STATEMENT-1: When a machine gun fires n bullets per second each with kinetic energy k, power of the gun is P = nk
STATEMENT-2: Power P = work done/time.
STATEMENT-3: For uniform circular motion the instantaneous power delivered by centripetal force is always zero.

Options

A. T T T
B. F F F
C. T T F
D. F T T

Answer

To analyze the statements provided in the question, let's break them down step by step.

### Step 1: Understanding the Power of the Machine Gun
When a machine gun fires \( n \) bullets per second, and each bullet has a kinetic energy \( k \), the total kinetic energy imparted by the gun in one second is given by:
\[ \text{Total Kinetic Energy} = n \times k \]
The power \( P \) of the gun can be defined as the rate at which work is done or energy is transferred. Therefore, we can express the power as:
\[ P = \text{Total Kinetic Energy per second} = n \times k \]
This confirms **Statement 1**: \( P = nk \).

### Step 2: Power Definition
Power is defined as the work done per unit time. Mathematically, this is expressed as:
\[ P = \frac{\text{Work Done}}{\text{Time}} \]
This is a fundamental definition of power and confirms **Statement 2**.

### Step 3: Instantaneous Power in Uniform Circular Motion
In uniform circular motion, the centripetal force acts towards the center of the circular path, and it does not do any work on the object moving in a circular path. This is because the direction of the force (centripetal force) is perpendicular to the direction of the displacement of the object at any instant. Since work done is defined as:
\[ \text{Work Done} = \text{Force} \times \text{Displacement} \times \cos(\theta) \]
where \( \theta \) is the angle between the force and displacement, in the case of uniform circular motion, \( \theta = 90^\circ \) and hence:
\[ \cos(90^\circ) = 0 \implies \text{Work Done} = 0 \]
Thus, the instantaneous power delivered by the centripetal force is always zero, confirming **Statement 3**.

### Conclusion
All three statements are correct based on the analysis provided above.