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#tension

12 public questions tagged with this topic.

Elastic Properties and Applications

A brass wire of length 1.9m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 2×10−4. If t

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×2×10−4 = 1.8×107N/m2. Force: F = Stress×A = 1.8×107×2×10−6 = 36N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 36N. 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 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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Young's Modulus and Elasticity

A steel wire of length 2.4m and cross-sectional area 4×10−6m2 is stretched by 0.6mm. If the Young's modulus of steel is

Strain: Strain = ΔLL = 0.6×10−32.4 = 2.5×10−4. Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 2×1011×2.5×10−4 = 5×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5×107N/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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Young's Modulus and Elasticity

A brass wire of length 2.6m and cross-sectional area 4×10−6m2 is stretched by a force of 400N. If the Young's modulus of

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 400×2.64×10−6×9×1010 = 10403.6×105≈2.89×10−3m = 2.89mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.89mm. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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Stress, Strain and Hooke's Law

A brass wire of length 2.0m and cross-sectional area 2×10−6m2 is stretched by a force of 300N. If the Young's modulus of

Stress: Stress = FA = 3002×10−6 = 1.5×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1.5×1089×1010≈1.67×10−3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.67×10−3. 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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Cell Communication and Cell Adhesion

Increased tension at adherens junctions recruits

Vinculin, is consistent with established principles of cell signaling, receptor pharmacology and cellular regulation. Experimental measurements of binding parameters, genetic loss-of-function studies and pharmacological interventions all converge on the same interpretation. Related options address neighboring concepts but do not satisfy the precise criterion stated in the question.

Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)

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WATER POTENTIAL+PYQS

Negative pressure potential in xylem is known as:

Tension (C) is correct here. This is core WATER POTENTIAL: once you know the definition or pathway step, Tension is the clear fit. The wrong ones are A) Turgor; B) Osmosis; D) Diffusion. If the topic is about gradients or potentials, water/solutes move from higher to lower of the relevant quantity.

Ref: Best CSIR NET Plant Physiology books: Master Unit 6 with Taiz & Zeiger and Salisbury & Ross. Crack Part C experimental questions with top textbooks.

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