A metallic bar 50 cm long has a uniform... | WAEC SSCE - Physics - 2014

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A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the value of Young's modulus

Answer

Definition. Young's modulus is the ratio of tensile stress to tensile strain within the elastic limit:

\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{F\,L}{A\,e} \]

Convert the data.

Original length \(L = 50\ \text{cm} = 0.50\ \text{m}\)
Cross-sectional area \(A = 4.0\ \text{cm}^{2} = 4.0\times10^{-4}\ \text{m}^{2}\)
Force \(F = 35\ \text{kN} = 35\,000\ \text{N}\)
Extension \(e = 0.25\ \text{mm} = 2.5\times10^{-4}\ \text{m}\)

Substitute.

\[ Y = \frac{35\,000\times0.50}{4.0\times10^{-4}\times2.5\times10^{-4}} = \frac{17\,500}{1.0\times10^{-7}} \] \[ Y = 1.75\times10^{11}\ \text{Pa} \]

Young's modulus of the bar is \(1.75\times10^{11}\ \text{Pa}\) (N m\(^{-2}\)).

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Answer Details

Definition. Young's modulus is the ratio of tensile stress to tensile strain within the elastic limit:

\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{F\,L}{A\,e} \]

Convert the data.

Original length \(L = 50\ \text{cm} = 0.50\ \text{m}\)
Cross-sectional area \(A = 4.0\ \text{cm}^{2} = 4.0\times10^{-4}\ \text{m}^{2}\)
Force \(F = 35\ \text{kN} = 35\,000\ \text{N}\)
Extension \(e = 0.25\ \text{mm} = 2.5\times10^{-4}\ \text{m}\)

Substitute.

\[ Y = \frac{35\,000\times0.50}{4.0\times10^{-4}\times2.5\times10^{-4}} = \frac{17\,500}{1.0\times10^{-7}} \] \[ Y = 1.75\times10^{11}\ \text{Pa} \]

Young's modulus of the bar is \(1.75\times10^{11}\ \text{Pa}\) (N m\(^{-2}\)).

Read lesson note on Heat Energy (WAEC)

Read lesson note on Elasticity (JAMB)

Heat Energy

Elasticity

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A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the va...

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A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the va...

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