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#brass rod

3 public questions tagged with this topic.

Specific Heat and Calorimetry

A brass rod of length 2.5m at 20∘C is heated to 220∘C. If its cross-sectional area increases by 0.018cm2, what was its o

Given: ΔT = 220−20 = 200∘C, ΔA = 0.018cm2, αl = 1.8×10−5K−1. Area expansion: ΔA = A0×2αlΔT. 0.018 = A0×2×1.8×10−5×200. 0.018 = A0×7.2×10−3⇒A0 = 0.0187.2×10−3 = 2.5cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 cm². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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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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