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A solid cylindrical rod of length 1.5 meters and diameter 0.6 cm is subjected to
a tensile force of 1200 N, causing it to elongate by 2.8 mm. The density of the
material of the rod is $8.0 \mathrm{~g} / \mathrm{cm}^3$ . Calculate the Young’s modulus of elasticity of
the material of the rod.
Option: 1
$3.1808 \times 10^9 \mathrm{~N} / \mathrm{m}^2$

Option: 2
$78.08 \times 10^9 \mathrm{~N} / \mathrm{m}^2$

Option: 3
$9.12 \times 10^9 \mathrm{~N} / \mathrm{m}^2$

Option: 4
$2.273 \times 10^9 \mathrm{~N} / \mathrm{m}^2$

Answers (1)

best_answer

Given data:
Original length (L) = 1.5 m
Diameter (d) = 0.6 cm
Tensile force (F) = 1200 N
Elongation (?L) = 2.8 mm
Density( $\rho$ ) = $8.0 \mathrm{~g} / \mathrm{cm}^3$

Step 1: Calculate the cross-sectional area (A) of the rod using its diameter:

$A=\pi\left(\frac{d}{2}\right)^2$

Convert diameter to meters and calculate area:

$\begin{gathered} A=\pi\left(\frac{0.6 \times 10^{-2} \mathrm{~m}}{2}\right)^2 \\ A \approx 2.827 \times 10^{-4} \mathrm{~m}^2 \end{gathered}$

Step 2: Calculate the stress (σ) experienced by the rod using the formula:

$\sigma=\frac{F}{A}$

Substitute the given values:

$\begin{gathered} \sigma=\frac{1200 \mathrm{~N}}{2.827 \times 10^{-4} \mathrm{~m}^2} \\ \sigma \approx 4.243 \times 10^6 \mathrm{~N} / \mathrm{m}^2(\mathrm{~Pa}) \end{gathered}$

Step 3: Calculate the strain (ε) experienced by the rod using the formula: $\varepsilon=\frac{\Delta L}{L}$

Convert elongation to meters and calculate strain:

$\begin{gathered} \varepsilon=\frac{2.8 \times 10^{-3} \mathrm{~m}}{1.5 \mathrm{~m}} \\ \varepsilon \approx 0.0018667 \end{gathered}$

Step 4: Calculate Young’s modulus (Y ) using the formula:

$Y=\frac{\sigma}{\varepsilon}$

Substitute the calculated values for stress and strain:

$\begin{aligned} & Y=\frac{4.243 \times 10^6 \mathrm{~N} / \mathrm{m}^2}{0.0018667} \\ & Y \approx 2.273 \times 10^9 \mathrm{~N} / \mathrm{m}^2 \end{aligned}$

Therefore, the Young’s modulus of elasticity of the material of the rod is
approximately $2.273 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ .

Therefore, the correct answer is D.

Posted by

Ajit Kumar Dubey

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