Q57 / 75 · JEE Main 2020 (03 Sep Shift 2) · Physics

A metallic sphere cools from $50 ° C$ to $40 ° C$ in $300 s$ . If atmospheric temperature around is $20 ° C$ , then the sphere's temperature after the nest $5$ minutes will be close to:

Solution

Using Newton's Law of cooling

$\frac{50 - 40}{5 \min} = K (\frac{50 + 40}{2} - 20)$ ......(i)

Next $5 \min .$

$\frac{40 - θ}{5} = K (\frac{40 + θ}{2} - 20)$ .....(ii)

Dividing (ii)/(i)

$\frac{40 - θ}{10} = \frac{40 + θ - 40}{50 + 40 - 40} = \frac{θ}{50}$

$40 - θ = \frac{θ}{5}$

$200 - 5 θ = θ$

$∴ θ = \frac{200}{6} = 33.3 ° C$

Hence, the correct answer is option: 'B'.