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tF = 95tC+32 = 95×75+32 = 135+32 = 167∘F. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 167°F. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Relation: T = tC+273.15. T = −23+273.15 = 250.15K≈250K. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 250 K. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Relation: TC = T−273.15. Given: T = 400K. TC = 400−273.15 = 126.85∘C≈127∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 127°C. 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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