Start Learning Free

Before You Begin

Question #42519

Difficulty: 8/10 | 🟢 Free

The Question

A solid block of volume V0V_0 and density dd is completely submerged in a liquid and suspended by a vertical string. The liquid has an initial density ρ0\rho_0 (where ρ0<d\rho_0 < d), mass mm, specific heat capacity cc, and coefficient of volume expansion β\beta. The block's thermal expansion is negligible. At t=0t=0, a heater starts supplying heat to the liquid at a constant power PP. The liquid loses heat to the surroundings at a rate λ(T−T0)\lambda(T - T_0), where TT is the instantaneous temperature of the liquid, T0T_0 is the constant ambient temperature, and λ\lambda is a positive constant. Assuming uniform temperature in the liquid and β(T−T0)≪1\beta(T-T_0) \ll 1, which of the following expressions represents the tension in the string as a function of time tt?

Options

  • A.

    V0g(d−ρ0)+V0ρ0gβPλ(1−e−λtmc)V_0 g (d - \rho_0) + \frac{V_0 \rho_0 g \beta P}{\lambda} \left(1 - e^{-\frac{\lambda t}{mc}}\right)

  • B.

    V0g(d−ρ0)−V0ρ0gβPλ(1−e−λtmc)V_0 g (d - \rho_0) - \frac{V_0 \rho_0 g \beta P}{\lambda} \left(1 - e^{-\frac{\lambda t}{mc}}\right)

  • C.

    V0g(d−ρ0)+V0ρ0gβPλ(1−e−Ptmcλ)V_0 g (d - \rho_0) + \frac{V_0 \rho_0 g \beta P}{\lambda} \left(1 - e^{-\frac{P t}{mc\lambda}}\right)

  • D.

    V0g(d−ρ0)+V0ρ0gβPtV_0 g (d - \rho_0) + V_0 \rho_0 g \beta P t

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Calorimetry with Heat Loss|Archimedes' Buoyancy Dynamics|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of thermal expansion with fluid mechanics and calorimetry.

A common trap is assuming the buoyant force remains constant or incorrectly applying the sign of the density change, leading to a decreasing tension (Option B).

Furthermore, students often fail to set up the non-separable differential equation for heat balance with Newton's law of cooling, or they misidentify the steady-state temperature limit, leading to incorrect exponential time constants (Option C) or linear approximations (Option D).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptCalorimetry with Heat Loss
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree