Q.28 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 28 Solution

Topic: Thermal Properties of Matter · Official NTA Booklet Code M1

Question

A cup of coffee cools from 908C to 808C in t minutes, when the room temperature is 208C. The time taken by a similar cup of coffee to cool from 808C to 608C at a room temperature same at 208C is:

✓

Verified Against NTA Final Answer Key

Option (2) : 13 / 5 t

📚 Thermal Properties of Matter Medium

📋 Given Parameters & Problem Setup

Initial Temperature 1 90°C

Final Temperature 1 80°C

Time 1 t minutes

Initial Temperature 2 80°C

Final Temperature 2 60°C

Surrounding Room Temperature (T_0) 20°C

⚡

The 30-Second Exam Hall Shortcut: ⚡ Average Temperature Difference Ratio

⏱️ Save ~25 seconds

Rate of cooling $= \frac{\Delta T}{\Delta t} = K (T_{\text{avg}} - T_0)$ . Case 1: $\frac{10}{t} = K(85 - 20) = 65 K$ . Case 2: $\frac{20}{t'} = K(70 - 20) = 50 K$ . Divide them: $\frac{10/t}{20/t'} = \frac{65}{50} \implies \frac{t'}{2t} = \frac{13}{10} \implies t' = \frac{13}{5}t$ ! Match Option (2)!

1 Newton's law average form:

\frac{T_i - T_f}{\Delta t} = K\left(\frac{T_i + T_f}{2} - T_0\right)

2 Case 1 (

90^\circ \to 80^\circ

\frac{10}{t} = K(85 - 20) = 65K

3 Case 2 (

80^\circ \to 60^\circ

\frac{20}{t'} = K(70 - 20) = 50K

4 Divide Case 1 by Case 2:

\frac{10/t}{20/t'} = \frac{65K}{50K} \implies \frac{t'}{2t} = \frac{13}{10}

5 Solve for

t'

t' = 2t \times \frac{13}{10} = \frac{13}{5}t

6 Match Option (2).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Newton's law of cooling for small to moderate temperature differentials.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulating Newton's Law of Cooling (Average Form)

For a body cooling in surroundings at constant ambient temperature $T_0$ , the average rate of fall of temperature over time interval $\Delta t$ is proportional to the excess of average temperature over surroundings:

\frac{T_1 - T_2}{\Delta t} = K \left(\frac{T_1 + T_2}{2} - T_0\right)

Step 2

Setting Up Equation for the First Cooling Process

For cooling from $90^\circ\text{C}$ to $80^\circ\text{C}$ in time $t$ with $T_0 = 20^\circ\text{C}$ :

\frac{90 - 80}{t} = K \left(\frac{90 + 80}{2} - 20\right) \implies \frac{10}{t} = K (85 - 20) = 65 K \quad \text{--- (1)}

Step 3

Setting Up Equation for the Second Cooling Process

For cooling from $80^\circ\text{C}$ to $60^\circ\text{C}$ in time $t'$ in the same surroundings:

\frac{80 - 60}{t'} = K \left(\frac{80 + 60}{2} - 20\right) \implies \frac{20}{t'} = K (70 - 20) = 50 K \quad \text{--- (2)}

Step 4

Dividing Equations to Determine t'

Divide equation (1) by equation (2) to eliminate the cooling constant $K$ :

\frac{10/t}{20/t'} = \frac{65 K}{50 K} \implies \frac{t'}{2t} = \frac{13}{10} \implies t' = \frac{13}{10} \times 2t = \frac{13}{5} t

⭐

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 1

✗ Incorrect

Verdict & Scientific Reason:

Forgets the factor of 2 arising from the temperature drop ratio ( $\Delta T_2 / \Delta T_1 = 20/10 = 2$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Writing

t'/t = 13/10

without accounting for the

20^\circ

drop vs.

10^\circ

drop.

2 2

✓ 100% Correct

Verdict & Scientific Reason:

Exact time derived from Newton's law of cooling: $t' = \frac{13}{5}t = 2.6t$ .

💡

Why this is airtight: None; rigorous thermal physics calculation.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Inverted ratio with missing factor of 2 (gives time less than $t$ , which is physically impossible for a larger temperature drop at lower temperature).

⚠️

Student Fallacy (Why candidates pick this trap): Inverting numerator and denominator.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Complete inversion of the correct ratio ( $5/13$ instead of $13/5$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Inverting final result.

⚠️

NTA Trap Alert: Double Temperature Drop Neglect Trap

Students frequently calculate the ratio of average temperature excess ( $65/50 = 13/10$ ) and immediately select Option (1) $\frac{13}{10}t$ , forgetting that the first drop was only $10^\circ\text{C}$ ( $90 \to 80$ ), while the second drop is TWICE as large ( $20^\circ\text{C}$ from $80 \to 60$ ).

🛡️ How to avoid the -1 mark: Always include the temperature drop in the rate:

\frac{\Delta T}{\Delta t}

. Here

\Delta T_2 = 20

, which is

2 \times \Delta T_1

📖

Core NEET Syllabus Standard

NCERT Physics · Thermal Properties of Matter Standard Benchmark

Core Scientific Principle Tested:

"The rate of cooling is directly proportional to the difference in temperature between the body and its surroundings: $\frac{dT}{dt} = -K(T - T_0)$ ."

Syllabus Focus: Newton's Law of Cooling

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 28 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Average Temperature Difference Ratio

Step-by-Step Rigorous Workflow & Mathematical Proof

Formulating Newton's Law of Cooling (Average Form)

Setting Up Equation for the First Cooling Process

Setting Up Equation for the Second Cooling Process

Dividing Equations to Determine t'

360° Option Elimination Engine Key Feature

NTA Trap Alert: Double Temperature Drop Neglect Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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