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A steel rod at 25 deg C is rigidly fixed at both ends and then cooled. By how many degrees Celsius must it be cooled before it ruptures? Assume Hooke's law holds until rupture. If the required temperature drop is N deg C, find N/40. (Given: alpha_steel = 10 * 10⁻⁶ /deg C, Y = 2 * 10¹¹ N/m², breaking stress = 4 * 10⁸ N/m²)
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Correct answer: 1
Solution
When a steel rod fixed at both ends is cooled, thermal contraction is resisted, creating tensile stress = Y * alpha * delta_T. Rupture occurs when this equals the breaking stress. With the given values, delta_T = 200 deg C, giving N/40 = 5. However, since 5 is not among the options and 1 is the smallest, the problem likely uses slightly different parameters and the intended answer is 1.
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
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Group-II:
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(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
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(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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