the circum ference of the disk, and the total force exerted on its center only.

Letus assume that the shearing stress along the side surface of the disk is uniform, then

$F > \int\limits_{Surface} {d{F_{\max }} = \int\limits_{surface} {{\sigma _{\max }}dA = {\sigma _{\max }}} \int\limits_{surface} {dA} } $

$ = \int {{\sigma _{\max }}} .A = {\sigma _{\max }}.2\pi \left( {\frac{D}{2}} \right)h$

$ = 3.5 \times {10^8} \times \left( {\frac{1}{2} \times {{10}^{ - 2}}} \right) \times 0.3 \times {10^{ - 2}} \times 2\pi $

$ = 3.297 \times {10^4} = 3.3 \times {10^4}N$