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Question: An aluminum rod (Young’s modulus = \(7 \times 10^9 N/m^2\)) has a breaking strain of 0.2%. The minim...

An aluminum rod (Young’s modulus = 7×109N/m27 \times 10^9 N/m^2) has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in order to support a load of 104N{10^4}N is
A) 1×10−2m21 \times {10^{ - 2}}{m^2}
B) 1⋅4×10−3m21 \cdot 4 \times {10^{ - 3}}{m^2}
C) 3⋅5×10−3m23 \cdot 5 \times {10^{ - 3}}{m^2}
D) 7⋅1×10−4m27 \cdot 1 \times {10^{ - 4}}{m^2}

Explanation

Solution

Strain is defined as the ratio of change in length to the original length and stress is defined as the ratio of force to the area. Young’s modulus is defined as the ratio of stress to the strain. Breaking strain is defined as the amount of strain which the body can withstand before it breaks is known as the breaking strain.

Formula used:
The formula of Young’s modulus is given by,
γ=σε\gamma = \dfrac{\sigma }{\varepsilon }
Where Young’s modulus is γ\gamma the stress is σ\sigma the strain is ε\varepsilon .
The formula of the strain is given by,
ε=Δll\varepsilon = \dfrac{{\Delta l}}{l}
Where strain is ε\varepsilon change in length is Δl\Delta l and the length of the body is ll.

Complete step by step answer:
It is given in the problem that an aluminum rod has a breaking strain of 0.2% and we need to find the minimum cross-sectional area of the rod in order to support a load of 104N{10^4}N.
The formula of Young’s modulus is given by,
γ=σε\gamma = \dfrac{\sigma }{\varepsilon }
Where Young’s modulus is γ\gamma the stress is σ\sigma the strain isε\varepsilon .
⇒γ=σε\Rightarrow \gamma = \dfrac{\sigma }{\varepsilon }
⇒γ=(FA)εBS\Rightarrow \gamma = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{{\varepsilon _{BS}}}}
⇒γ=FA⋅εBS\Rightarrow \gamma = \dfrac{F}{{A \cdot {\varepsilon _{BS}}}}
⇒A=Fγ⋅εBS\Rightarrow A = \dfrac{F}{{\gamma \cdot {\varepsilon _{BS}}}}
Since the force is equal to 104N{10^4}N the Young’s modulus is 7×109Nm27 \times {10^9}\dfrac{N}{{{m^2}}} and the breaking strain is 0.2% therefore we get,
⇒A=Fγ⋅εBS\Rightarrow A = \dfrac{F}{{\gamma \cdot {\varepsilon _{BS}}}}
⇒A=104(7×109)⋅(21000)\Rightarrow A = \dfrac{{{{10}^4}}}{{\left( {7 \times {{10}^9}} \right) \cdot \left( {\dfrac{2}{{1000}}} \right)}}
⇒A=7.1×10−4m2\Rightarrow A = 7.1 \times {10^{ - 4}}{m^2}

The minimum area of the cross sectional area is equal to A=7⋅1×10−4m2A = 7 \cdot 1 \times {10^{ - 4}}{m^2}. The correct answer for this problem is option D.

Note:
The breaking strain is the maximum value of the strain and if the strain is increased then this body may break easily. For the maximum strain to happen the change in length of the body has to be in comparable size to the length of the body’s length.