Two strips of metal are riveted together at their ends by four rivets, each of diameter 6mm. What is the maximum tension that can be exterted by the riveted strip if the shearing stress on the rivet is not to exceed `6.9 xx 10^(7) Pa`? Assume that each rivet is to carry one quarter of the load .

Options

A. `2xx10^(3)N`
B. `3.9xx10^(3)N`
C. `7.8xx10^(3)N`
D. `15.6xx10^(3)N`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the radius of the rivet
Given the diameter of the rivet is 6 mm, we can calculate the radius (r) as follows:

\[ r = \frac{D}{2} = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the shearing area of one rivet
The shearing area (A) for one rivet can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 \]

Substituting the radius we found:

\[ A = \pi (3 \times 10^{-3})^2 = \pi (9 \times 10^{-6}) \approx 2.827 \times 10^{-5} \, \text{m}^2 \]

### Step 3: Calculate the maximum force that one rivet can withstand
Using the shearing stress formula:

\[ \tau = \frac{F}{A} \]

Where:
- \(\tau\) is the shearing stress (given as \(6.9 \times 10^7 \, \text{Pa}\)),
- \(F\) is the maximum force,
- \(A\) is the area calculated in Step 2.

Rearranging the formula to find the maximum force \(F\):

\[ F = \tau \cdot A \]

Substituting the values:

\[ F = (6.9 \times 10^7) \cdot (2.827 \times 10^{-5}) \approx 1949.94 \, \text{N} \]

### Step 4: Calculate the total maximum tension in the riveted strip
Since there are four rivets and each rivet carries one quarter of the total load, the total maximum tension (T) can be calculated as:

\[ T = 4 \cdot F = 4 \cdot 1949.94 \approx 7799.76 \, \text{N} \]

### Step 5: Convert the total maximum tension to scientific notation
To express the total tension in scientific notation:

\[ T \approx 7.8 \times 10^3 \, \text{N} \]

### Final Answer
The maximum tension that can be exerted by the riveted strip is approximately \(7.8 \times 10^3 \, \text{N}\).

---