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MHT CET 2019 (2 May, Shift 1) · Thermometry & Thermal Expansion

Thermometry & Thermal Expansion question from MHT CET 2019 (2 May, Shift 1)

Q62MHT CET 2019 · 2 MAY · SHIFT 1Difficult

Thermometry & Thermal Expansion

A clock pendulum having coefficient of linear expansion. α=9×10−7/∘C−1\alpha = 9 \times 10^{-7}/{}^{\circ}\text{C}^{-1} has a period of 0.5 s at 20∘C20^{\circ}\text{C}. If the clock is used in a climate, where the temperature is 30∘C30^{\circ}\text{C}, how much time does the clock lose in each oscillation? ( g=g = constant)

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Answer: (d)

  1. ΔTT=12αΔθ\dfrac{\Delta T}{T} = \dfrac{1}{2}\alpha\Delta\theta
  2. ΔT=12×9×10−7×10×0.5=2.25×10−6\Delta T = \dfrac{1}{2} \times 9\times10^{-7} \times 10 \times 0.5 = 2.25\times10^{-6} s
  3. So the answer is (d).
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