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#stress

24 public questions tagged with this topic.

ZOOLOGY

Which hormone is secreted by the adrenal medulla during emergency situations?

Adrenaline (epinephrine) is secreted by the adrenal medulla in response to stress, increasing heart rate and alertness. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Biology Textbook for Class XI and XII (Zoology section), Chapter: Human Physiology - Digestion, Respiration, Circulation, Neural Control and Chemical Coordination, Topic: Organ system functions and hormonal regulation.

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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 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.

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Elastic Properties and Applications

A brass wire of length 1.7m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 4×10−4. If t

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×4×10−4 = 3.6×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6×107N/m2. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Elastic Properties and Applications

A copper wire of length 2.5m and cross-sectional area 2×10−6m2 is stretched by a force of 200N. If the Young's modulus o

Stress: Stress = FA = 2002×10−6 = 1×108N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1×108N/m2. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Elastic Properties and Applications

A copper wire of length 1.2 m and cross-sectional area 1.5 × 10-6 m2 is stretched by a force producing a stress of 2 × 1

Young's modulus: Y = Stress / Strain. Strain: Strain = Stress / Y = (2 × 107) / (1.1 × 1011) ≈ 1.82 × 10-4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82 × 10-4. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Elastic Properties and Applications

An aluminium wire of length 1.5m and cross-sectional area 2×10−6m2 is stretched by a force of 140N. If the Young's modul

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 140×1.52×10−6×7×1010 = 2101.4×105 = 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 C is scientifically correct.

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

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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 rod of radius 0.01m and length 1.5m is subjected to a tensile force producing a stress of 4×107N/m2. What is the

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.01)2 = 3.14×10−4m2. Force: F = Stress×A = 4×107×3.14×10−4 = 1.256×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.256×104N. 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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Bulk Modulus and Compressibility

A brass rod of radius 0.008m and length 1.0m is subjected to a tensile force producing a stress of 5×107N/m2. What is th

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.008)2 = 3.14×6.4×10−5≈2.01×10−4m2. Force: F = Stress×A = 5×107×2.01×10−4≈1.005×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.005×104N. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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