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Physics Question on Stress and Strain

There is some change WW length when a 33000 N 33000 \, N tensile force is applied on a steel rod of area of cross-section 10−3 m210^{-3} \, m^2. The change of temperature required to produce the same elongation, if the steel rod is heated, is (The modulus of elasticity is 3×1011N/m23 \times 10^{11} N/m^2 and the coefficient of linear expansion of steel is 1.1×10−5/∘C) 1.1 \times 10^{-5} /^{\circ} C).

A

20∘C20^{\circ} C

B

15∘C15^{\circ} C

C

10∘C10^{\circ} C

D

0∘C0^{\circ} C

Answer

10∘C10^{\circ} C

Explanation

Solution

Modulus of elasticity
= Force  Area ×lΔl=\frac{\text { Force }}{\text { Area }} \times \frac{l}{\Delta l}
3×1011=3300010−3×lΔl3 \times 10^{11}=\frac{33000}{10^{-3}} \times \frac{l}{\Delta l}
Δll=3300010−3×13×1011\frac{\Delta l}{l}=\frac{33000}{10^{-3}} \times \frac{1}{3 \times 10^{11}}
=11×10−5=11 \times 10^{-5}
Change in length, Δll=αΔT\frac{\Delta l}{l}=\alpha \Delta T
11×10−5=1.1×10−5×ΔT11 \times 10^{-5}=1.1 \times 10^{-5} \times \Delta T
⇒ΔT=10 K\Rightarrow \Delta T=10\, K or 10∘C10^{\circ} C