A Metal Ball Cools from 64 °C to 50 °C in 10 Minutes and to 42 °C in Next 10 Minutes. the Ratio of Rates of Fall of Temperature During the Two Intervals is _______. - Physics

MCQ

A metal ball cools from 64 °C to 50 °C in 10 minutes and to 42 °C in next 10 minutes. The ratio of rates of fall of temperature during the two intervals is _______.

• 4/7

• 7/4

• 2

• 2.5

Solution

7/4

Rate of fall of temperature in first 10 minutes = R1 = (64 - 50)/10.°C/min = 1.4°C/min

Rate of fall of temperature in next 10 minutes = R2 = (50 - 42)/10.°C/min = 0.8°C/min

⇒ R_1/R_2 = 7/4

Concept: Heat and Temperature
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