A metal ball cools from `64^(@)C` to `50^(@)C` in 10 minutes and to `42^(@)C` in the next 10 minutes. The ratio of the rates of fall of temperature during the two intervals is

Options

A. `4/7`
B. `7/4`
C. 2
D. 2.5

Answer

B

Explanation

In the first interval, the rate of fall of temperature
`((d theta)/(dt))_(1)=(64-50)/10=14/10=7/5" "...(1)`
and in the second interval
`((d theta)/(dt))_(2)=(50-42)/10=8/10=4/5" "...(2)`
`(((d theta)/(dt))_(1))/(((d theta)/(dt))_(2))=(7/5)/(4/5)=7/5xx5/4=7/4`