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Two uniform steel rods P and Q are made of the same material. Rod P has length l and radius 2r; rod Q has length 2l and radius r. Both rods are heated by the same temperature increment. What is the ratio of the increase in length of P to that of Q?
- 1: 1
- 1: 2
- 2: 1
- 1: 4
Correct answer: 1: 2
Solution
Linear thermal expansion formula: DeltaL = L * alpha * DeltaT. Since both rods are made of the same steel, they have the same coefficient of linear expansion alpha. Same temperature change DeltaT. DeltaL_P = l * alpha * DeltaT. DeltaL_Q = 2l * alpha * DeltaT. Ratio DeltaL_P: DeltaL_Q = l: 2l = 1: 2. The cross-sectional radius has no effect on linear thermal expansion.
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]
Group-I:
(P) 2000 K
(Q) 3000 K
(R) 5000 K
(S) 10000 K
Group-II:
(1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV.
(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
(4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K.
(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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