Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. The maximum tension that can be exerted by the riverted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa. Assume that each rivet is to carry one quarter of the load. 

Diameter = 6.0 mm

maximum stress $=6.9\times {10}^{7}Pa$

The radius of the river,

$\mathrm{r}=\frac{6}{2}\mathrm{mm}=3\mathrm{mm}=3\times {10}^{-3}\mathrm{m}$

$\mathrm{Maximum} \mathrm{load} \mathrm{on} \mathrm{a} \mathrm{river} = \mathrm{Maximum} \mathrm{stress} \times \mathrm{Area} \mathrm{of} \mathrm{cross}-\mathrm{section}$$\mathrm{Maximum} \mathrm{load} \mathrm{on} \mathrm{a} \mathrm{river}=6.9\times {10}^7\times \mathrm{\pi }{(3\times {10}^{-3})}^2=1950\mathrm{N}$

Maximum tension a river strip can withstand

$\mathrm{Tension}=4\times 1950\mathrm{N}=7800\mathrm{N}$

Hence the correct answer is 7800N.