A metal ball cools from $64^\circ$C to $50^\circ$C in 10 minutes and to $42^\circ$C in next 10 minutes. The ratio of rates of fall of temperature during the two intervals is _______.

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KollegeApply · Accepted Answer

$\frac{7}{4}$ Rate of fall of temperature in first 10 minutes = R1 = (64 - 50)/10.°C/min = 1.$4^\circ$C/min Rate of fall of temperature in next 10 minutes = R2 = (50 - 42)/10.°C/min = 0.$8^\circ$C/min $\Rightarrow$ R_1/R_2 = $\frac{7}{4}$

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