AP EAMCET · PHYSICS · Mechanical Properties of Solids
One end of the steel rod is clamped to the roof and the other end is attached to mass of \(1000 \mathrm{~kg}\) as shown in the figure. The length of the rod is \(50 \mathrm{~cm}\) and its cross-sectional area is \(1000 \mathrm{~mm}^2\). The change in the length of the rod due to the weight of the mass is (Young's modulus of steel \(=2 \times 10^{11} \mathrm{Nm}^{-2}\) and acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(0.025 \mathrm{~mm}\)
- B \(0.10 \mathrm{~mm}\)
- C \(0.050 \mathrm{~mm}\)
- D \(0.075 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(A) \(0.025 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
As, young's modulus of a rod under tension, \[ Y=\frac{F / A}{\Delta l / l}=\frac{F l}{A \cdot \Delta l} \] We have, change in length \(\Delta l=\frac{F . l}{A . Y}\) Here, force on rod, \(F=1000 \times 10=10,000 \mathrm{~N}\) Length of rod,…
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