The maximum tension a rope can withstand is 60 kg.wt. The ratio of maximum acceleration with which two boys of masses 20 kg and 30 kg can climb up the rope at the same time is

Options

A. `1:2`
B. `2:1`
C. `2:3`
D. `3:2`

Answer

To solve the problem, we need to determine the ratio of maximum accelerations with which two boys of masses 20 kg and 30 kg can climb up a rope that can withstand a maximum tension of 60 kg.wt.

### Step-by-Step Solution:

1. **Convert Tension to Newtons**:
The maximum tension that the rope can withstand is given as 60 kg.wt. We need to convert this to Newtons using the conversion factor \(1 \text{ kg.wt} = 9.8 \text{ N}\).
\[ T = 60 \text{ kg.wt} \times 9.8 \text{ N/kg} = 588 \text{ N} \]

2. **Identify Masses**:
Let \(m_1 = 20 \text{ kg}\) (mass of the first boy) and \(m_2 = 30 \text{ kg}\) (mass of the second boy).

3. **Apply Newton's Second Law for Each Boy**:
For the first boy (mass \(m_1\)):
\[ T - m_1 g = m_1 a_1 \]
Rearranging gives:
\[ a_1 = \frac{T - m_1 g}{m_1} \]

For the second boy (mass \(m_2\)):
\[ T - m_2 g = m_2 a_2 \]
Rearranging gives:
\[ a_2 = \frac{T - m_2 g}{m_2} \]

4. **Calculate Acceleration for Each Boy**:
- For \(m_1 = 20 \text{ kg}\):
\[ a_1 = \frac{588 \text{ N} - 20 \text{ kg} \times 9.8 \text{ N/kg}}{20 \text{ kg}} = \frac{588 - 196}{20} = \frac{392}{20} = 19.6 \text{ m/s}^2 \]

- For \(m_2 = 30 \text{ kg}\):
\[ a_2 = \frac{588 \text{ N} - 30 \text{ kg} \times 9.8 \text{ N/kg}}{30 \text{ kg}} = \frac{588 - 294}{30} = \frac{294}{30} = 9.8 \text{ m/s}^2 \]

5. **Calculate the Ratio of Accelerations**:
Now, we find the ratio of the maximum accelerations:
\[ \text{Ratio} = \frac{a_1}{a_2} = \frac{19.6}{9.8} = 2 \]

### Final Answer:
The ratio of maximum acceleration with which the two boys can climb up the rope is \(2:1\).