an aluminium rod has a breaking strain 0.2%.The minimum cross section area of he rd in m^2 in order to support a load of 10^4N is if young's modulus is 7*10^9 N/m^2
ans:
1.) 7.1*10^-4
To determine the minimum cross-sectional area of the aluminum rod needed to support a load of 10^4 N, we need to take into account the breaking strain and Young's modulus.
The breaking strain of 0.2% indicates that the rod can withstand a strain of up to 0.2% before it breaks. Strain is defined as the change in length divided by the original length. In mathematical terms, strain (ε) is given by:
ε = ΔL / L₀
Where ΔL is the change in length and L₀ is the original length.
Now, let's assume that the rod is of length L₀ and has a cross-sectional area A. When a load of 10^4 N is applied to the rod, it will experience a stress (σ). Stress is defined as the force applied per unit area. In mathematical terms:
σ = F / A
Where F is the applied force and A is the cross-sectional area.
According to Hooke's law, stress (σ) is directly proportional to strain (ε) through Young's modulus (E). Mathematically, this can be expressed as:
σ = E * ε
Combining the equations for stress and strain, we have:
F / A = E * ΔL / L₀
Rearranging the equation, we can calculate the cross-sectional area (A) as:
A = F * L₀ / (E * ΔL)
Substituting the given values:
F = 10^4 N
L₀ = (unknown)
E = 7 * 10^9 N/m²
ΔL = breaking strain = 0.2% = 0.002
Now, let's solve for the cross-sectional area (A):
A = 10^4 N * L₀ / (7 * 10^9 N/m² * 0.002)
Simplifying further:
A = 1.42857 * 10^(-4) L₀ m²
To find the minimum cross-sectional area, we can set L₀ to the minimum length that will support the load. Since the length of the rod is not provided, we assume that it is fully supported (i.e., there is no additional length beyond the supported length). Therefore, L₀ will be the same as the length of the rod (unknown).
Thus, the minimum cross-sectional area (A) required to support the load of 10^4 N is:
A = 1.42857 * 10^(-4) m²
Rounding to the appropriate significant figures, the answer is approximately:
1.) 7.1 * 10^(-4) m²