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 _______.

#### Options

`4/7`

`7/4`

2

2.5

Advertisement Remove all ads

#### 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`

Concept: Heat and Temperature

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads