Question

Steel ruptures when a shear of $3 \cdot 5 \times {10^8}N - {m^{ - 2}}$ is applied. The force needed to punch a 1 cm diameter hole in a steel 0.3 cm thick is nearly:
A) $3 \cdot 3 \times {10^4}N$.
B) $1 \cdot 4 \times {10^4}N$.
C) $1 \cdot 1 \times {10^4}N$.
D) $2 \cdot 7 \times {10^4}N$.

Answer 1

The stress is defined as the ratio of force and area of cross section .The shear stress is defined as ratio of applied force parallel to the area per unit area. Steel ruptures at a certain force means it is the maximum permissible force on the stress.

Formula used: The formula of the shear stress is given by,
$\Rightarrow \tau = \dfrac{F}{A}$
Where $\tau$is shear stress the force on an object is F and the area on which the force is acting is$A$.

Complete step by step solution:
It is given in the problem that steel ruptures when a shear of $3 \cdot 5 \times {10^8}N - {m^{ - 2}}$ is applied and we need to find the force required to punch a 1 cm diameter hole in a steel 0.3 cm thick.
The formula of the shear stress is given by,
$\Rightarrow \tau = \dfrac{F}{A}$
Where $\tau$is shear stress the force on an object is F and the area on which the force is acting.
$\Rightarrow F = \tau \times A$
$\Rightarrow F = \left( {3 \cdot 5 \times {{10}^8}} \right) \times \left( A \right)$………eq. (1)
The area is equal to,
$\Rightarrow A = 2 \cdot \pi \cdot r \cdot l$
$\Rightarrow A = 2 \cdot \pi \cdot \left( {0 \cdot 5 \times {{10}^{ - 2}}} \right) \cdot \left( {0 \cdot 3 \times {{10}^{ - 2}}} \right)$
$\Rightarrow A = 0 \cdot 943 \times {10^{ - 4}}{m^2}$………eq. (2)
Replacing the area of cross section from equation (2) into equation (1) we get,
$\Rightarrow F = \left( {3 \cdot 5 \times {{10}^8}} \right) \times \left( A \right)$
$\Rightarrow F = 3 \cdot 5 \times {10^8} \times 0 \cdot 943 \times {10^{ - 4}}$
$\Rightarrow F = 3 \cdot 3 \times {10^4}N$.
The shear force is equal to$F = 3 \cdot 3 \times {10^4}N$.

The correct option for this problem is option A.

Additional information: Shear force is the force parallel to the area of cross section. While walking on the road the stress that a person applies on the ground is shear stress.

Note: It is advisable to students to remember the formula of the shear stress as it is very helpful in solving these types of problems. In shear stress the force is applied parallel to the area of the cross section. The SI unit of shear stress is$N - {m^{ - 2}}$.