In a second ODI match between England and India Bhuwnesh kumar bowled his medium fast bowling. He starts his run up from the distance of `50m` from the stumps (bowling end) and at the time of bowling a ball,his speed is `36(km)//(hr)` and the speed of ball is `144(km)//(hr)`.
If his mass is `70 kg` and mass of the ball is `(1)/(4)kg` and the heat produce in his body is one tenth of the total work done by him, then
Energy spent by bowler in throwing one ball is `:-`

Options

A. `3500` `J`
B. `200` `J`
C. `4070` `J`
D. `3700` `J`

Explanation

To solve the problem, we need to calculate the energy spent by Bhuvnesh Kumar in throwing one ball. We will follow these steps:

### Step 1: Convert Speeds from km/hr to m/s
- The speed of Bhuvnesh Kumar (V1) is given as 36 km/hr.
- To convert this to m/s, we use the conversion factor:
\[ V1 = 36 \times \frac{1000}{3600} = 10 \text{ m/s} \]

- The speed of the ball (V2) is given as 144 km/hr.
- Similarly, we convert this to m/s:
\[ V2 = 144 \times \frac{1000}{3600} = 40 \text{ m/s} \]

### Step 2: Calculate the Work Done by the Bowler
The work done (W) can be calculated using the formula:
\[ W = \frac{1}{2} m_1 V1^2 + \frac{1}{2} m_2 V2^2 \]
Where:
- \( m_1 = 70 \text{ kg} \) (mass of the bowler)
- \( m_2 = \frac{1}{4} \text{ kg} \) (mass of the ball)

Substituting the values:
\[ W = \frac{1}{2} \times 70 \times (10)^2 + \frac{1}{2} \times \frac{1}{4} \times (40)^2 \]

Calculating each term:
- For the bowler:
\[ W_1 = \frac{1}{2} \times 70 \times 100 = 3500 \text{ J} \]

- For the ball:
\[ W_2 = \frac{1}{2} \times \frac{1}{4} \times 1600 = \frac{800}{4} = 200 \text{ J} \]

Thus, the total work done is:
\[ W = 3500 + 200 = 3700 \text{ J} \]

### Step 3: Calculate the Total Energy Spent by the Bowler
According to the problem, one-tenth of the total work done is lost as heat. Therefore, the energy spent (E) can be calculated as:
\[ E = W + \frac{1}{10} W \]
This can be rewritten as:
\[ E = W \left(1 + \frac{1}{10}\right) = W \times \frac{11}{10} \]

Substituting the value of W:
\[ E = 3700 \times \frac{11}{10} = 4070 \text{ J} \]

### Final Answer
The energy spent by the bowler in throwing one ball is:
\[ \boxed{4070 \text{ J}} \]