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Q.173 Physics ⭐ 360° Option Elimination Active
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NEET 2020 Physics Question 173 Solution

Topic: Current Electricity · Official NTA Booklet Code E1
NEET 2020 Physics Question 173: Calculation of Carrier Mobility from Drift Velocity and Electric Field

Question

A charged particle having drift velocity of 7.5×10⁻⁴ m s⁻¹ in an electric field of 3×10⁻¹⁰ V m⁻¹, has a mobility in m² V⁻¹ s⁻¹ of:

✓
Verified Against NTA Final Answer Key
Option (2) : 2.5 x 10^6
📚 Current Electricity Easy
📋 Given Parameters & Problem Setup
Drift Velocity (vd) 7.5 × 10^-4 m s^-1
Electric Field (E) 3 × 10^-10 V m^-1
⚡

The 30-Second Exam Hall Shortcut: ⚡ "Mobility mu = vd / E = 7.5 × 10^-4 / (3 × 10^-10) = 2.5 × 10^6 m^2 V^-1 s^-1!"

⏱️ 20 seconds

Mobility is defined as drift velocity acquired per unit electric field: mu = vd / E. Directly divide: mu = (7.5 × 10^-4) / (3 × 10^-10) = (7.5 / 3) × 10^(-4 - (-10)) = 2.5 × 10^6 m^2 V^-1 s^-1. Positive exponent +6 immediately rules out negative powers! Mark Option (2) in 5 seconds.

1 Formula: μ=vdE\mu = \frac{v_d}{E}.
2 Substitute given values: vd=7.5×10−4 m/sv_d = 7.5 \times 10^{-4}\text{ m/s} and E=3×10−10 V/mE = 3 \times 10^{-10}\text{ V/m}.
3 Divide coefficients: 7.53=2.5\frac{7.5}{3} = 2.5.
4 Divide exponents: 10−410−10=10−4−(−10)=10+6\frac{10^{-4}}{10^{-10}} = 10^{-4 - (-10)} = 10^{+6}.
5 Result: μ=2.5×106 m2V−1s−1\mu = 2.5 \times 10^6\text{ m}^2\text{V}^{-1}\text{s}^{-1}.
6 Select Option (2).
🛡️
Why You Can 100% Trust This Shortcut: NCERT Physics · Current Electricity (Drift of Electrons and Mobility)
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Definition of Carrier Mobility

Conductivity in materials arises from mobile charge carriers. Mobility (μ\mu) is defined as the magnitude of the drift velocity per unit electric field: μ=∣vd∣E\mu = \frac{|v_d|}{E} It has SI units of m2V−1s−1\text{m}^2 \text{V}^{-1}\text{s}^{-1}.

Step 2

Insert Given Parameters

- Drift velocity: vd=7.5×10−4 m s−1v_d = 7.5 \times 10^{-4}\text{ m s}^{-1} - Electric field: E=3×10−10 V m−1E = 3 \times 10^{-10}\text{ V m}^{-1}.

Step 3

Evaluate Numerical Division

μ=7.5×10−43×10−10=(7.53)×10−4−(−10)=2.5×106 m2V−1s−1\mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = \left(\frac{7.5}{3}\right) \times 10^{-4 - (-10)} = 2.5 \times 10^6\text{ m}^2\text{V}^{-1}\text{s}^{-1}.

Step 4

Match with Options

Matches Option 2.

⭐

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
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; arbitrary calculation unrelated to vd / E.

⚠️
Student Fallacy (Why candidates pick this trap): Multiplying or squaring parameters.
2 B
✓ 100% Correct
Verdict & Scientific Reason:

Accurate value; mu = vd / E = 7.5 × 10^-4 / (3 × 10^-10) = 2.5 × 10^6 m^2 V^-1 s^-1.

💡
Why this is airtight: None; exact application of mobility definition.
3 C
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; sign error in exponent (-4 - 10 = -14 or inverting division as E / vd).

⚠️
Student Fallacy (Why candidates pick this trap): Inverting ratio or sign error in exponent subtraction.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; product vd × E = 22.5 × 10^-14.

⚠️
Student Fallacy (Why candidates pick this trap): Multiplying velocity by field instead of dividing.
⚠️

NTA Trap Alert: Exponent Subtraction Rule

Dividing by 10^-10 adds +10 to the exponent: -4 - (-10) = +6. Students often hastily write -14 or -6.

🛡️ How to avoid the -1 mark: Remember: 10−a10−b=10−a+b\frac{10^{-a}}{10^{-b}} = 10^{-a + b}.
📖

Core NEET Syllabus Standard

Physics · Current Electricity Standard Benchmark
Core Scientific Principle Tested:

""

Syllabus Focus: Calculation of Carrier Mobility from Drift Velocity and Electric Field
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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