6 public questions tagged with this topic.
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.
At the boiling point of water under standard pressure, the Celsius scale reads 100∘C, while the Fahrenheit scale reads 212∘F, reflecting different interval sizes and zero points. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Celsius is 100∘, Fahrenheit is 212∘. 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.
The Kelvin scale is offset from the Celsius scale by 273.15, so T = tC+273.15 (Section 10.4), where T is Kelvin and tC is Celsius. As per NCERT, applying relevant law/formula with correct units and sign convention leads to T\=tC+273.15. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
At the ice point (freezing of water at standard pressure), Celsius is 0∘C and Fahrenheit is 32∘F (Section 10.3). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Celsius is 0∘, Fahrenheit is 32∘. This satisfies dimensional consistency and physical conditions given, so option D 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.
