Two strips of metal are riveted together at their ends by four rivets,each of diameter $6.0\; mm$. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed $6.9 \times 10^{7}\; Pa$? Assume that each rivet is to carry one quarter of the load.

(C) Diameter of the rivet,$d = 6.0\; mm = 6.0 \times 10^{-3}\; m$.
Radius of the rivet,$r = d/2 = 3.0 \times 10^{-3}\; m$.
Maximum shearing stress,$\tau_{max} = 6.9 \times 10^{7}\; Pa$.
Area of cross-section of one rivet,$A = \pi r^{2} = \pi \times (3.0 \times 10^{-3})^{2} = 9\pi \times 10^{-6}\; m^{2}$.
Maximum force that one rivet can withstand,$F_{rivet} = \tau_{max} \times A = 6.9 \times 10^{7} \times 9\pi \times 10^{-6} \approx 1950.6\; N$.
Since each rivet carries one-quarter of the total load $F$,the total tension is $F = 4 \times F_{rivet}$.
$F = 4 \times 1950.6 = 7802.4\; N$ (using $\pi \approx 3.14159$).
Rounding to significant figures,the maximum tension is approximately $7.8 \times 10^{3}\; N$.