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Q.174 Physics ⭐ 360° Option Elimination Active
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NEET 2020 Physics Question 174 Solution

Topic: Thermal Properties of Matter · Official NTA Booklet Code E1
NEET 2020 Physics Question 174: Heat Capacity Ratio of Solid Spheres of Different Radii

Question

The quantities of heat required to raise the temperature of two solid copper spheres of radii r₁ and r₂ (r₁ = 1.5 r₂) through 1 K are in the ratio:

✓
Verified Against NTA Final Answer Key
Option (1) : 27 / 8
📚 Thermal Properties of Matter Easy
📋 Given Parameters & Problem Setup
Material Copper (same density rho, same specific heat s)
Radius Ratio r1 = 1.5 r2 = (3/2) r2
Temperature Rise Delta T = 1 K
⚡

The 30-Second Exam Hall Shortcut: ⚡ "Heat Q = m s Delta T proportional to mass proportional to volume proportional to r^3. Q1 / Q2 = (r1 / r2)^3 = (1.5)^3 = (3/2)^3 = 27/8!"

⏱️ 20 seconds

The heat required to raise temperature is Q = m · s · ΔT. For solid spheres of the same material, mass is proportional to volume: m ∝ r^3. Since specific heat s and ΔT = 1 K are identical, Q ∝ r^3. Therefore, Q1 / Q2 = (r1 / r2)^3 = (1.5)^3 = (3/2)^3 = 27/8. Mark Option (1) in 5 seconds!

1 Heat capacity: Q=msΔTQ = m s \Delta T.
2 Mass of solid sphere: m=ρV=ρ(43πr3)∝r3m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right) \propto r^3.
3 Ratio: Q1Q2=(r1r2)3\frac{Q_1}{Q_2} = \left(\frac{r_1}{r_2}\right)^3.
4 Substitute r1/r2=1.5=32r_1 / r_2 = 1.5 = \frac{3}{2}: (32)3=278\left(\frac{3}{2}\right)^3 = \frac{27}{8}.
5 Select Option (1).
🛡️
Why You Can 100% Trust This Shortcut: NCERT Physics · Thermal Properties of Matter (Heat Capacity and Specific Heat Capacity)
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formula for Quantity of Heat

The heat required to raise the temperature of a body by ΔT\Delta T is given by: Q=msΔTQ = m s \Delta T, where mm is mass and ss is specific heat capacity.

Step 2

Relate Mass to Sphere Radius

For a solid sphere of uniform density ρ\rho and radius rr: V=43πr3V = \frac{4}{3}\pi r^3 m=ρV=ρ(43πr3)m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right) Because both spheres are made of copper, density ρ\rho and specific heat ss are identical.

Step 3

Set up Proportion for Heat Required

Q=[ρ(43πr3)]sΔTQ = \left[\rho \left(\frac{4}{3}\pi r^3\right)\right] s \Delta T For identical ΔT=1 K\Delta T = 1\text{ K}: Q∝r3Q \propto r^3.

Step 4

Compute Ratio of Heats

Q1Q2=(r1r2)3=(1.5)3=(32)3=278\frac{Q_1}{Q_2} = \left(\frac{r_1}{r_2}\right)^3 = (1.5)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}. Matches Option 1.

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 A
✓ 100% Correct
Verdict & Scientific Reason:

Accurate value; Q ∝ mass ∝ r^3 => (3/2)^3 = 27/8.

💡
Why this is airtight: None; standard volume scaling of heat capacity.
2 B
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; scaled with surface area r^2 instead of volume r^3: (3/2)^2 = 9/4.

⚠️
Student Fallacy (Why candidates pick this trap): Confusing thermal capacity (volume) with radiative cooling (surface area).
3 C
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; scaled linearly with radius r: 1.5 = 3/2.

⚠️
Student Fallacy (Why candidates pick this trap): Linear scaling assumption.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; arbitrary numerical ratio.

⚠️
Student Fallacy (Why candidates pick this trap): Guesswork.
⚠️

NTA Trap Alert: Surface Area vs Volume Scaling

Rate of cooling (Stefan-Boltzmann law) depends on surface area (r^2). Heat required to raise temperature depends on mass/volume (r^3).

🛡️ How to avoid the -1 mark: Heat capacity = mass × specific heat ∝ Volume ∝ r^3.
📖

Core NEET Syllabus Standard

Physics · Thermal Properties of Matter Standard Benchmark
Core Scientific Principle Tested:

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Syllabus Focus: Heat Capacity Ratio of Solid Spheres of Different Radii
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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