ t_1^2 = 1 \quad \Rightarrow \quad t_1 = 1 \, \text{s} \]

### Step 5: Calculate the total time of ascent
Since the body accelerates for the first half and decelerates for the second half, the total time \( T \) of ascent is:
\[ T = t_1 + t_1 = 1 \, \text{s} + 1 \, \text{s} = 2 \, \text{s} \]

### Conclusion
The least time of ascent is \( 2 \, \text{s} \).