Question

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

• A. $\frac{4}{7}$
• B. $\frac{7}{4}$
• C. $2$
• D. $2.5$

Answer 1

Answer: (B)

$\frac{7}{4}$

Rate of fall of temperature in first 10 minutes=$R_1=\frac{64-50}{10}{.}^{\circ }C/min$

$={1.4}^{\circ }C/min$

Rate of fall of temperature in next 10 minutes=$R_2=\frac{50-42}{10}{.}^{\circ }C/min$

$={0.8}^{\circ }C/min$

$⟹\frac{R_1}{R_2}=\frac{7}{4}$