Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times `t_A` and `t_B`, then the ratio `t_A/t_B` is

Options

A. `1/2`
B. 2
C. `2/5`
D. `5/6`

Explanation

To solve the problem, we need to find the ratio of the times \( t_A \) and \( t_B \) for two bodies A and B, which have masses of 20 kg and 5 kg respectively, and are acted upon by a force of 4 kg wt.

### Step-by-step Solution:

1. **Convert the Force to Newtons:**
The force given is 4 kg wt. To convert this to Newtons, we use the conversion factor \( 1 \text{ kg wt} = 9.8 \text{ N} \).
\[ F = 4 \text{ kg wt} \times 9.8 \text{ N/kg wt} = 39.2 \text{ N} \approx 40 \text{ N} \]

**Hint:** Remember to convert kg wt to Newtons using the acceleration due to gravity.

2. **Calculate the Acceleration of Each Body:**
Using Newton's second law, \( F = ma \), we can find the acceleration of each body.
- For body A:
\[ a_A = \frac{F}{m_A} = \frac{40 \text{ N}}{20 \text{ kg}} = 2 \text{ m/s}^2 \]
- For body B:
\[ a_B = \frac{F}{m_B} = \frac{40 \text{ N}}{5 \text{ kg}} = 8 \text{ m/s}^2 \]

**Hint:** Use the formula \( a = \frac{F}{m} \) to find the acceleration of each body.

3. **Relate Kinetic Energy and Velocity:**
The kinetic energy \( KE \) for both bodies is given by:
\[ KE_A = \frac{1}{2} m_A v_A^2 \quad \text{and} \quad KE_B = \frac{1}{2} m_B v_B^2 \]
Since \( KE_A = KE_B \), we can set the equations equal to each other:
\[ \frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2 \]
This simplifies to:
\[ m_A v_A^2 = m_B v_B^2 \]

**Hint:** Remember that kinetic energy is proportional to the mass and the square of the velocity.

4. **Express Velocity in Terms of Acceleration and Time:**
The velocity of each body can be expressed as:
\[ v_A = a_A t_A \quad \text{and} \quad v_B = a_B t_B \]
Substituting these into the kinetic energy equation gives:
\[ m_A (a_A t_A)^2 = m_B (a_B t_B)^2 \]

**Hint:** Use the kinematic equation \( v = at \) to express velocity in terms of acceleration and time.

5. **Rearranging the Equation:**
Substituting the values of \( m_A \), \( m_B \), \( a_A \), and \( a_B \):
\[ 20 (2 t_A)^2 = 5 (8 t_B)^2 \]
Simplifying this gives:
\[ 20 \cdot 4 t_A^2 = 5 \cdot 64 t_B^2 \]
\[ 80 t_A^2 = 320 t_B^2 \]
Dividing both sides by 80:
\[ t_A^2 = 4 t_B^2 \]

**Hint:** Simplify the equation step by step to isolate \( t_A \) and \( t_B \).

6. **Finding the Ratio of Times:**
Taking the square root of both sides:
\[ \frac{t_A}{t_B} = \sqrt{4} = 2 \]

**Hint:** Remember that taking the square root of both sides will give you the ratio of the times.

### Final Answer:
The ratio \( \frac{t_A}{t_B} \) is \( 2 \).