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

Answer 1

`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`