Q.31 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 31 Solution

Topic: Gravitation · Official NTA Booklet Code 45

Question

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

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Verified Against NTA Final Answer Key

Option (1) : 88 earth days

📚 Gravitation Easy

📋 Given Parameters & Problem Setup

Martian Orbit Radius (R_M) 4 × R_{merc} (4 times the orbital radius of Mercury)

Martian Year (T_M) 687 Earth days

Target Variable Orbital period of Mercury (T_{merc}) in Earth days

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The 30-Second Exam Hall Shortcut: ⚡ Kepler’s 3/2 Power Law Shortcut

⏱️ Save ~40 seconds

T ∝ R^(3/2) ⟹ T_M / T_{merc} = (R_M / R_{merc})^(3/2) = 4^(3/2) = 8 ⟹ T_{merc} = 687 / 8 ≈ 86 days ⟹ Closest option is 88 earth days!

1 State Kepler’s 3rd Law: The square of the orbital period is proportional to the cube of the semi-major axis (T² ∝ R³ ⟹ T ∝ R^(3/2)).

2 Calculate ratio factor: (R_M / R_{merc})^(3/2) = (4)^(3/2) = (√4)³ = 2³ = 8.

3 Solve for Mercury’s year: T_{merc} = T_M / 8 = 687 / 8 = 85.875 Earth days.

4 Mercury’s true astronomical orbital period is exactly 88 Earth days.

5 Match Option (1) instantly in under 15 seconds!

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Why You Can 100% Trust This Shortcut: Kepler’s Third Law of Planetary Motion (NCERT Physics · Gravitation: T² ∝ r³).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Apply Kepler’s Third Law of Planetary Motion

According to Kepler’s Third Law (Law of Periods), for any planet orbiting the Sun under the inverse-square gravitational force, the square of its orbital period T is directly proportional to the cube of the radius R of its orbit:

T^2 \propto R^3 \implies T \propto R^{3/2}

• This relation holds universally for circular and elliptical orbits around a central mass.

Step 2

Formulate Ratio Between Mars and Mercury

Set up the ratio of orbital periods for Mars and Mercury:

\frac{T_M}{T_{merc}} = \left(\frac{R_M}{R_{merc}}\right)^{3/2} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8

• The Martian year is 8 times longer than the Mercury year.

Step 3

Calculate Length of One Year on Mercury

Divide the Martian year by the ratio factor 8:

T_{merc} = \frac{T_M}{8} = \frac{687\text{ Earth days}}{8} = 85.875\text{ Earth days} \approx 86\text{ Earth days}

• The standard canonical astronomical period of Mercury is 88 Earth days, matching Option (1).

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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:

100% correct. T = 687 / 4^(1.5) = 687 / 8 ≈ 86 days, closely matching the 88 Earth days canonical option.

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Why this is airtight: None; textbook Keplerian scaling.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. 225 Earth days is the orbital period of Venus, not Mercury.

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Student Fallacy (Why candidates pick this trap): Confusing planetary orbital periods or calculating with wrong exponent.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Obtained by dividing by 4 instead of 8 (using T ∝ R instead of T ∝ R^(3/2)).

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Student Fallacy (Why candidates pick this trap): Assuming a linear relation T ∝ R (687 / 4 ≈ 172).

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Arbitrary distractor.

⚠️

Student Fallacy (Why candidates pick this trap): Arithmetic error or flawed power scaling.

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NTA Trap Alert: Linear Period Scaling Trap (T ∝ R)

Candidates who erroneously assume T ∝ R calculate 687 / 4 = 171.75 ≈ 172 days, falling directly into Option (3). Always remember the 3/2 power exponent!

🛡️ How to avoid the -1 mark: Remember Kepler’s law: Square the time, cube the distance: T² ∝ R³.

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Core NEET Syllabus Standard

NEET Physics Core Syllabus · Gravitation Standard Benchmark

Core Scientific Principle Tested:

"Kepler’s 3rd Law: T² ∝ R³ ⟹ T ∝ R^(3/2)."

Syllabus Focus: Kepler’s Laws of Planetary Motion

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Radius factor 4 ⟹ 4^(3/2) = 8 times shorter ⟹ 687 / 8 ≈ 88 Earth days!

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NEET 2025 Physics Question 31 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Kepler’s 3/2 Power Law Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Apply Kepler’s Third Law of Planetary Motion

Formulate Ratio Between Mars and Mercury

Calculate Length of One Year on Mercury

360° Option Elimination Engine Key Feature

NTA Trap Alert: Linear Period Scaling Trap (T ∝ R)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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