KCET · Physics · Mechanical Properties of Solids
A metallic rod breaks when strain produced is \(0.2 \%\). The Young's modulus of the material of the \(\operatorname{rod} 7 \times 10^9 \mathrm{~N} / \mathrm{m}^2\). The area of crosssection to support a load of \(10^4 \mathrm{~N}\) is
- A \(7.1 \times 10^{-6} \mathrm{~m}^2\)
- B \(7.1 \times 10^{-4} \mathrm{~m}^2\)
- C \(7.1 \times 10^{-2} \mathrm{~m}^2\)
- D \(7.1 \times 10^{-8} \mathrm{~m}^2\)
Answer & Solution
Correct Answer
(B) \(7.1 \times 10^{-4} \mathrm{~m}^2\)
Step-by-step Solution
Detailed explanation
Given, Youngs' modulus, \(Y=7 \times 10^9 \mathrm{~N} / \mathrm{m}^2\) Load, \(F=10^4 \mathrm{~N}\)
We know that.
\(Y =\frac{F l}{A \Delta l} \)
\(\Rightarrow A =\frac{F l}{Y \Delta l}=\frac{F}{Y} \times \frac{1}{\left(\frac{\Delta l}{l}\right)} \)
\(=\frac{10^4}{7 \times 10^9} \times \frac{1}{0.2}=7.1 \times 10^{-4} \mathrm{~m}^2\)
We know that.
\(Y =\frac{F l}{A \Delta l} \)
\(\Rightarrow A =\frac{F l}{Y \Delta l}=\frac{F}{Y} \times \frac{1}{\left(\frac{\Delta l}{l}\right)} \)
\(=\frac{10^4}{7 \times 10^9} \times \frac{1}{0.2}=7.1 \times 10^{-4} \mathrm{~m}^2\)
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