Question

Steel ruptures when a shear of $3.5 \times {10^8}{{ N}}{{{m}}^{ - 2}}$ is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly
A) $1.4 \times {10^4}{{ N}}$
B) $2.7 \times {10^4}{{ N}}$
C) $3.3 \times {10^4}{{ N}}$
D) $1.1 \times {10^4}{{ N}}$

Answer 1

Hint: The shear value for the rupturing of steel is given. We know, shear= $\dfrac{{force}}{{area}}$. We have to punch a hole of 1 cm diameter and 0.3 cm thickness. So, we have the area to be taken as the product of the thickness of the sheet and the circumference of the hole. Therefore we can find the required force.

Formula used:
Shear = horizontal stress = $\dfrac{{force}}{{area}}$=$\dfrac{F}{A}$

Complete step by step solution:
Shear is known as the horizontal stress. Mathematically, shear is given by $\dfrac{F}{A}$, where force F is applied perpendicularly over area A.
Given that, when a shear of $3.5 \times {10^8}{{ N}}{{{m}}^{ - 2}}$ is applied, steel ruptures. The sheet is 0.3 cm (0.003 m) thick and the diameter of the hole to be punched is 1 cm (0.01 m).
Here, the area A is to be taken as the product of the thickness of the sheet and the circumference of the hole.
Now, shear $3.5 \times {10^8}{{ N}}{{{m}}^{ - 2}} = \dfrac{F}{{\pi \times 1 \times {{10}^{ - 2}} \times 0.003}}$
$ \Rightarrow F = \pi \times 1 \times {10^{ - 2}} \times 3 \times {10^{ - 3}} \times 3.5 \times {10^8}$
$ \Rightarrow F = 10.5\pi \times {10^3}$
$ \Rightarrow F = 3.297 \times {10^4}$ (taking $\pi = 3.14$)
$ \Rightarrow F \approx 3.3 \times {10^4}$

Hence, the correct answer is option (C), $3.3 \times {10^4}{{ N}}$.

Note: Shear is also known as the horizontal stress. Mathematically, shear is given by $\dfrac{F}{A}$, where force F is applied perpendicularly over area A. Here, the area A is to be taken as the product of the thickness of the sheet and the circumference of the hole.