Q97 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
A body cools from $100^\circ\text{C}$ to $60^\circ\text{C}$ in 20 minutes, the temperature of the surroundings being $20^\circ\text{C}$. The total time taken (in minutes) for the body to cool down to $40^\circ\text{C}$ is ...
- A$60$
- Correct answerB$40$
- C$50$
- D$30$
Solution
Using Newton's law of cooling:
$\ln\left(\dfrac{T_1 - T_s}{T_2 - T_s}\right) = kt$
For the first interval, the body cools from $100^\circ\text{C}$ to $60^\circ\text{C}$ in $20$ minutes with $T_s = 20^\circ\text{C}$:
$\ln\left(\dfrac{100 - 20}{60 - 20}\right) = k(20)$
$\ln\left(\dfrac{80}{40}\right) = 20k$
$\ln(2) = 20k$
For the total time $t$ to cool from $100^\circ\text{C}$ to $40^\circ\text{C}$:
$\ln\left(\dfrac{100 - 20}{40 - 20}\right) = kt$
$\ln\left(\dfrac{80}{20}\right) = kt$
$\ln(4) = kt$
$2\ln(2) = kt$
Substituting $20k$ for $\ln(2)$:
$2(20k) = kt$
$t = 40$ minutes
Answer: $40$
$\ln\left(\dfrac{T_1 - T_s}{T_2 - T_s}\right) = kt$
For the first interval, the body cools from $100^\circ\text{C}$ to $60^\circ\text{C}$ in $20$ minutes with $T_s = 20^\circ\text{C}$:
$\ln\left(\dfrac{100 - 20}{60 - 20}\right) = k(20)$
$\ln\left(\dfrac{80}{40}\right) = 20k$
$\ln(2) = 20k$
For the total time $t$ to cool from $100^\circ\text{C}$ to $40^\circ\text{C}$:
$\ln\left(\dfrac{100 - 20}{40 - 20}\right) = kt$
$\ln\left(\dfrac{80}{20}\right) = kt$
$\ln(4) = kt$
$2\ln(2) = kt$
Substituting $20k$ for $\ln(2)$:
$2(20k) = kt$
$t = 40$ minutes
Answer: $40$
Hence, the correct answer is option: 'B'.