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A cylindrical rod with one end in a steam chamber (100 deg C) and the other end in ice (0 deg C) causes melting of 0.1 g of ice per second. If the original rod is replaced by another rod with half the length and double the radius (same material, same temperature difference), find the rate of melting of ice in grams per second with the new rod.
- 0.4 g/s
- 0.8 g/s
- 1.6 g/s
- 3.2 g/s
Correct answer: 0.8 g/s
Solution
Fourier's law: dQ/dt = k * A * delta_T / L. For the new rod: A_new = pi * (2r)² = 4 * pi * r² = 4*A, and L_new = L/2. So (dQ/dt)_new / (dQ/dt)_old = (A_new / L_new) / (A_old / L_old) = (4A / (L/2)) / (A/L) = (8A/L) / (A/L) = 8. Original melting rate = 0.1 g/s. New melting rate = 8 * 0.1 = 0.8 g/s. The answer is 0.8 g/s.
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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