80 railway wagons all of same mass `5xx10^(3)`kg are pulled by an engine with a force of `4xx10^(5)`N. The tension in the coupling between 30 th and st 31st wagon from the engine is :-

Options

A. `400xx10^(4)N`
B. `32xx10^(4)N`
C. `20xx10^(4)N`
D. `25xx10^(4)N`

Answer

To find the tension in the coupling between the 30th and 31st wagon from the engine, we can follow these steps:

### Step 1: Calculate the total mass of the wagons
Given:
- Number of wagons = 80
- Mass of each wagon = \(5 \times 10^3 \, \text{kg}\)

Total mass \(M\) of all wagons:
\[ M = \text{Number of wagons} \times \text{Mass of each wagon} = 80 \times (5 \times 10^3) = 400 \times 10^3 \, \text{kg} = 4 \times 10^5 \, \text{kg} \]

### Step 2: Calculate the acceleration of the train
Given:
- Force \(F\) applied by the engine = \(4 \times 10^5 \, \text{N}\)

Using Newton's second law:
\[ F = M \cdot a \implies a = \frac{F}{M} \]
Substituting the values:
\[ a = \frac{4 \times 10^5 \, \text{N}}{4 \times 10^5 \, \text{kg}} = 1 \, \text{m/s}^2 \]

### Step 3: Calculate the mass of the wagons being pulled by the tension
To find the tension in the coupling between the 30th and 31st wagons, we need to consider the mass of the wagons from the 30th to the 80th wagon (i.e., 51 wagons).

Mass of the wagons from the 30th to the 80th:
\[ \text{Number of wagons from 30th to 80th} = 80 - 30 = 50 \]
\[ \text{Total mass} = 50 \times (5 \times 10^3) = 250 \times 10^3 \, \text{kg} = 2.5 \times 10^5 \, \text{kg} \]

### Step 4: Calculate the tension in the coupling
Using Newton's second law again for the mass being pulled by the tension:
\[ T = \text{mass} \times \text{acceleration} = (2.5 \times 10^5 \, \text{kg}) \times (1 \, \text{m/s}^2) = 2.5 \times 10^5 \, \text{N} \]

### Final Answer
The tension in the coupling between the 30th and 31st wagon from the engine is:
\[ \boxed{2.5 \times 10^5 \, \text{N}} \]

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