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Mechanical Properties of Solids appeared 21 times across 3 years — 2.4% of Physics. This question is from Hooke's Law and Young's Modulus.

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Questions 4 9 8 21
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A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

Solution & Explanation

Related Formula

Young's modulus Y is defined as:

Y = StressStrain = (F / A)/(l / L) = (F L)/(A l)

Solving for the extension l:

l = (F L)/(A Y) = (F L)/(π r² Y)

Since the material does not change, Y remains constant.

Core Logic

Let the original parameters be F, r, L, l. The extension is:

l = (F L)/(π r² Y) (i)

When the force and radius are both halved, keeping original length L and Young's Modulus Y constant:

  • New Force, F' = (F)/(2)
  • New Radius, r' = (r)/(2)

    • The new extension l' is:

l' = (F' L)/(π (r')² Y) = ((F/2) L)/(π (r/2)² Y) l' = (F L / 2)/(π r² Y / 4) = 2 ( (F L)/(π r² Y) )
Step 1: Calculate Final Extension

Comparing this with equation (i):

l' = 2l

Thus, the increase in length becomes 2 times its original value.

Pattern Recognition

From the formula l ∝ (F)/(r²), if F is scaled by (1)/(2) and r by (1)/(2), the scaling factor for l is (1/2)/((1/2)²) = 2 directly.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

Reference Study Guides

Mechanical Properties Of Solids

More Mechanical Properties of Solids Previous-Year Questions — Page 5

Q58 jee_main_2024_31_jan_morning Bulk Modulus
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is ________ m. (Take density of sea water = 10³ kgm⁻³, Bulk modulus of rubber = 9 × 10⁸ Nm⁻², and g = 10 ms⁻²)
Numerical Answer. Answer: 18 to 18

Solution

Related Formula
β = (-Δ P)/((Δ V)/(V)) Δ P = ρ g h
Core Logic

The change in pressure Δ P is the hydrostatic pressure at depth h.

Δ P = -β (Δ V)/(V) ρ g h = -β (Δ V)/(V)
Step 2: Calculation

Given values: ρ = 10³ kg/m³ g = 10 m/s² β = 9 × 10⁸ N/m² (Δ V)/(V) = -0.02% = -(0.02)/(100)

Substitute into the equation:

10³ × 10 × h = - (9 × 10⁸) × (-(0.02)/(100)) 10⁴ × h = 9 × 10⁸ × 2 × 10⁻⁴ 10⁴ h = 18 × 10⁴ h = 18 m
Chapter Mix

Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids

More Mechanical Properties of Solids Questions — jee_main_2024_29_january_evening

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)