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NEET Physics · Properties of Solids and Liquids

Young's modulus, bulk modulus and modulus of rigidity

Taught in NCERT Class 11 Physics, Chapter 8: Mechanical Properties of Solids.

3 verified questions: Easy 0 · Medium 2 · NEET level 1

Sample questions

Sample 1Medium

A figure shows a steel wire 2 m long, of cross-sectional area 1 × 10⁻⁶ m², clamped at the top and hanging vertically, with a 100 N load hooked at its lower end. A scale beside it shows the wire stretched by 1 mm. What is the Young's modulus of the steel?

  1. A

    2 × 10¹¹ Pa

  2. B

    1 × 10¹¹ Pa

  3. C

    2 × 10⁸ Pa

  4. D

    4 × 10¹¹ Pa

Show the answer
The answer is A.

Y = (F/A)/(ΔL/L) = (100 / 10⁻⁶) / (10⁻³ / 2) = 10⁸ / (5 × 10⁻⁴) = 2 × 10¹¹ Pa.

  • B.

    1 × 10¹¹ Pa leaves out the factor L = 2 m.

  • C.

    2 × 10⁸ Pa treats the extension as 1 m instead of 1 mm.

  • D.

    4 × 10¹¹ Pa takes the extension as 0.5 mm; the wire stretched by 1 mm.

NCERT: Class 11 Physics, Chapter 8

Sample 2NEET level

A steel wire of length 2 m and cross-sectional area 1 mm² hangs vertically and a load of 10 kg is attached to its lower end. Young's modulus of steel is 2 × 10¹¹ N/m² and g = 10 m/s². The extension of the wire is

  1. A

    0.01 mm

  2. B

    0.1 mm

  3. C

    1 mm

  4. D

    10 mm

Show the answer
The answer is C.

ΔL = FL/(AY) = (100 N × 2 m)/(1 × 10⁻⁶ m² × 2 × 10¹¹ N/m²) = 1 × 10⁻³ m = 1 mm. The force is the weight mg = 100 N and 1 mm² = 10⁻⁶ m².

  • A.

    Takes 1 mm² as 10⁻⁴ m² (a cm² conversion), making the area 100 times too large.

  • B.

    Takes the force as 10 N, using the mass instead of the weight.

  • D.

    Takes 1 mm² as 10⁻⁷ m², making the area 10 times too small.

NCERT: Class 11 Physics, Chapter 8 (Mechanical Properties of Solids), Young's modulus

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