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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 Elongation of String.

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Questions 4 9 8 21
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The length of a light string is 1.4m when the tension on it is 5N . If the tension increases to 7N , the length of the string is 1.56m . The original length of the string is ______ m .

Numerical Answer Type:
Enter a numerical value Answer: 1 to 1 +4 marks

Solution & Explanation

Related Formula

By Hooke's Law, the tension (T) in a stretched elastic string is proportional to its change in length:

T = k · Δ L = k(L - L₀)

where: k = spring constant of the string L = current stretched length L₀ = natural original length

Core Logic

We can set up two algebraic equations using the given conditions:

  • Case 1: Tension T₁ = 5 N produces stretched length L₁ = 1.4 m:
5 = k(1.4 - L₀) (1)
  • Case 2: Tension T₂ = 7 N produces stretched length L₂ = 1.56 m:
7 = k(1.56 - L₀) (2)
Step 1: Solve for natural length

Divide equation (1) by equation (2) to eliminate the spring constant k:

(5)/(7) = (1.4 - L₀)/(1.56 - L₀)

Cross-multiply and solve for L₀:

5(1.56 - L₀) = 7(1.4 - L₀) 7.8 - 5 L₀ = 9.8 - 7 L₀

Rearrange to isolate the variable:

2 L₀ = 9.8 - 7.8 = 2 L₀ = 1 m

Thus, the original length of the string is 1 ~m.

Pattern Recognition

Sees: Elastic stretching of a string/wire under variable load. Trap: Attempting to resolve details like Young's modulus or cross-sectional area. Taking ratios allows the spring constant k to cancel cleanly, saving computational effort. Shortcut: Use the ratio equation (T₁)/(T₂) = (L₁ - L₀)/(L₂ - L₀) directly. Solving the linear equation yields L₀ = 1 m.

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

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