#### 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

#### Solution

`7/4`

Rate of fall of temperature in first 10 minutes = R_{1} = `(64 - 50)/10`.°C/min = 1.4°C/min

Rate of fall of temperature in next 10 minutes = R_{2} = `(50 - 42)/10`.°C/min = 0.8°C/min

⇒ `R_1/R_2 = 7/4`

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Solution 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 _______. Concept: Temperature and Heat.