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Question
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 _______.
Options
`4/7`
`7/4`
2
2.5
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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`
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