The force required to punch a 25 mm hole in a mild steel plate 10 mm thick, when ultimate shear stress of the plate is 500 N/mm$^2$ will be nearly
- A. 78 kN
- B. 393 kN
- C. 98 kN
- D. 158 kN
Answer: B. 393 kN
Explanation: Force $F = \text{Perimeter} \times t \times \tau = \pi (25) \times 10 \times 500 \approx 393$ kN.