#steel wire

Transverse and Longitudinal Waves

A steel wire of length 4 m and mass 0.08 kg is under tension. If a transverse wave takes 0.02 s to travel its length, wh

**Wave classification** depends on particle vibration relative to propagation. Longitudinal waves have particle oscillation parallel to propagation, creating compressions and rarefactions as in sound in air; transverse have perpendicular oscillation. Tuning fork generates longitudinal sound because air cannot sustain shear. Speed: v = (length/time) = (4/0.02) = 200 m/s . μ = (0.08/4) = 0.02 kg/m . v = √((T/μ)) ⇒ 200 = √((T/0.02)) ⇒ 200² = (T/0.02) . T = 40000 × 0.02 = 800 N . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 800 N, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Transverse and Longitudinal Waves

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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.

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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.

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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.

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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.

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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.

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Elastic Properties and Applications

A steel wire of length 2.8m and cross-sectional area 3×10−6m2 is stretched by 0.7mm. If the Young's modulus of steel is

Strain: Strain = ΔLL = 0.7×10−32.8 = 2.5×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5×10−4. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Shear Modulus and Rigidity

A steel wire of length 3.0m and cross-sectional area 2×10−6m2 is stretched by a force of 200N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 200×3.02×10−6×2×1011 = 6004×105 = 1.5×10−3m = 1.5mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.5mm. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Shear Modulus and Rigidity

A steel wire of length 3.2m and cross-sectional area 5×10−6m2 is stretched by 0.8mm. If the Young's modulus of steel is

Young's modulus: Y = FLAΔL. Rearrange: F = YAΔLL. Substitute: ΔL = 0.8×10−3m. F = 2×1011×5×10−6×0.8×10−33.2 = 8003.2 = 250N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 250N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Shear Modulus and Rigidity

A steel wire of length 2.5m and cross-sectional area 3.5×10−6m2 is stretched by a force of 350N. If the elongation is 0.

Young's modulus: Y = FLAΔL. Substitute: ΔL = 0.2×10−3m. Y = 350×2.53.5×10−6×0.2×10−3 = 8757×10−10≈1.25×1012N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.25×1012N/m2. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Bulk Modulus and Compressibility

A steel wire of length 2.6m and cross-sectional area 2×10−6m2 is stretched by a force of 400N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 400×2.62×10−6×2×1011 = 10404×105 = 2.6×10−3m = 2.6mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6mm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Bulk Modulus and Compressibility

A steel wire of length 2.7m and cross-sectional area 2×10−6m2 is stretched by 0.54mm. If the Young's modulus of steel is

Strain: Strain = ΔLL = 0.54×10−32.7 = 2×10−4. Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 2×1011×2×10−4 = 4×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4×107N/m2. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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