Question

Steel ruptures when a shear of $ 3.5 \times 10^8 \,Nm^{-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

The Correct Option is C

To calculate the force required to punch a hole in a steel sheet, we need to consider the shear stress at which steel ruptures. Given:

• Shear stress \( \sigma = 3.5 \times 10^8 \, \text{Nm}^{-2} \) (the shear stress at which steel ruptures)
• Diameter of the hole \( d = 1 \, \text{cm} = 0.01 \, \text{m} \)
• Thickness of the sheet \( t = 0.3 \, \text{cm} = 0.003 \, \text{m} \)

The force required can be calculated using the formula for shear force:

F = \sigma \times A

where \( A \) is the area to be sheared. In the case of a circular hole, the area to be sheared is the circumference of the hole times the thickness of the sheet:

A = \pi \times d \times t

Substitute the given values to find \( A \):

A = \pi \times 0.01 \, \text{m} \times 0.003 \, \text{m} = 9.42 \times 10^{-5} \, \text{m}^2

Now, using the force formula:

F = 3.5 \times 10^8 \, \text{Nm}^{-2} \times 9.42 \times 10^{-5} \, \text{m}^2

On calculating,

F = 3.297 \times 10^4 \, \text{N}

Therefore, the force required is approximately 3.3 \times 10^4 \, \text{N}.

Thus, the correct answer is:

3.3 \times 10^4 \, N

This matches the given correct option.