Skip to content
Login Register

#Young's modulus

69 public questions tagged with this topic.

Wave Equation and Displacement Relation

A steel rod of length 2 m and density 7800 kg/m³ has a longitudinal wave speed of 5000 m/s. What is its Young’s modulus?

**Displacement relation** encodes λ = 2π/k and f = ω/2π. Comparing given equation y = a sin(kx - ωt) with standard form yields k and ω, hence λ = 2π/k and v = ω/k, essential for identifying propagation characteristics and phase. Speed: v = √((Y/rho)) . 5000 = √((Y/7800)) ⇒ 5000² = (Y/7800) . Y = 5000² × 7800 = 1.95 × 10¹¹ Pa . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1.95 × 10¹¹ Pa, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Equation and Displacement Relation

View Details
Wave Equation and Displacement Relation

A copper rod has a density of 8900 kg/m³ and a longitudinal wave speed of 3560 m/s. What is its Young’s modulus?

**Wave equation** y(x,t) = A sin(kx - ωt + φ) describes displacement of progressive harmonic wave, where k = 2π/λ wave number (rad/m), ω = 2πf angular frequency (rad/s), v = ω/k wave speed (m/s). Sign of ωt indicates direction, amplitude A is maximum displacement. v = √((Y/rho)) . 3560 = √((Y/8900)) ⇒ 3560² = (Y/8900) . Y = 3560² × 8900 ≈ 1.13 × 10¹¹ Pa . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1.13 × 10¹¹ Pa, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Equation and Displacement Relation

View Details
Sound Waves, Reflection and Characteristics

A steel rod of density 7800 kg/m³ has a Young’s modulus of 2 × 10¹¹ Pa. What is the speed of a longitudinal wave in the

**Sound wave reflection** at rigid wall behaves like string fixed end, displacement inverted. Equation of reflected wave includes sign change and direction reversal, amplitude unchanged but sign may flip, explaining standing wave formation with incident wave. Speed: v = √((Y/rho)) = √((2 × 10¹¹/7800)) ≈ √(2.56 × 10⁷) ≈ 5060 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 5060 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Sound Waves, Reflection and Characteristics

View Details
PHYSICS

A steel wire of length 2.0 m and cross-sectional area 2.5 × 10⁻⁶m² is stretched by a force of 250 N . If the Young's mod

Given: A steel wire of length 2.0 m and cross-sectional area 2.5 × 10⁻⁶m² is stretched by a force of 250 N . If the Young's modulus of steel is 2 × 10¹¹N/m², what is the strain produced? These values define the system as per NCERT data. Formula: Stress: Stress = F/A = frac2502.5 × 10⁻⁶= 1 × 10⁸N/m². This is standard NCERT relation. Substitution & Calculation: Young's modulus: Y = fracStressStrain . Strain: Strain = fracStressY = frac1 × 10⁸² × 10¹¹= 5 × 10⁻⁴. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details
Stress, Strain and Hooke's Law

An aluminium wire of length 1.8m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 2×10−4.

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 7×1010×2×10−4 = 1.4×107N/m2. Force: F = Stress×A = 1.4×107×2×10−6 = 28N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 28N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

View Details
Stress, Strain and Hooke's Law

A steel wire of length 2.6m and cross-sectional area 2×10−6m2 is stretched by 0.52mm. If the Young's modulus of steel is

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.52×10−3m. F = 2×1011×2×10−6×0.52×10−32.6 = 2082.6 = 80N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 80N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details
Stress, Strain and Hooke's Law

A copper wire of length 2.8m and cross-sectional area 2×10−6m2 is stretched by a force of 160N. If the Young's modulus o

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 160×2.82×10−6×1.1×1011 = 4482.2×105≈2.04×10−3m = 2.04mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.04mm. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

View Details
Stress, Strain and Hooke's Law

A steel wire of length 2.0m and cross-sectional area 2.5×10−6m2 is stretched by a force of 250N. If the Young's modulus

Stress: Stress = FA = 2502.5×10−6 = 1×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1×1082×1011 = 5×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details
Universal Law of Gravitation

A steel wire and a copper wire have the same length and cross-sectional area. Both are stretched by the same force. If Y

Elongation: ΔL = (F L) / (A Y). Ratio: (ΔLsteel)/(ΔLcopper) = Ycopper / Ysteel = (1.1 × 1011) / (2 × 1011) = 11/20 = 0.55. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.55. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details
Stress, Strain and Hooke's Law

A copper wire of length 1.8m and cross-sectional area 3×10−6m2 is stretched by a force producing a strain of 3×10−4. If

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 1.1×1011×3×10−4 = 3.3×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.3×107N/m2. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

View Details
Elastic Properties and Applications

A steel wire of length 2.8m and cross-sectional area 4×10−6m2 is stretched by a force producing a strain of 1.5×10−4. If

Strain: Strain = ΔLL. Rearrange: ΔL = Strain×L = 1.5×10−4×2.8 = 4.2×10−4m = 0.42mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.42mm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details
Elastic Properties and Applications

A steel wire of length 2.0m and cross-sectional area 3×10−6m2 is stretched by 0.4mm. If the Young's modulus of steel is

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.4×10−3m. F = 2×1011×3×10−6×0.4×10−32 = 2402 = 120N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 120N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

View Details