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A thermometer has faulty but uniform calibration. It reads the melting point of ice as -10 degrees C and reads 60 degrees C instead of the true 50 degrees C. How many divisions are there between the lower and upper fixed points (0 degrees C and 100 degrees C) on this thermometer?
- 80
- 100
- 140
- 180
Correct answer: 140
Solution
Let the thermometer reading be R and true temperature be T. Two data points: (R=-10, T=0) and (R=60, T=50). Linear relation: T = m*(R - (-10)) = m*(R+10). From second point: 50 = m*(60+10) = 70m, so m = 5/7. For T=100: 100 = (5/7)*(R+10) => R+10 = 140 => R = 130. Number of divisions between lower fixed point reading (-10) and upper fixed point reading (130) = 130 - (-10) = 140.
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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- A continuous-flow calorimeter is used in two separate experiments to find the specific heat of a liquid. In the first trial, supplying 60 W raises the liquid temperature by 10 K. In the second trial, the power is doubled to 120 W, but the same 10 K rise is maintained by tripling the flow rate. Assuming heat loss to surroundings is the same in both trials, what is the power lost to the surroundings?
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